rupture: a personal or social separation (as between opposing factions); "they hoped to avoid a break in relations"
Breach is an EP of 5 unreleased songs including 2 cover songs by Shivaree, released by Zoë Records in 2004.
Whales exhibit various types of behaviour when they surface. This article describes the different behaviours commonly observed at sea and the possible reasons for the behaviour.
(Breached) When a law, right or duty is broken.
(breaches) a rarely used term for pauses or breaks in dialogue: "At the breaches in this speech following, the other two Boys interrupt him" and "At the breaches he takes his tobacco" (Cynthia's Revels, Induction.38–43, 121–4); see also Epicoene, 2.1.10–14 and Noble Stranger, E3v.
(breaching) a spectacular behavior exhibited by whales in which the animal propels itself out of the water and clears the surface with at least two-thirds of its body.
(BREACHING) Failure of the BEACH HEAD or a DIKE allowing flooding by tidal action.
(breaching) A behavior of cetaceans that jumps or leaps out of the water and reenters on its side or back with a big splash.
The breaking of a law or contract.


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HANNAH & KAYLEE "HoW to PrOtEcT YoUrSelF from MoSqUiToeS"
Hannah Hughes and Kaylee Patrick talk about how to protect yourself from Louisianas state bird "the mosquito"!!
This video looks at several of Louisiana's state symbols.
STATE BIRD OF LOUISIANA: BROWN PELICAN!!! TIME LAPSE DRAWING!!!
Protecting the pelican: Plans rise to restore Louisiana state bird habitat.
Contemporary Solo to Bird Song entered in 2012 American All Star State Junior High Solo Competition in Lafayette, Louisiana.
SEE THIS SURPRISE ON THE LOUISIANA STATE QUARTER!
birds, nature, wildlife, sparrows, waterbirds, eagles, hawks, songbirds, tanagers, grosbeaks, herons, egrets, hawks, raptors, warblers, pelicans, cormorants.
view of baton rouge, louisiana from the 27th floor observation deck on the louisiana state capitol building.
Members of Tri-State Bird Rescue work to clean oil-soaked pelicans after the Gulf oil spill. Video by Steven Johnson / Miami Herald.
A Brown Pelican covered in oil was found on Storm Island. The bird is being cared for and fed by Tri-State Bird Rescue in Plaquemines Parish, Louisiana.
Behold this rare bird TIP US HERE; http://www.gofundme.com/thespellfamily https://www.gofundme.com/SteveSpell we lost everything in the flood in Louisiana.
Artwork painted on pelicans in Slidell, Louisiana.
Flowers, Blooming Flowers, Nesting Birds during early spring in South Louisiana.


text:
Little Hills 158A is an Indian reserve of the Lac La Ronge Indian Band in Saskatchewan. It is 6 miles south-west of La Ronge, and on the north bank of the Little Hills River.

References

Indian reserves in Saskatchewan
Division No. 18, Saskatchewan

redpajama_set_name: RedPajamaWikipedia


Are you a plumber looking for a job in Toronto? Here's a list of the most recent plumbing companies who are currently hiring plumbers like you.
$99 Clears any drain! Get your plumbing fixed right by professional, highly skilled technicians. We don't "learn" on your system; we fix it.


text:
New Zealand Sevens 2000 – pierwsza edycja wchodzącego w skład IRB Sevens World Series turnieju New Zealand Sevens przeznaczonego dla męskich reprezentacji narodowych w rugby 7. Odbyła się w dniach 4–5 lutego 2000 roku na Westpac Stadium w Wellington będąc piątym turniejem sezonu 1999/2000.

Informacje ogólne
Rozegrane na nowo zbudowanym Westpac Stadium zawody były piątym turniejem sezonu 1999/2000 IRB Sevens World Series i wzięło w nich udział szesnaście reprezentacji. Drużyny rywalizowały w pierwszym dniu systemem kołowym podzielone na cztery czterozespołowe grupy, po czym w drugim dniu osiem najlepszych awansowało do ćwierćfinałów, a pozostała ósemka zmierzyła się w walce o Bowl. Ogłoszenie składów grup i terminarza rozgrywek nastąpiło 21 stycznia 2000 roku.

Zawody przebiegały pod dyktando liderów cyklu. Zespoły te spotkały się w finale, w którym w obecności 30 tysięcy widzów triumfowali reprezentanci Fidżi. Graczem turnieju został zawodnik gospodarzy, Tony Monaghan.

Faza grupowa

Grupa A

Grupa B

Grupa C

Grupa D

Faza pucharowa

Cup

Plate

Bowl

Klasyfikacja końcowa

Przypisy

Bibliografia

IRB Sevens World Series (1999/2000)
New Zealand Sevens
2000 w Nowej Zelandii


If you had the money, would U be a sugar momma? Renee Zellwegger is!
Plus, Megyn Kelly's in a real bad place!
NeNe Leakes is in a worse one!
The shocking allegations against George Clooney. More on the ones against Ben Affleck.
Taylor Swift's secret album sessions!
The very latest on Harvey Weinstein. And much more.
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Listen To This: Where Things Go Naturally!


text:
Why we all hate Syria so much. Do we?
Wed 1:14 pm +00:00, 11 Sep 2013 3
Why the US, UK, EU & Israel hate Syria
Adrian Salbuchi is a political analyst, author, speaker and radio/TV commentator in Argentina.
A young, soft-spoken girl living the Syrian tragedy spells it out with far more common sense, truth and honesty than powerful Western governments and their money-controlled mass media puppets.
Identifying herself only as "Syrian, Patriot, anti-Neocon, anti-NWO, anti-Zionist", early last year she set up her own YouTube Channel (YouTube/User/SyrianGirlpartisan).
In a short (nine-minute) video she explains "eight reasons why the NWO (New World Order) hates Syria." We would all do well to listen in…
Her 'Top Eight Reasons Why They Hate Us' is an excellent wrap-up, applicable to just about every self-respecting country in the world: no Rothschild-controlled Central Bank, no IMF debt, no genetically modified foods, oil and pipelines, anti-secret societies, anti-Zionism, secularism and nationalism.
Her brief message unravels as a sort of common sense manual which explains why the United States of America, the United Kingdom, the European Union (especially France) and Israel are so keen on destroying Syria, a country whose leadership just won't bow down to the New World Order elites embedded deep inside the Western powers' own public (government) and private (corporate/banking) power structures.
She describes these eight reasons succinctly and convincingly, giving the world much food for thought and should hopefully inspire deep soul-searching. Especially amongst the people of the US, UK, EU and Israel who are the only populations that can put very direct pressure on their elected politicians in Washington, London, Paris, Tel Aviv and other Western capitals, forcing them to stop behaving like global criminals gone berserk, and to start heeding the word of We the People, in a responsible and democratic manner.
8 reasons the New World Order hates Syria
1) Syria's Central Bank is state-owned & controlled – In other words, it manages its national currency so that it serves the Syrian people and not the Rothschild-controlled global bankers operating from their New York, London, Frankfurt, Tel Aviv, Basel and Paris hideouts.
This means that the volume of currency it issues is in proper sync with the true needs of real economy of work, labor, production and all that is useful to Syria's people, instead of being in sync with parasitic, usurious, speculative foreign financiers. The latter seek to control local central banks so they can artificially limit the volume of currency available for genuine economic needs, especially the no-interest credit needed to finance useful things in the real economy: power plants, roads, gas works, housing, private enterprise and initiatives. This forces productive players – public and private – to have to resort to deadly interest and usury-based private banking loans whereupon the eternal debt chain starts to grow and grow as the so-called'sovereign debt crises' that hit country after country throughout decades of time eloquently show.
By artificially distorting the volume of 'public currency' issued by sovereign central banks that generates no interest, countries are thus forced to resort to high interest bearing 'private currency' (loans) handled by the monopolistic private bankster cabal in the hands of Rothschild, Rockefeller, Warburg, Goldman Sachs, HSBC, CitiCorp, JP Morgan Chase interests.
Clearly, a very good reason for these parasitic banksters to want to take out Syria.
2) Syria has no IMF (International Monetary Fund) debt. This means that Syria's leadership understands that the IMF – a public multilateral agency of member governments – is controlled by the global mega-bankers, and acts as their auditor and debt collection police whenever one of its weaker member states runs into sovereign debt trouble, which is another way of saying when those countries reach a point where they cannot siphon enough money out of their real economies – the work, toil and labor of its people – to hand it over to the parasitic private global bankers.
In a sense, the IMF's real job is to act as the global power elite's tax office – its 'IRS' so to speak – only that it does not tax people directly, but rather through proxy government and nation-states' tax offices. Are we starting to understand the real roots of the 'debt crises' hitting Greece, Cyprus, Ireland, Argentina, Spain, Italy, UK, US, Portugal, France?
Global slavery couldn't have been better thought out and planned!
Actually, true Islamic nations rightly reject banking fractional lending and interest practices as being immoral. That's what Libya's Gaddafi did, and what Syria and Iran presently do.
Clearly, a very good reason for parasitic banksters to want to take out Syria, just as they took out Libya and now target Iran.
3) Syria has banned genetically modified (GMO) seeds – Bashar Assad banned GMO's in order "to preserve human health," knowing full well that the Monsantos of this world are out to control the world's entire food supply, because coming global crises will not only be about oil, but about how much food countries will be able to put on their people's tables.
That's why after invading Iraq, the US ordered that only Monsanto seeds should to be used. That's why submissive client states like Argentina are poisoning their own soil and people by bowing down really low to Monsanto's demands.
Clearly, a very good reason for Monsanto to want to take out Syria.
4) Syria's population is well informed about the New World Order – Its media and universities openly debate the global power elite's influence in things. This means that they fully grasp the fact that real power in the West lies not in the White House, 10 Downing Street, Congress or Parliament, but rather with the complex and powerful grid of elite think-tanks led by New York's Council on Foreign Relations, the Bilderberg Conference, Trilateral Commission, Americas Society, World Economic forum and London's Royal Institute of International Affairs, which interact with mega-bankers, media, universities, the military, multinationals and the corporate over-world.
As our young friend aptly explains, Syrians dare to talk about secret societies like Freemasonry or Yale University's Skull & Bone Lodge whose members include top cats like former President George W Bush and current Secretary of State John Kerry.
Clearly, a very good reason for those top cats to order their errand boy Obama to take out Syria.
5) Syria has massive oil and gas reserves – here we go again! Every time the West goes to war to protect "freedom, human rights and democracy,"there's always the nauseous stench of oil; whether in Iraq, Libya, Kuwait, the Falklands, Afghanistan… Syria has offshore and onshore oil and gas reserves, and is helping to build a massive pipeline together with Iran, but without Western oil giants' control. Clearly, the full militarization of all oil production and reserve zones, and the militarization of transportation routes to 'bring oil home' from everywhere in the world, is a key on-going joint US/UK strategy.
Clearly, a very good reason for BP, Exxon, Royal Dutch Shell, Texaco, Total, Repsol and Chevron to want to take out Syria.
6) Syria clearly and unequivocally opposes Zionism and Israel – Israel practices criminal racist apartheid against the occupied Palestinians. Syria's leadership has no qualms in accusing Israel of being what it is: a racist, imperialist, genocidal entity, as the Wall of Hate Israel erected around Palestine clearly shows. Israel manages what can only be described as an Auschwitz-like mega-concentration camp in Palestine with millions of ill-treated, often-assassinated and humiliated prisoners.
Such geopolitical clarity of mind was shared by Gadhafi's Libya and Saddam's Iraq, and today also by Iran, China, Russia and India.
Clearly a very good reason for political juggernauts like AIPAC (American-Israeli Public Affairs Committee), the World Jewish Congress, the ADL (Anti-Defamation League), Likud, Kadima and Netanyahu/Lieberman to want to take out Syria.
7) Syria is one of the last secular Muslim states in the Middle East, whilst Zionist Jewish supremacists – in line with born-again-Israel-First-Bushite'Christian' kooks in the West – need for everybody to align to the will of their dark demiurge god which has its own 'chosen people'.
The Global Power Elite's implicit order is clear: everybody must believe in Israeli superiority, whilst our young Syrian friend aptly points out that Syria, like Saddam's Iraq, Gaddafi's Libya and Iran just could not be convinced of that.
She adds that in Syria, "asking about religion is not polite," because Syria has bred many of mankind's prime religions for thousands of years, and those millennia have taught Syrians to be sensitive, tolerant and respectful of all creeds. Something we clearly do not see in the pro-West Arabian sheikdoms, nor in the US, UK and EU with its anti-Islamic paranoia, and where laws are passed imposing the most blatant cultural, political and historical lies demanded by religious bigots who insist that their god will only accept their own holocaust offerings.
Clearly, another very good reason for neocon fanatics and their Orwellian Thought Police to want to take out Syria.
8) Syria proudly maintains and protects its political and cultural national identity – she stresses how Syria "holds on to its uniqueness," whilst respecting the uniqueness of others. The standardized coming world government simply abhors anybody standing up to its imposed standardization of thought, behavior and 'values', where the West's global megabrands, shopping malls, and fashion & style dictatorships "makes every place look pretty much the same, which leads to a very boring world."
Today, revolutionary thought in the West even amongst the young, boils down to choosing between Coke and Pepsi.
Clearly, a very good reason for Coke, Pepsi, McDonalds, Levis, Lauder, Planet Hollywood and Burger King to want to take out Syria.
Our young Syrian friend ends her message by reminding us that "if Syria falls, it could be the tipping point the ends in victory for the New World Order," adding that today "Syria is the frontline against the New World Order."
Wise words from a young lady who understands the catastrophic failure of the Western powers' political class, who have now completely turned our world upside down; where the very worst and most malignant criminals have infected governments and private power structures, be it in Washington, New York, London, and Paris, or in Berlin, Rome, Bogota, Madrid, Tokyo, Seoul, Amsterdam, Buenos Aires or Riyadh.
If sometimes Hollywood serves as a showcase that reveals the darkest recesses of the Western power elite's sick group psyche, we might even say that they are playing our the 'Planet of the Apes' saga, where a weird and hellish genetic inversion places horrendously destructive animals in places of world power, whilst noble vanquished humans are enslaved and thrown into cages.
Is this today's metaphor best describing the US against Syria drama?
The eight points mentioned above are as good a guide to get all our countries back on course as we can muster in today's troubled, out-of-control world.
Whether American, European, Arab, Muslim, Christian, Jewish, Buddhist, Hindu or Shinto, the time has come for 'We the People' to make our voices heard on the streets, with neighbors, family and friends, work and school colleagues, through social networks, demanding that Western so-called'democratically-elected' governments – all of which are the direct result of the will of money-sloshing elites that financed their climb to top government posts through their favorite lie they call 'democracy' – stop doing what they are doing, and start doing what we demand they do. Now; immediately: we must take our countries back.
Our young Syrian friend has certainly set an example for all of us to follow.
http://rt.com/op-edge/syria-world-hate-message-604/
3 Responses to "Why we all hate Syria so much. Do we?"
If there are any Syrians reading this blog, please don't think it is us as in we.
It is only the Zionists and Politicians that hate everyone.
We don't hate you.
the hating started with the jews,
they hate everyone, and in using us
the way they have, they have made us dislike them
jew riff raff.
My father was a paramedic, he is now retired due to poor health forcing him to stop working,
when he was called to duty he did not ask "Did he cause his own injury? Did he cause somebody
else's injury? Will he cause trouble if i help him?" to judge, but to determine what to do and
how to do it to HELP the patients health so he could transport him to where he would be made
well (ie hospital). He told me that when he is on shift at whatever ambulance depot he's at,
they put on the ambulance sirens and lights according to what priority the call is. For
example, a priority 1 – the lights flash bright and fast, the siren is loud and its pattern
can be described as frenetic, it will stay like that the whole journey. The lights and sirens
on an ambulance are like a car horn used to indicate you are going somewhere very important,
and how that "horn" "honks" indicates how important your journey is. No one questions WHY they
should "honk" the "horn", other drivers on the road behave according to what's valuable, and
how much they value it. It's a four letter word. Starts with L, ends with E, has an "if" in the
middle. Life is pretty important, and everyone generally respects the ambulance is off to
protect and save lives. Afterall, that could be them someday.
When a paramedic arrives on the scene – a priority 1 in this example – their first thoughts
are along the lines of; what the situation is eg 'is there a dog?' or 'are the police
involved?' or 'do i have a clear view of the scene of the incident?' and other such. Next
they'll determine where the patient is and ask what's their condition, then they figure out how
to enter/approach, after that what is appropriate to do exactly, and if it applies they consider
is it safe to approach or are they able to approach at all. "Who" is an indicator of where to
apply medical assistance, not a judge of character. The person most injured is aided first, that's
who is most in need of medical attention, the aim is to stabilise them. From there the next worst
off is attended to, again, to stabilise them. After each person is treated with the highest
quality of care the paramedic can provide (believe me it's a stressful job) the person(s) to be
transported to hospital are then loaded into an ambulance and taken to the nearest hospital
available (according to distance and beds available for type of condition).
Was anyone in there discriminated against for being wrong?
If a paramedic does this, the person whose blame interfered with their duties is… well one
way or another the situation is rectified. Sometimes there has be room to help someone through
being human, that shouldn't interfere though, most often someone is incompetent because the job
is too much for them or they just don't keep up with what's required of them.
Unless you're a soldier, I recommend becoming a paramedic. You learn how to help people get to
their place of healing. I hope that you remember nothing of my ramblings, and instead consider
who you value as important to you is worth your focus, and so are those whom you don't value.
Occams on Get ready for the INSURRECTION ACT
166,71,190,49,266,920,1149,1008,934,1064,1068,880,719,965,787,706,566,544,613,609,468,707,786,981

redpajama_set_name: RedPajamaCommonCrawl


text:
function toDense(sparse){
    return sparse.filter(function(el){return el!==null && typeof el!=="undefined";});
}

redpajama_set_name: RedPajamaGithub


text:
British Columbia's new Arbitration Act and the VANIAC
By David Wotherspoon, Rachel Howie, and Chelsea Wilson
British Colombia made substantial changes to the Arbitration Act that came into force on September 1, 2020. Our authors discuss how these changes impact the new dynamic and their effects on the appointment of arbitrators and the arbitration process.
Arbitration Act, British Columbia, procedure, VANIAC
About David Wotherspoon
David Wotherspoon is a partner in and lead of our Litigation and Dispute Resolution group in Vancouver. With a focus on corporate and commercial disputes, David's practice encompasses all areas of litigation and dispute resolution.
About Rachel Howie
Rachel is a partner in the Litigation and Dispute Resolution Group and co-leader for Dentons Canada's national ADR and Arbitration group. Her clients are primarily in the energy and natural resources industries, where she advises on complex matters that have an international or multi-jurisdictional aspect.
About Chelsea Wilson
Chelsea is a senior litigation associate in Dentons' Vancouver office and is a member of the Construction, Infrastructure and PPP and Litigation and Dispute Resolution practice groups. Chelsea provides strategic advice to clients during all stages of the procurement and construction process and has a breadth of experience advising developers and contractors on a variety of disputes including negligence claims, builders liens, enforcement of restrictive covenants, as well as residential and commercial real estate development.
Expanding options for efficiency: UNCITRAL's Expedited Arbitration Rules
By Rachel Howie, Emily McMurtry, and Maggie Sullivan
Probate application process made easier for "Small Estates" by Ontario's Smart and Stronger Justice Act
By Kathryn McCulloch


Last month I revisited the facts regarding buy to let and how it still offers a great return on investment, despite the recent changes to tax and profit calculations. To avoid the risk of having a blog which scrolled down the whole length of the page I didn't add the details of how Strawberry take it a step further and look to achieve the best possible return for our Landlords right from the very beginning, even when finding the property itself.
The roots of Strawberry are very much entrenched in the worlds of buy to let and relocation and our main strength has always been our knowledge of specific areas, not only in terms of what amenities lie within them but also where properties are selling well and letting easily for high rents. In short we get 'ahead of the pack', find the places which are 'on the up' and find the properties within them which will attract tenants easily and achieve good rental values.
When a new or existing Landlord comes to us with a budget in mind we will apply our many years of experience to come up with some options where yields can be maximised and tenants are in abundance. What's more, provided a Strawberry management package is put in place for the first 12 months (and with the many benefits to be had why wouldn't it be??) then this service is entirely free of charge.
Starting with the right property, and with great management in place, our Landlords can still see the kind of returns that make good financial sense and can minimise their day-to-day stress at the same time.


text:
1918 Spanish Flu Pandemic Pictures
Fads & Fashions
by Jennifer Rosenberg
Jennifer Rosenberg is a historian, history fact-checker, and freelance writer who writes about 20th-century history topics.
From the spring of 1918 to the early months of 1919, the Spanish flu pandemic ravaged the world, killing an estimated 50 million to 100 million people. It came in three waves, with the last wave being the most deadly.
This flu was unusual in that it was both extremely deadly and seemed to target the young and healthy, being particularly deadly to 20 to 35-year-olds. By the time the flu had run its course, it had killed upwards of five percent of the world's population.
Included below is a fantastic collection of pictures from the deadly 1918 Spanish flu pandemic, including tent hospitals, people wearing preventative masks, a sick child, no spitting signs, and much more.
A Nurse Wearing a Mask While Filling a Pitcher From a Fire Hydrant
During the 1918 Spanish Flu Pandemic Nurse wearing a mask as protection against influenza. (September 13, 1918).
Underwood Archives / Getty Images
Medical Personnel Giving Treatment to an Influenza Patient
During the 1918 Spanish Flu Pandemic Masked medical personnel giving treatment to an influenza patient. U.S. Naval Hospital, New Orleans, Louisiana. (Circa autumn 1918). Picture courtesy U.S. Naval History and Heritage Command.
A Letter Carrier Wearing a Mask for Protection
During the 1918 Spanish Flu Pandemic Letter carrier in New York wearing a mask for protection against influenza. New York City. (October 16, 1918). Picture courtesy National Archives at College Park, MD.
A Sign Warning Theater Goers Not to Enter if They Have a Cold
During the 1918 Spanish Flu Pandemic "Influenza frequently complicated with pneumonia is prevalent at this time throughout America. This theatre is cooperating with the Department of Health. You must do the same. If you have a cold and are coughing and sneezing do not enter this theatre.". Picture courtesy of the History of Medicine (NLM).
A Doctor Spraying the Throat of His Patient in an Attempt to Prevent Influenza
During the 1918 Spanish Flu Pandemic Preventive treatment against influenza, spraying throat. A.R.C. (American Red Cross). Love Field, Texas. (November 6, 1918). Picture courtesy of the National Museum of Health and Medicine.
Boxing Match on a Ship With Spectators Wearing Masks
During the 1918 Spanish Flu Pandemic Boxing match on the U.S.S. Siboney's forecastle, while she was at sea in the Atlantic Ocean, transporting troops to or from France in 1918-1919. Spectators are wearing masks as a precaution against the spread of influenza. Picture courtesy the U.S. Naval History and Heritage Command.
Rows of Beds Separated by Sneeze Screens in a Hospital
During the 1918 Spanish Flu Pandemic Naval Training Station, San Francisco, California. Scene in "D" Ward of the Station Hospital, showing sneeze screens erected around beds. Picture courtesy the U.S. Naval History and Heritage Command.
A Typist Wearing a Mask
During the 1918 Spanish Flu Pandemic Typist wearing mask, New York City. (October 16, 1918).
PhotoQuest / Getty Images
A Crowded Barracks With Beds Seaparated by Sneeze Screens
During the 1918 Spanish Flu Pandemic Naval Training Station, San Francisco, California. Crowded sleeping area extemporized on the Drill Hall floor of the Main Barracks, with sneeze screens erected as a precaution against the spread of influenza. Picture courtesy of the U.S. Naval History and Heritage Command.
A Sign Warning People Not to Spit on the Floor
During the 1918 Spanish Flu Pandemic Naval Training Station, San Francisco, California. "Do Not Spit On The Floor, To Do So May Spread Disease" sign on the balcony edge of the Drill Hall floor of the Main Barracks, which is in use as an extemporized sleeping area. Picture courtesy of the U.S. Naval History and Heritage Command.
A Child Sick With the Spanish Flu
During the 1918 Spanish Flu Pandemic Public Health: A child with influenza, her mother, and a visiting nurse from a local Child Welfare Association. Picture courtesy of the History of Medicine (NLM).
Sign Stating Number of Cases and Deaths at the Naval Aircraft Factory
During the 1918 Spanish Flu Pandemic Mounted on a wood storage crib at the Naval Aircraft Factory, Philadelphia. As the sign indicates, the Spanish Influenza was then extremely active in Philadelphia. Note the sign's emphasis on the epidemic's damage to the war effort. (October 19, 1918). Picture courtesy of the U.S. Naval History and Heritage Command.
Policemen in Seattle Wearing Masks
During the 1918 Spanish Flu Pandemic Policemen in Seattle wearing masks made by the Red Cross, during the influenza epidemic. (December 1918.). Picture courtesy of the National Archives at College Park, MD.
A Street Car Conductor Not Allowing Passengers Aboard Without a Mask
During the 1918 Spanish Flu Pandemic Street car conductor in Seattle not allowing passengers aboard without a mask. (1918). Picture courtesy of the National Archives at College Park, MD.
The Interior of an Influenza Ward at a U.S. Army Field Hospital
During the 1918 Spanish Flu Pandemic U.S. Army Field Hospital No. 127, Rengsdorf, Germany Interior view - Influenza Ward. Picture courtesy of the History of Medicine (NLM).
A Sign That States: Careless Spitting, Coughing, Sneezing Spreads Influenza
During the 1918 Spanish Flu Pandemic A sign that states: Prevent Disease, Careless Spitting, Coughing, Sneezing Spread Influenza and Tuberculosis. Picture courtesy of the History of Medicine (NLM).
A U.S. Army Tent Hospital for Influenza Patients
During the 1918 Spanish Flu Pandemic U.S. Army Base Hospital, Camp Beauregard, Louisiana. Tents for influenza patients.
Hulton Archive / Getty Images
An Influenza Ward at a U.S. Army Camp Hospital
During the 1918 Spanish Flu Pandemic U. S. Army Camp Hospital No. 45, Aix-les-Bains, France. Influenza ward No. 1. Picture courtesy of the History of Medicine (NLM).
Army Hospital Patients Wearing Masks at a Moving Picture Show
During the 1918 Spanish Flu Pandemic U. S. Army Hospital Number 30, Royat, France. Patients at moving picture show wearing masks because of an influenza epidemic. Picture courtesy of the History of Medicine (NLM).
Patients in Bed in an Army Field Hospital's Influenza Ward
During the 1918 Spanish Flu Pandemic U.S. Army Field Hospital No. 29, Hollerich, Luxembourg. Interior view - Influenza ward. Picture courtesy of the History of Medicine (NLM).
Naked Man Getting Inoculated Against the Spanish Flu
During the 1918 Spanish Flu Pandemic Embarkation Camp, Genicart, France. Flu and pneumonia inoculation being administered. Picture courtesy of the National Museum of Health and Medicine.
Liberty Loan Parade at Philadelphia
During the 1918 Spanish Flu Pandemic Liberty Loan Parade at Philadelphia, Pennsylvania. This parade, with its associated dense gatherings of people, contributed significantly to the massive outbreak of influenza which struck Philadelphia a few days later. (September 28, 1918). Picture courtesy the U.S. Naval History and Heritage Command.
A Cartoon That Shows a Mask as a Countermeasure
During the 1918 Spanish Flu Pandemic Cartoon by E. Verdier, published as cover art for the "Ukmyh Kipzy Puern," the magazine of the U.S Naval Cable Censor Office. The cartoon, and the face mask drawn in upper right, may reflect countermeasures against the 1918-19 influenza epidemic. Picture courtesy the U.S. Naval History and Heritage Command.
The 1918 Spanish Flu Pandemic That Killed Millions
Discover the Difference Between the Homophones Flew, Flu, and Flue
Medical School Personal Statement Examples
'One Flew Over the Cuckoo's Nest' Summary
'One Flew Over the Cuckoo's Nest' Overview
'One Flew Over the Cuckoo's Nest' Themes
'One Flew Over the Cuckoo's Nest' Characters
Protective Gear and Safety Equipment Photo Gallery
The Fascinating History of Korean Masks
5 Classic Hollywood Stars Who Passed for White
How to (Safely) Perform a the Thermite Reaction
The Pre-Med Student's Guide to Shadowing a Doctor
Respiratory System and How We Breathe
'One Flew Over the Cuckoo's Nest' Quotes Explained
What Happened in the 1910s? A Timeline of This Decade in History
Medical Schools in Virginia


What is Beckham law and its advantages for foreigners?
Home / Blog Offshore Company Formation / Recoverable taxes / What is Beckham law and its advantages for foreigners?
Beckham law was approved in 2005 but not all foreigners who move to Spain know about it and or are familiar with the advantages they can take from it. If you are an Expat in Spain and came to work to this country, you might be interested in knowing what it is about. In this article we will try to explain how you can apply for this law and what are the requirements that one has to fulfill in order to be authorized and access it.
Before moving to a foreign country, it is very important to get informed about all the laws and regulations that can affect your working or everyday life. This knowledge can prevent you from uncomfortable situation or offer you the benefits that you were not able to enjoy in your home country. One of the laws that affect all the foreigners who move to Spain is a Beckham law and its favorable tax regime.
The Beckham law allows the foreigners to take advantage of a special income tax regime by being taxed as non-residents for first 6 years of their residence. As a result, instead of paying 45-48% of tax the foreigners are only subject to 19.5% (in some cases 24%) tax rate. Being authorized by this law allows a foreigner to save some money for larger investments without this implying significant tax payment or set up of a new business with the benefit of this favorable tax regime. It is very important to note, that higher your income is, more money you can save of tax.
Why is this law called Beckham law?
The original name of this law is attributed to a former player of Real Madrid, one of the main football Spanish teams, David Beckham. He was the first one to take advantage of this law when he moved to this country and since them more and more football players and foreign investors apply to access this law. Even though this law can be applied to a majority of the foreigners who move to the territory of Spain, there are still some very important requirements that one should fulfill in order to benefit from it fully.
The first and the most important requirement that one should comply with is moving to Spain with working contract or a plan to register a new company.
The second requirement is not having been a resident in Spain during last ten years before the day apply for a Beckham law. Otherwise, you will have to wait for these 10 years to pass in order to be authorized by tax authorities.
The third condition is that prior to the application as a resident you must not have generated any income in any form of business on the territory of Spain.
If you comply with these three requirements you can apply for a Beckham law with a right to be authorized.
How to apply for Beckham law?
In order to apply for a Beckham law, one has to go to Public Treasury to request and fill in a model 149. Along the first year of your residence in Spain when you have to present annual tax returns you apply for a favorable tax regime by filling in a model 151. It is recommendable to go through this process during the first year of your residence in Spain, however, you can do it any moment during following 5 years.


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Why Dems Are Happy About the Primaries and I'm Not
The early spin on last night's primaries is that it was a good night for the Democrats, largely because Republican voters in four states rallied behind candidates who, in the words of Politico, are "tarnished by scandal, gaffes or some other significant vulnerability." And when you think about the likes of Linda McMahon, the World Wrestling tycoon, running for the Senate from Connecticut, the spin makes a lot of sense.
This morning McMahon was on "Good Morning America" and host George Stephanopoulos asked her whether she regretted anything she did during her tenure at WWE. As McMahon was dodging the question, by talking up the charity work the company had done, ABC rolled a clip of her in the wrestling ring, kicking a contestant in the shins. Throw in the misogyny of pro wrestling and I just don't see how Connecticut votes for her. (Click here to subscribe to The New Republic.)
But something else on the GMA broadcast got my attention--and not in a good way.
Stephanopoulos' other guest was Michael Bennet, the Democratic senator from Colorado, who'd just beaten insurgent Democrat Andrew Romanoff. Stephanopoulos put to Bennet the questions Republicans are already asking--namely, how does he defend his votes for health care reform, the stimulus, and other items on the Obama agenda? Bennet responded by dodging the question every bit as brazenly as McMahaon had hers, instead talking about the Republicans and their lack of a positive agenda.
Now, this was a brief interview, apparently taped last night not long after the results came in. Given time, perhaps Bennet would have made a more explicit and direct defense of those votes. And maybe this isn't so indicative of the way he's run, or will run, his campaign. Honestly, I haven't followed the Colorado race very closely. (Also on TNR: Can We Trust People Who Find God on Their Deathbeds?)
But Bennet's answer seems to be of a piece with what I understand to be the basic Democratic strategy for the midterms: Make this about the Republicans and their lack of fitness for office.
And that worries me for two reasons. First, it's a sign of how weak Democrats seem to be politically right now. It's just not a good sign when candidates have to run, as far and as fast as they can, from their party's agenda. And while I'd expect Democrats to do that in the deep South, Colorado is a swing state that's been trending Democratic and voted for Obama in 2008.
Second, I just don't have a lot of faith in this strategy. When a candidate's argument is all about why the other guy is worse, it doesn't exactly build up voter confidence. And if the candidate is an incumbent, I fear, a lot of voters will eventually decide some kind of change is better than the status quo.
This is the place where I usually remind readers that I could be totally wrong--that I am a lot less comfortable second-guessing strategy than I am second-guessing policy. If Bennet and other Democrats are making the campaign all about the Republicans, presumably that's because strategist who know a heck of a lot more than I do have run the different scenarios and determined this is the best way to win.
But over the last two years, in particular, I've come to realize that political strategists don't always know what they're talking about. What would happen if Bennet answered the Stephanopoulos question differently--what if he said, yes, I voted for the stimulus because it created a bunch of jobs and kept us from falling into a full-blown depression, while giving Colorado a ton of good public works? What if he said, yes, I voted for health care because it means that it means middle class people won't have to worry so much about their medical bills--or fear the possibility they'd get sick with no way of paying for treatment?
The strategists would say those are no-no's, because the laws are unpopular and voters hate spending. I'm not so sure a smart argument like that wouldn't win over at least some skeptics, and put Democrats in better standing.
Or maybe not. Like I said, I could be totally wrong. The bottom line is that Democrats are still in a lot of trouble, notwithstanding last night's results.
The one silver lining, I think, is that Republicans seem to be running a mirror strategy, making it all about the Democrats. The cliche in politics is that you can't beat something with nothing. But maybe you can beat nothing with nothing.
America, Andrew Romanoff, Jonathan Cohn, ABC, Linda McMahon, Michael Bennet, George Stephanopoulos, Colorado, Senate, Good Morning America, Connecticut


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A's Continue Marvelous Run Of Quality Starts, Tepid Tribe Loses Fifth Straight As Offense Sleeps Again
By AncientMariner Aug 25, 2010, 9:09pm CDT
Share All sharing options for: A's Continue Marvelous Run Of Quality Starts, Tepid Tribe Loses Fifth Straight As Offense Sleeps Again
Cleveland, OH (Sports Network) – Trevor Cahill pitched Oakland's 17th straight quality start, and the Athletics shut down the Indians in a 6-1 win at Progressive Field.
Cahill (14-5) continued his sensational season by pitching seven innings, allowing seven hits — tying a season-high — and one unearned run with two walks and two strikeouts. In his last six starts, Cahill has a 5-1 record and a 0.77 earned run average.
Kevin Kouzmanoff hit a three-run double to punctuate a five-run first, as the A's won for the sixth time in eight games.
Asdrubal Cabrera drove in the lone run for the Indians, who have lost five straight. Mitch Talbot (8-11) extended his winless skid to eight starts, allowing six runs, six hits and three walks in six innings.
Oakland took command of the game quickly, battering Talbot for five runs in the opening inning.
Coco Crisp grounded out leading off, but the next six batters reached base. Daric Barton walked, and Kurt Suzuki was hit by a pitch. Jack Cust followed with a single to plate Barton, and Mark Ellis drew a free pass to load the bases.
Rajai Davis' RBI single made it 2-0 and kept the bases loaded for Kouzmanoff, who scored all three runners with a double off the top of the wall in center.
The Indians finally got on the board on Cabrera's RBI single in the fifth, but Gabe Gross answered for Oakland in the sixth with a sacrifice fly.
Cleveland never threatened the rest of the way.
The 16 straight quality starts is the most by the franchise since the Philadelphia A's recorded 18 straight in 1927…A's starters have thrown six or more innings in 23 straight games…Oakland has not allowed more than five runs in 23 straight games, the longest streak in the AL since 1989…The A's lead the season series, 6-2…Shin-Soo Choo went 3-for-3 with a walk for the Indians.
Indians Go Quietly Into The Night; Five-Run First Carries A's to 6-1 Win
Indians Continue Series Against A's With Mitch Talbot On The Mound


text:
\section{Introduction}

Ramsey theory refers to a large body of deep results, which roughly say that any sufficiently large structure is guaranteed to have a large well-organised substructure. Its inception dates back to 1929 and the celebrated theorem of Ramsey \cite{ramsey1929problem} which states that any sufficiently large graph must contain a clique or an independent set of arbitrarily large size. In terms of quantitative results in 1935 Erd\H{o}s and Szekeres \cite{E-S} showed that any graph on $n$ vertices contains a clique or an independent set of size $0.5\log n$. On the other hand in what was one of the first applications of the now indispensable probabilistic method Erd\H{o}s \cite{ramsey-random} has shown that in a random graph $\mathcal{G}(n,1/2)$ w.h.p. there are no cliques or independent sets of size $2 \log n.$ Despite considerable effort \cite{explicit-ramsey-1,explicit-ramsey-2,explicit-ramsey-3,frankl-wilson} there are still no known non-probabilistic constructions which match the random graph.

We say a graph is \textit{$k$-Ramsey} if it contains neither a clique nor an independent set of size $k$. In general an $n$-vertex graph is said to be a \textit{Ramsey graph} if it is $k$-Ramsey for some $k$ ``close'' to $\log n.$ Over the years there has been a wide body of work studying properties of Ramsey graphs. In particular, based on the apparent difficulty of finding non-probabilistic Ramsey graphs, it is widely believed that with an appropriate definition of ``close'' any Ramsey graph must be random-like. While there is a vast number of results (see \cite{random-ramsey-property-1,random-ramsey-property-2,random-ramsey-property-3,random-ramsey-property-4,random-ramsey-property-5,random-ramsey-property-6,random-ramsey-property-7, random-ramsey-property-8} and references within) showing that indeed Ramsey graphs need to satisfy, to an extent, various properties usually associated with random graphs, our understanding of Ramsey graphs is still far from sufficient to consider this claim in any way settled.

Given an integer $k$ let $G$ be a $k$-Ramsey graph with the largest number of vertices. Observe that $G$ must contain \textit{both} a clique and an independent set of size $k-1,$ as otherwise we can add a new vertex joined to all or none of the vertices of $G$ to find a larger $k$-Ramsey graph. This shows that if we have a good Ramsey graph the largest clique and largest independent set should be of similar size. For example, if we consider the random graph, which is the best known Ramsey graph, it will with high probability contain both a clique and an independent set of size a $1$ or $2$ less than the largest $k$ for which it is $k$-Ramsey. Furthermore, Ramsey graphs satisfy a similar property locally as well. Given a $k$-Ramsey graph, since it has no clique or independent set of size $k$, we know that any subset consisting of $R(k,\ell)$\footnote{$R(k,\ell)$ denotes the off-diagonal Ramsey number, defined as the minimum number of vertices in a graph needed to guarantee there is either a clique of order $k$ or an independent set of order $\ell$.} vertices contains \textit{both} a clique and an independent set of size $\ell$.

The so called local-global principle, stating that one can obtain global understanding of a structure from having a good understanding of its local properties, or vice versa, has been ubiquitous in many areas of mathematics and beyond for many years \cite{local-global-1, local-global-2, local-global-3, local-global-4, local-global-5}. Keeping this in mind the following problem of Erd\H{o}s and Hajnal \cite{erdos-problems} seems to be very relevant to understanding Ramsey graphs. Given a $k$ (which might be a function of $n$) and an $n$-vertex graph $G$ they ask what is the smallest $m$ for which any $m$ vertex subset of $G$ contains both a clique and an independent set of size $k$? We denote the answer by $m_G(k)$ and say that a graph is \textit{\lramsey{m}{k}} if $m \ge m_G(k)$. To see the relation to Ramsey graphs first observe that being \lramsey{m}{2} is equivalent to being $m$-Ramsey. Secondly, more interestingly if we can find an $n\ge 3m$ vertex \lramsey{m}{k+1} graph it can contain at most $k-1$ vertex disjoint cliques of size $m/k,$ as otherwise they would give us a set of $m$ vertices in which there is no independent set of size $k+1$. The same clearly applies for independent sets. So if we remove a maximal collection of such cliques and independent sets we are left with a graph on at least $n/3$ vertices which is $m/k$-Ramsey in addition to still being \lramsey{m}{k+1}.

This means that understanding the behaviour of $m_G(k)$ very well could lead us to better understanding of Ramsey graphs, as well as interesting new examples of Ramsey graphs. Let us first consider what happens with $m_G(k)$ for the random graph $G\sim \mathcal{G}(n,1/2).$ If $k$ is small compared to $n$ we have that w.h.p.\ $m_G(k)= \Theta(k \log n)$ (see Section \ref{sec:small-k}) which as we will see, and as one might expect since for $k=2$ this is the standard Ramsey problem, is actually smallest possible among all graphs. On the other hand we also have by Erd\H{o}s's results \cite{ramsey-random} that $m_G(k)\ge 2^{k/2}$ so as $k$ becomes larger than $\log \log n$ the bound deteriorates quickly. For example, one needs at least $\sqrt{n}$ size sets to guarantee to be able to find both cliques and independent sets of size $\log n$.

A natural question is whether one can do better.
In fact, Erd\H{o}s \cite{erdos-problems} singled out the case of $k=\log n$ and asked if such graphs exist with $m_G(k)=(\log n)^3.$ If the answer were positive this would give rise to $(\log n)^2$-Ramsey graphs which are very different than $\mathcal{G}(n,1/2)$, since they would still be \lramsey{(\log n)^3}{\log n} which is very far from being true in the random graph. This question remains open. However, the first and the third author \cite{local-independence-numbers} show that $m_G(\log n) \ge \Omega ((\log n )^3/\log \log n),$ which perhaps validates Erd\H{o}s' intuition behind asking the question with the parameters he chose. The authors in fact essentially resolve an analogous local Tur\'an type problem which they use to obtain the above-mentioned bound. This problem, for various choices of parameters was also studied in \cite{erdos-problems, LIN-linial, LIN-us, LIN-kriv, LIN-kost}.

On the other hand, in terms of upper bounds nothing better than the one mentioned above, coming from random graphs, namely that there is a graph $G$ for which $m_G(\log n) \le O(\sqrt{n}),$ was known, leaving the possibility that no significantly different Ramsey graphs arise this way. Perhaps surprisingly our main result shows this is not the case, giving a twofold sub-polynomial improvement over the above bound.
\begin{thm}\label{thm:intro-log-n}
There exists an $n$-vertex graph $G$ for which $$m_G(\log n) \le 2^{2^{(\log \log n)^{1/2+o(1)}}}.$$
\end{thm}

As discussed above this gives rise to Ramsey graphs which, while being worse than the random graph, are significantly better than the classical explicit construction of Frankl and Wilson \cite{frankl-wilson} and even the recent breakthrough explicit construction of Barak, Rao, Shaltiel and Wigderson \cite{explicit-ramsey-1}. While unfortunately our construction does use randomness, it still gives rise to somewhat weaker Ramsey graphs which are very different from $\mathcal{G}(n,1/2)$ in the sense that they are \lramsey{2^{2^{(\log \log n)^{1/2+o(1)}}}}{\log n}.

So far we have restricted attention to the case of $k=\log n$ for simplicity and to allow for easier comparison between results. We do find examples in the general case as well.

\begin{restatable}{thm}{maingeneral}
\label{thm:intro-main-general}
For any $n \ge 4$ and $k \ge \log n$ there exists an $n$-vertex graph $G$ with $$\log \log m_G(k) \le 6\sqrt{ \log \log n \log \log k}.$$
\end{restatable}
Finally, we prove a simple proposition which determines $m_n(k)$, defined as the minimum of $m_G(k)$ over all $n$ vertex graphs $G,$ up to a constant factor, provided $k$ is small enough compared to $n$.

\begin{prop}\label{prop:small-r}
Provided $n$ is sufficiently large compared to $k\ge 2$ we have $m_n(k)=\Theta(k \log n).$
\end{prop}

\textbf{Notation.}
We denote by $K_k$ the complete graph on $k$ vertices and by $I_k$ the independent set of $k$ vertices. We denote by $\omega(G)$ the clique number of $G$. All our logarithms are in base $2$. When saying a graph $G$ is \lramsey{m}{r} we do not require either $m$ or $r$ to be integers, we want that any set of at least $m$ vertices contains a clique and an independent set of size at least $r.$

\section{Locally Ramsey graphs and lexicographic products}\label{sec:prelim}
We begin with a proposition which provides us with a starting point for our further constructions.

\begin{prop}\label{prop:random-graph}
For any $n$ there exists an $n$-vertex graph which is \lramsey{2^{r+8}\log n}{r} for all $r$.
\end{prop}
\begin{proof}
Let us first fix an $r\ge 2$ and set $m=\ceil{2^{r+8}\log n}$.
The chance that an $m$ vertex induced subgraph of $\mathcal{G}(n,1/2)$ does not contain $K_r$ (or $I_r$) is equal to the chance that $\omega(\mathcal{G}(m,1/2))<r.$ Using Janson's inequality (as in Section 10.3 in \cite{alon-spencer}) implies that this probability is at most $e^{-\frac{\mu^2}{2(\mu+\Delta)}}$ where $\mu=\binom{m}{r}2^{-\binom{r}{2}}$ and $\Delta=\mu^2\cdot \sum_{i=2}^{r-1} \binom{r}{i}\binom{m-r}{r-i}2^{\binom{i}{2}}/\binom{m}{r}.$ In our case $\Delta \le \mu^2 \binom{m}{r}^{-1}\binom{r}{2}\binom{m-r}{r-2}\cdot 2\sum_{i=2}^{r-1}2^{2-i}\le 2\mu^2r^4/m^2.$ It is easy to check that $\mu > m^2/r^4$ and hence $\mu +\Delta \le 2\Delta.$ Thus, $\frac{\mu^2}{2(\mu+\Delta)} \ge \frac{m^2}{8r^4}$ and therefore $\P(\omega(\mathcal{G}(m,1/2))<r) \le e^{-\frac{m^2}{8r^4}}.$ Finally, by a union bound, the probability that there exists a set of $m$ vertices in $\mathcal{G}(m,1/2)$ which does not have $K_r$ or $I_r$ is at most
$$ \binom{n}{m} \cdot 2 e^{-\frac{m^2}{8r^4}}\le 2e^{m(\log n-m/(8r^4))}\le 2e^{-m\log n/8} \le \frac{1}{n^2}.$$
Here in the second to last inequality we used $2^{r+8}\ge 9r^4.$ Now taking a union bound over all $r \le \log n$ we deduce that the desired graph exists.
\end{proof}

Note that in fact we proved that $\mathcal{G}(n,1/2)$ is \lramsey{2^{r+8}\log n}{r} with high probability. This bound can be slightly improved (see Section \ref{sec:small-k}) but we would gain little in our applications since $m_G(r)$ is going to ``go within a log'' so we chose for simplicity to show the above bound.

As already mentioned in the introduction, the random graph performs close to best possible when $r$ is very small. Our next construction already does much better when $r\ge \log n$, it serves as a basis and an illustration for our main construction presented in the following section.

\begin{lem}\label{lem:illustration}
For any integer $N\ge 4$ there exists an $N$-vertex \lramsey{m}{r} graph for any $m,r$ which satisfy $\log r \le \frac{(\log m)^2}{2^{9}\log N}$.
\end{lem}

Upon inverting we obtain a graph $G$ for which $m_G(r)\le 2^{16\sqrt{2\log N \log r}}.$ In particular, when $r=\log N$ this is already significantly better compared to about $\sqrt{N}$ in case of the random graph.

The example we use to prove the lemma is the lexicographic product of a random graph $\mathcal{G}(n,1/2)$ with itself multiple times. The lexicographic product $G\times H$ of two graphs $G$ and $H$ is defined as the graph on the vertex set $V(G)\times V(H)$ in which two vertices $(v,u)$ and $(x,y)$ are adjacent iff $v \sim_G x$ or $v=x$ and $u \sim_H y$. We write $G^{\ell}$ for the lexicographic product of $G$ with itself $\ell$ times. The main property of the lexicographic product which makes them natural candidates for our graphs is that clique and independence numbers are multiplicative (see \cite{lex-product-property}). Let us give some intuition as to why this is useful. Let $G \sim \mathcal{G}(n,1/2)$ and let us compare $G^\ell$ with the random graph on the same number of vertices $G' \sim \mathcal{G}(n^\ell,1/2).$ If we take an induced subgraph $H$ of $G$ on $m$ vertices then $H^{\ell}$ gives us a subset of $m^\ell$ vertices of $G^\ell$ which by the multiplicative property above contains both a clique and an independent set of size about $(2\log m)^\ell.$ On the other hand a subset of $m^\ell$ vertices of $G'$ w.h.p. does not have cliques (or independent sets) of size $2\log (m^\ell) = 2\ell \log m.$ This means that, at least if we restrict our attention to subsets of $G^\ell$ arising in this product fashion, $G^\ell$ is a much better \lramsey{m}{r} graph than $G'$ for most choices of the parameters. Of course one may not just restrict attention to such sets. The following lemma allows one to show that even in arbitrary subsets of $G^{\ell}$ it is possible to find big cliques (and independent sets).

The statement of the next lemma is somewhat technical, one of the reasons for this is that we want to state it in a very general form since we want to use it twice with very different choices of parameters. Second reason is that we believe it might be useful in improving other constructions people might come-up with in the future, as well as possibly for other problems involving subgraphs of lexicographic products.

Let us sketch the proof idea. The lemma starts with a graph $G$ on $n$ vertices in which for some $2\le r_2<r_3< \ldots <r_k$ we know that any $m_t$ vertices contain both $K_{r_t}$ and $I_{r_t}$ for all $t \ge 2$. We now take a subset $S$ of $G^{\ell}$ in which we want to find a big clique (the argument for an independent set will be analogous). For every vertex $v\in G$ we denote by $S_v$ the subset of $S$ consisting of all elements having $v$ as their first coordinate. We then look at $m_t$ vertices with highest $|S_v|.$  For some $t$ all these vertices need to actually have a reasonably large $|S_v|,$ say at least $m',$ as we know that $\sum_{v \in G} |S_v|=|S|.$ We now use the information that any $m_t$ vertices in $G$ have a clique of size $r_t,$ so in particular among our top $m_t$ vertices some $r_t$ make a clique, say $1,\ldots, r_t$. Now for any two elements of $S$ if their first coordinates are adjacent in $G$ then they are also adjacent in $G^\ell$. So if we look at the sets $S_{i}, i \le r_t$ all the edges between $S_{i}$ and $S_j$ for $i \neq j$ exist. In particular, if we find a clique in each of $S_i$ we may take a union of these cliques to obtain a clique in $S$. Since all vertices in $S_i$ share the first vertex and $|S_i| \ge m'$, finding a clique reduces to finding a clique in a subset of size $m'$ of $G^{\ell-1}$ for which we may use induction. For an example, if we work with the assumption that $G$ is \lramsey{2\log n}{2}, which we can get from the random graph, then $r_2=2, m_2=2\log n$ and let $|S|=m.$ We split in $2$ cases, either some vertex has $|S_v| \ge m/(4\log n)$ or there are $2\log n$ vertices which all have $|S_v| \ge m/(2n)$, since otherwise $|S|=\sum_{v \in G} |S_v|<(2 \log n)\cdot m/(4\log n)+n \cdot m/(2n)=m$. In the former case we take a vertex $v$ with $|S_v|\ge m/(4 \log n)$ and look for a clique in $S_v$. This reduces the task to looking for a clique in a subset of size $m/(4\log n)$ of $G^{\ell-1}$ which we do by induction. In the latter case by our assumption on $G$ among $2\log n$ vertices there must exist an edge $vu$ of $G.$ This means that we can find a clique of twice the size we are guaranteed in a subset of size $m/(2n)$ of $G^{\ell-1},$ which we once again do by induction. Optimising the choice of parameters will already bring us close to the bound in \Cref{lem:illustration}.

It will be more convenient to work with the inverse of $m_{G^\ell}(r).$ So, let $\beta_G(m,\ell)$ denote the largest $r$ such that in any $m$-vertex subset of $G^{\ell}$ we can always find both $K_r$ and $I_r$.

\begin{lem}\label{lem:lex-product}
Let $G$ be an $n$-vertex graph. Suppose that for some $2 \le r_2 < \ldots <r_k$ we know that $ m_G(r_t)\le m_t,$ for $2\le t\le k$. Then
\begin{equation}\label{eq:beta}
\log \beta_G(m,\ell) \ge \frac{\log m-\ell \log(2m_2)}{\max\limits_{2 \le t \le k} \left( \frac{\log (m_{t+1}/m_2)+t}{\log r_t}\right)},
\end{equation}
for any choice of $m_{k+1}\ge \min(n,m)+1.$
\end{lem}

\begin{proof}
First note that if for some $i<j$ we have $m_i>m_j$ we may decrease $m_i$ to be equal to $m_j$, since $r_i<r_j$ implies $m_j \ge m_G(r_j) \ge m_G(r_i)$ and doing this can only increase the target function. So we may assume that $m_i$ is increasing.

We will prove the claim by induction on $\ell$ for every $m$ satisfying $\min(n,m)+1 \le m_{k+1}$ (where we are treating $G,r_i$'s and $m_i$'s as fixed parameters).
For the base case of $\ell=1$ since $G^{\ell}=G$ we have $m \le n$ so $m_{k+1}\ge m+1$. We also have $m\ge 2m_2$ as otherwise $\beta(m,\ell)<0$ and the claim is trivial. So there exists some $2\le t\le k$ such that $m_t \le m < m_{t+1}.$ Since $m_t \ge m_G(r_t)$ it is sufficient to show that $\beta_G(m,1)\le r_t$. This indeed holds since $\log m - \log (2m_2)  < \log m_{t+1}-\log m_2= \log (m_{t+1}/m_2).$

Now assume $\ell \ge 2$ and that the claim holds for $\ell-1$ and any $m$ for which $\min(n,m)+1\le m_{k+1}.$ Let $S$ be a set of $m$ vertices in $G^{\ell}$. For any vertex $v\in G$ let us denote by $s(v)$ the number of elements in $S$ which have $v$ as their first coordinate. Let us also set $m_1=m_G(1)=1$ and $r_1=1$ for convenience.
\begin{claim*} For some $1 \le t \le k$ there are $m_t$ vertices $v \in G$ with $|S_v| \ge m/(2^t m_{t+1})$.
\end{claim*}

\begin{proof}
Let $v_1, \ldots, v_{s}$ be the vertices of $G$ which have $s(v_i)>0$ ordered so that $s(v_i)$ is decreasing in $i$. Note that $s \le \min(|G|,|S|) \le m_{k+1}-1.$ If $s(v_{m_i})
\ge  m/(2^i m_{i+1})$ for some $i\le k$ we may take $t=i$ and $\{v_1,\ldots, v_{m_t}\}$ all have $s(v_i) \ge  m/(2^t m_{t+1})$ so we are done. Hence, we may assume that $s(v_{m_i}) <  m/(2^i m_{i+1})$ for all $i\le k$. But this implies that $$|S|=\sum_{i=1}^s s(v_i)=\sum_{i=1}^{k}\sum_{j=m_i}^{m_{i+1}-1}s(v_j) \le \sum_{i=1}^{k} (m_{i+1}-m_i)\cdot m/(2^i m_{i+1})<m\sum_{i=1}^{k} \frac{1}{2^i}<m,$$ which is a contradiction.
\end{proof}

Since $m_t \ge m_G(r_t)$ among the $m_t$ vertices given by the claim there must be $r_t$ forming a $K_{r_t}$ in $G$. Let us denote by $S_1,\ldots,S_{r_t}\subseteq S$ the sets of elements in $S$ whose first coordinate is the $i$-th vertex of this clique.
By definition of the lexicographic product if $2$ elements in $S$ have adjacent first coordinates they are adjacent in $G^\ell$ as well. This means that we can combine the cliques we find in each of $S_i$ into a single clique in $S$. Furthermore, since all elements in $S_i$ have the same first coordinate we can delete their first coordinate when looking for a clique, which leaves us with a subset of at least $m'=m/(2^{t} m_{t+1})$ elements of $G^{\ell-1}$ to which we can apply induction\footnote{Note that $m'<m$ so $\min(n,m')\le \min (n,m)\le m_{k+1}-1$ as required by the assumption.} to find a clique. This means we can find a clique in $G^{\ell}$ of size at least $r_t \cdot \beta_G(m',\ell-1)$. Repeating the argument for independent sets we can find an independent set of this size as well. Therefore:
\begin{align*}
    \log \beta(m,\ell) &\ge \log \left (r_t \cdot \beta\left(m',\ell-1\right)\right) = \log r_t + \frac{\log m' -(\ell-1)\log (2m_2)}{C} \\
    & = \frac{C\log r_t+\log m-\log (2^{t} m_{t+1})-\ell \log (2m_2)+\log (2m_2)}{C} \\
    & \ge \frac{\log m-\ell \log (2m_2)}{C}
\end{align*}
where $C=C(m_2,\ldots,m_{k+1},r_2,\ldots,r_k)=\max\limits_{2 \le t \le k} \left( \frac{\log (m_{t+1}/m_2)+t}{\log r_t}\right)$ and the last inequality follows since we get an equality if $t=1$ and since $C \log r_t \ge \log m_{t+1}-\log m_2+t$ if $t \ge 2.$ As this is precisely the RHS of \eqref{eq:beta} this completes the proof.
\end{proof}

\Cref{lem:illustration} now follows as a corollary upon making the appropriate choice for the parameters.

\begin{proof}[ of \Cref{lem:illustration}]
Let $G$ be the $n$-vertex graph which is \lramsey{2^{t+8}\log n}{t} for all $t$ provided by \Cref{prop:random-graph}. This implies we can apply \Cref{lem:lex-product} with $r_t=t$ and $m_t:=2^{t+8}\log n \ge m_G(t)$ for $2 \le t \le k \le \log n,$ where $k$ is the largest integer such that $m_{k}\le n,$ and $m_{k+1}:=2^{k+9}\log n>n$. The lemma implies that
\begin{equation}\label{eq:1}
    \log \beta_G(m,\ell) \ge  \frac{\log m-\ell\log \log n-11\ell}{2\log n/\log \log n},
\end{equation}
since $\log m_2=10+\log \log n,$ $\log (m_{t+1}/m_2)=t-1$ and $(2t-1)/\log t$ is increasing in $t.$

We claim that $G^\ell$ with an appropriate choice of parameters $n$ and $\ell$ (in terms of $N$ and $m$) provides us with the desired graph. Note that we may assume that $m \le N$ (as otherwise the claim is vacuous) and that $\log m \ge 2^4 \sqrt{ 2\log N}\ge 2^4\log \log N$ as otherwise the claimed bound holds trivially. Let $n$ be the smallest integer for which $\log n / \log \log n \ge 32\log N /\log m$ and let $\ell:= \ceil{\log m/(32 \log \log n)}.$  With this choice of $n$ and $\ell$ we get a graph on $n^\ell \ge N$ vertices, since $$\ell \log n\ge \frac{\log m}{32 \log \log n} \cdot \frac{32\log N \log \log n}{\log m} =\log N.$$ Furthermore, by \eqref{eq:1} any set of size $m$ contains both cliques and independent sets of size at least $2$ to the power
$$\frac{\log m - 12 \ell \log \log n}{2\log n/ \log \log n} \ge \frac{\log m - \frac{12}{16}\log m }{2\cdot 33\log N /\log m} = \frac{(\log m)^2}{33\cdot 8\log N}$$
where in the denominator we used that $\log n / \log \log n < 33\log N /\log m$ (which holds since $\log n /\log \log n$ grows slower than $n$ and $\log N/\log m \ge 1$) and in the numerator $\ell \le \log m/(16 \log \log n)$ (which holds since $\log m/(32 \log \log n)\ge 1/2$ and $\ceil{x} \le 2x$ for any $x \ge 1/2$).
\end{proof}

\section{Locally Ramsey graphs and scrambling}
It might be tempting to try to reiterate the argument used in the previous section by starting with our better construction in place of the random graph. Notice however, that all our examples are in fact already powers of the random graph so doing this would only provide us with higher powers of the random graph which are already considered by our argument. This idea however has some merit when combined with a further twist. If we start with a high power of the random graph it will be a much better \lramsey{m}{r} than the random graph for some fixed value of $r$ but perform comparatively poorly for small values of $r.$ If we now scramble this graph a little bit, in the sense that we flip every edge and non-edge with some small probability this will improve the performance of our graph when $r$ is small while only slightly decreasing performance for larger $r$. Taking the lexicographic powers of this graph in place of the random graph is how we obtain our improved construction.

Let us define the $p$-scramble $\mathcal{G}(G,p)$ of a graph $G$ to be the graph obtained by independently removing every edge of $G$ and adding every non-edge of $G$ with probability $p$. The following lemma makes formal the above idea that by taking a $p$-scramble of an \lramsey{m}{r} graph $G$ we obtain a graph which is close to being as good a Ramsey graph as $\mathcal{G}(n,p),$ meaning it has no cliques or independent sets of size about $\log n/p$ or in other words is \lramsey{\log n/p}{2} but is in addition still close to being \lramsey{m}{r}.

\begin{lem}\label{lem:randomize}
If there exists an $n$-vertex \lramsey{m}{r} graph then there exists a graph which is both \lramsey{m}{\frac{r}{17 \log n}} and \lramsey{r/2}{2}, provided $r \ge 16 \log n$.
\end{lem}

\begin{proof}

Let $G$ be an \lramsey{m}{r} graph on $n$ vertices. Let $G'\sim \mathcal{G}(G,p)$ with $p:=\frac{8\log n}{r}\le 1/2.$

By an immediate coupling, the probability that $G'$ contains a $K_{r/2}$ (or $I_{r/2}$) is at most the probability that $\mathcal{G}(n,1-p)$ contains a $K_{r/2}.$ This probability is, by a union bound, at most $\binom{n}{r/2}(1-p)^{\binom{r/2}{2}}\le 2^{r/2 \cdot \log n-pr^2/8}=n^{-r/2} \le \frac14$ since $\frac r2=\frac{4\log n}{p}$. So $G'$ contains neither $K_{r/2}$ nor $I_{r/2},$ or in other words is \lramsey{r/2}{2} with probability more than $1/2$.

Given a set of $r$ vertices forming a clique in $G$ the probability that in $G'$ this set  still contains a clique of size $t$ is equal to the probability that $\mathcal{G}(r,1-p)$ has a $K_t$. Note that the expected number of missing edges is $\mu=\binom{r}{2}p=4(r-1)\log n$ so by Chernoff's inequality (see Appendix A of \cite{alon-spencer}) the probability that there are more than $2\mu$ edges is at most $e^{-\mu/3}$. If we have less edges then by Turan's theorem (see \cite{alon-spencer}) there is a clique of size at least $\frac{r^2}{4\mu+r}\ge \frac{r}{16r\log n+r} \ge \frac{r}{17 \log n}$. This means that with probability $1- \binom{n}{r}e^{-\mu/3}\ge 1-n^{r-\frac{4}{3}(r-1)}\ge 1/4$ any clique of size $r$ in $G$ contains a clique of size at least $\frac{r}{17 \log n}$ in $G'$. Repeating for the independent sets we conclude that $G'$ is \lramsey{m}{\frac{r}{17 \log n}} with probability at least $1/2$. Therefore, with positive probability the desired graph exists.
\end{proof}

Being more careful one can improve $\frac{r}{17 \log n}$ to $\Omega\left(\frac{r\log r}{\log n}\right)$ but not more (since $\omega(\mathcal{G}(r,1-p))=\Theta\left(\frac{\log r}{p}\right)$ w.h.p.) However, this improvement seems to be negligible in our applications so we opted for the above simpler argument. The following lemma gives our main construction. We obtain it by starting with our construction from the previous section, scrambling it using \Cref{lem:randomize} then taking an appropriate lexicographic power using \Cref{lem:lex-product} and repeating with this new graph. The parameter $t$ will control the number of iterations that we do. We also, for now, add an assumption that the clique/independent set size $r$ we are looking for is not too small.

\begin{thm}\label{thm:main-general}
For any $t \ge 2$ there exists a \lramsey{m}{r} graph on $N\ge 4$ vertices, provided $\log m \ge
t^{2t}(\log r)^{t} (\log N)^{1/t}$ and $\log r \ge t\log \log N$.
\end{thm}

\begin{proof}
Let us define $\log m(N,r,t):=t^{2t}(\log r)^{t} (\log N)^{1/t}.$ We will prove by induction on $t$ that for any $N \ge 4$ and $\log r\ge t \log \log N$ there exists an \lramsey{m(N,r,t)}{r} graph on $N$ vertices. The base case of induction for $t=2$ follows (with room to spare) from \Cref{lem:illustration}.

Let us take an \lramsey{m'}{r'} graph $G$ on $n \ge 4$ vertices
(with parameters $r',n$ satisfying $r' \ge 16 \log n,$ $\log r' \ge t \log \log n$ and $n \ge 4$, to be chosen later)
given by the inductive assumption for some $t\ge 2$, so with $m':=m(n,r',t)$. Let $G'$ be the scrambled graph given by \Cref{lem:randomize} applied to $G.$ So in particular $G'$ is \lramsey{r'/2}{2} and \lramsey{m'}{r''} where we write $r'':=\ceil{{r'}/{(17 \log n)}},$ note that we are using $r' \ge 16 \log n$ so that the lemma applies.

We now take lexicographic products of $G'.$ \Cref{lem:lex-product} allows us to use these locally Ramsey properties of $G'$ to give bounds on the locally Ramsey properties of $G'^\ell$ which will be our actual example for iteration $t$.
So in particular we may apply \Cref{lem:lex-product} with $r_2=2,m_2=r'/2 \ge m_{G'}(2); r_3=r'', m_3=m' \ge m_{G'}(r'')$ and $m_4=m+1$. The lemma implies
\begin{equation}\label{eq:2}\log \beta_{G'}(m,\ell) \ge \frac{\log m-\ell \log r'}{\max\left(\log m', \log m / \log r''\right)},
\end{equation}
since $\log (2m'/r')+2\le \log m'$ (since $r' \ge 8$) and $\log (2(m+1)/r')+3 \le m$ (since $r' \ge 16 \log n \ge 32$).

We note that at this point what remains to be done is to choose the parameters and use \eqref{eq:2} to show the induction step holds. The rest of the proof is somewhat technical and it might help the reader to at first ignore various constants and floors and ceils.
Let us now choose all our parameters in terms of $N,r$ and $t.$ Let $$m:=m(N,r,t+1),\:\:\:\: \log r':=\left(1+\frac{2}{t}\right)\log r, \:\:\:\: \ell:=\floor{\frac{\log m}{(t+1)\log r'}} \:\: \text{ and } \:\: \log n:=\ceil{\frac{\log N}{\ell}}.$$ Our goal is to show that with this choice of parameters the RHS of \eqref{eq:2} is at least $\log r.$ This would give us a graph on $n^\ell \ge N$ (by definition of $n$) vertices which is \lramsey{m}{r} so we obtain the desired graph by taking a subgraph consisting of exactly $N$ vertices. We first show the following easy inequalities.

\begin{claim*}
We have $16 \le n \le N,$ $\ell \ge 64$ and $\log r' \ge (1+1/t)(\log r+\log (17\log n)).$
\end{claim*}
\begin{proof}
Note that since $\log m=\log m(N,r,t+1)= (t+1)^{2(t+1)}(\log r)^{t+1} (\log N)^{1/(t+1)}>(t+1)^{2(t+1)}\log r$ so in particular $\frac{\log m}{(t+1)\log r'}= \frac{\log m}{(t+1)(1+2/t)\log r} \ge (t+1)^{2(t+1)-1}/(1+2/t) > 64,$ which in turn implies $\ell=\floor{\frac{\log m}{(t+1)\log r'}} \ge 64.$ This together with $N \ge 4$ and the definition of $n$ imply $n < N.$ If $m>N$ then there are no subsets of size at least $m$ in $G'^\ell$ so the induction step is vacuously true, therefore we may assume $m \le N$. Using this we get $\log n \ge \frac{\log N}{\ell} \ge \frac{\log N \cdot (t+1)\log r'}{\log m} \ge (t+1)\log r'>3,$ (where we used $r'>1$ and $t \ge 2$). This in particular implies that $\log n \ge 4$ and $\log n \le 4/3 \cdot \log N /\ell \le \log N / 32$ (using $\ell \ge 64$). This in turn implies $\log r' = (1+2/t)\log r\ge (1+1/t)\log r+\log \log N \ge (1+1/t)\log r+ \log (17\log n),$ where we are using $1/t \cdot \log r \ge \log \log N \ge \log (32 \log n).$
\end{proof}

This immediately implies the required inequalities on $n,r'$ and $t,$ indeed $ r' \ge 17 \log n$ and $n \ge 4$
while $\log r \ge (t+1)\log \log N$ implies $\log r' \ge \log r \ge (t+1)\log \log N \ge t \log n.$

Let us now turn to the main inequalities. Observe that
\begin{equation}\label{eq:3} \frac{\log m- \ell \log r'}{\log m / \log r''} \ge \left(\frac{\log m-\log m/(t+1)}{\log m}\right)\log r'' \ge \left(1-\frac1{t+1}\right)(\log r'-\log (17\log n))
\ge \log r,
\end{equation}
where in the last inequality we used the main inequality from the claim. This shows that one of the two desired inequalities that we need to show to conclude that RHS of \eqref{eq:2} is at least $\log r$ holds. The second inequality we need is equivalent to $\log m- \ell \log r' \ge \log m' \log r$ and is implied by $\log m \ge (1+1/t)\log m' \log r$ (since $\ell \le \log m /((t+1)\log r')$). Let us now show this inequality holds (recall that we have chosen $m'=m(n,r',t)$):
\begin{align*}
 (1+1/t)\log r \log m'  & = (1+1/t)\log r \cdot t^{2t}(\log r')^{t} (\log n)^{1/t}\\
  & \le (t+1)t^{2t-1}\log r \cdot (\log r')^{t+1/t} \cdot 2 \left(\frac{\log N}{\log m}\right)^{1/t}\\
  &\le (t+1)^{2(t+1)} (\log r)^{t+1+1/t} (\log N)^{1/t}\cdot (\log m)^{-1/t}\\
  & \le (\log m)^{1+1/t}\cdot (\log m)^{-1/t} =\log m,
\end{align*}
where in the first inequality we used $\log n \le \frac{4\log N}{3 \ell}\le \frac{4(t+1)\log r'\log N}{3 \log m}$ (following since $n \ge 16$ so $\log n \ge 4$ and definitions of $n$ and $\ell$) and $(4(t+1)/3)^{1/t}\le 2$ (since $t\ge 2$). In the second inequality we used $\log r' = (1+2/t)\log r$ and $(1+2/t)^{t+1/t}\le (1+1/t)^{2(t+1/t)}\le (1+1/t)^{2t+1}.$ The third inequality follows as $m=m(N,r,t+1)=(t+1)^{2(t+1)}(\log r)^{t+1}(\log N)^{1/(t+1)}$ since $\left(t+1+1/t\right)\cdot t/(t+1)=t+1/(t+1) \le t+1.$
Together with \eqref{eq:3} this shows that the RHS of \eqref{eq:2} is at least $\log r$ completing the proof.
\end{proof}
We were relatively lax with various estimates in the argument above for the sake of simplifying the inequalities as much as possible. For example, the optimal exponent of $\log r$ (which one may obtain using exact same parameters as we did above) is $(t+1)/2-2/t.$ We also note that the assumption $r \ge (\log n)^t$ was also made for the sake of simplicity, since for smaller values of $r$ the above argument would give barely any improvement over just using the above bound with $r=(\log n)^t$ and monotonicity of $m_G(r)$ in $r$. Let us now optimise over $t$ and obtain \Cref{thm:intro-main-general} as a corollary. Recall the statement of \Cref{thm:intro-main-general}.

\maingeneral*
\begin{proof}
If $\log N \le k \le (\log N)^t$ we use \Cref{thm:main-general}, with $r=(\log N)^t$ to obtain a graph $G$ with $\log m_G(k) \le \log m_G((\log N)^t)\le t^{3t}(\log \log N)^t(\log N)^{1/t}\le t^{3t}(\log k)^t(\log N)^{1/t}.$ If $k>(\log N)^t$ we may use \Cref{thm:main-general} directly with $r=k$ to conclude that for any $k \ge \log N$ there is a graph $G$ with $\log m_G(k) \le  t^{3t}(\log k)^t(\log N)^{1/t}.$

We now choose $t=\floor{ \sqrt{\frac{\log \log N}{\log \log k}}}$. If $t < 2$ we obtain that the desired inequality requires $m>N$ making the claim vacuous. Hence the above inequality gives us
$$\log \log m_G(k) \le 3t \log t+ t\log \log k+\frac{1}{t}\cdot \log \log N \le 4t\log \log k+\frac{1}{t}\cdot \log \log N\le 6\sqrt{ \log \log N \log \log k}$$
where we used $t \le \log \log N \le \log k$ and $t\ge 2.$
\end{proof}

\Cref{thm:intro-log-n} follows by simply plugging in $k= \log N$ in \Cref{thm:intro-main-general}.

\textbf{Remark.} By following the argument used in \Cref{prop:random-graph} it is not hard to show that $\mathcal{G}(n,p)$ is w.h.p.\ \lramsey{(1/p)^{r+8}\log n}{r} for all $r$, assuming $p \le 1/2$. Using this in \Cref{lem:randomize} would give us a graph which is \lramsey{(1/p)^{r+8}\log n}{r} for all $r$ in addition to being \lramsey{m}{\frac1{3p}}, provided $1/p\ge r/(8\log n)$. We can then use this extra information for small values of $r$ similarly as we did in the proof of \Cref{lem:illustration} to obtain an improvement in \eqref{eq:2}. Ultimately, this would lead to an improvement in \Cref{thm:main-general} in which we divide by roughly a $(\log \log N)^{(t-3)/2}$ factor which would only slightly improve the $o(1)$ term in \Cref{thm:intro-log-n}.

\section{Small cliques and independent sets}\label{sec:small-k}
In this section we show \Cref{prop:small-r}. We begin with the lower bound.
\begin{prop}\label{prop-lb}
Provided $n\ge 4r \log n$ and $r\ge 2$ we have $m_n(r)\ge (0.5+o(1))r \log n.$
\end{prop}
\begin{proof}
Let us start with the lower bound. Given an $n$-vertex graph $G,$ by the standard bound on Ramsey numbers (see e.g. \cite{E-S}), $G$ must contain a clique or an independent set of size at least $0.5 \log n$. If we remove this set and repeat $2r - 3$ times we get either $r - 1$ vertex disjoint cliques or $r - 1$ vertex disjoint independent sets of size at least $0.5  \log (n/2)$ as at each step we are left with at least $n-2r\log n \ge n/2$ vertices. The union of these sets give us a set of $(0.5+o(1))r \log n$ vertices in which we can not find a clique (if the sets were independent) or an independent set (if the sets were cliques) of size $r.$ This shows that $m_G(r)\ge (0.5+o(1))r \log n.$
\end{proof}

\textbf{Remark.} The above bound applies for essentially the whole range but is beaten by the approach in \cite{local-independence-numbers} as soon as $r$ is bigger $\log \log n$. They show that provided there is an $I_r$ in every subset of size $s$ then one can find an independent set of size $\Omega(r \log (n/s) /\log (s/r))$. Using this for our graphs and finding $r$ copies of this big independent set as in the proof above would show $m_n(r) \ge \Omega(r^2 \log n /\log \log n),$ provided $r$ is at most polylogarithmic in $n$ which beats the bound in \Cref{prop-lb} when $r\gg \log \log n$.

Let us now turn to the upper bound. Perhaps not too surprisingly since we are working with ``small'' values of $r$ the example is the random graph $G \sim \mathcal{G}(n,1/2)$. Since $n$ is much bigger than $r$ the argument we used to prove \Cref{prop:random-graph}, even when done more carefully, would only give us $m_G(r) \le O(r^4 \log n)$ so we make use of a slightly different approach.

\begin{prop}
Provided $n$ is sufficiently large compared to $r$ we have for $G\sim \mathcal{G}(n,1/2)$ that w.h.p.\ $m_G(r)= \Theta(r \log n).$
\end{prop}

\begin{proof}
The lower bound follows from the previous proposition. So we focus on the upper bound.
It is well-known (see \cite{chi-clique}) that almost all graphs without $K_r$ are
$r - 1$ colourable. Hence, provided $m \to \infty $ as $n \to \infty$ we have $\P(\omega(\mathcal{G}(m,1/2))<r)\le (1+o(1))\P(\chi(\mathcal{G}(m,1/2))<r) \le (1+o(1))r^{m}2^{-(1+o(1))m^2/(2r)}.$ Here, the last inequality follows since there are $(r-1)^m$ many ways to assign $r-1$ colours to $m$ vertices and given a colouring there are at least $(r-1)\binom{m/(r-1)}{2}\ge (1+o(1))m^2/(2r)$ pairs of vertices assigned the same colour which are not allowed to appear as edges.
The same estimate holds for the probability that a graph on $m$ vertices contains no independent set of size $r$. Thus in $G$ the expected number of sets of size $m$ which contain no clique of size $r$ or no independent set of size $r$ is at most
$$\binom{n}{m}\cdot 2(1+o(1))r^{m}2^{-(1+o(1))m^2/(2r)}$$
which tends to $0$ for $m = (2 + o(1))r \log n,$ completing the proof.
\end{proof}
The fact that almost all $m$-vertex $K_r$-free graphs are $r-1$ colourable has recently been shown to be true for $r$ up to $\log m/(10 \log \log m)$ in \cite{containers}. Since our sets have size $m$ which is roughly $r \log n$ this means that $n$ sufficiently large in the above result may be replaced with $r \le O(\log \log n/ \log \log \log n).$

\section{Concluding remarks}
In this paper we study the function $m_G(r)$ with particular interest in how small it can be. The function $m_n(r)$ defined as the minimum of $m_G(r)$ over all $n$-vertex graphs $G$ was introduced by Erd\H{o}s and Hajnal almost 30 years ago. Combined with the lower bound obtained in \cite{local-independence-numbers} we obtain
$$\frac{(\log n)^3}{\log \log n}\le m_n(\log n)\le 2^{2^{(\log \log n)^{1/2+o(1)}}}.$$

In general it would be very interesting to get better bounds on $m_n(\log n)$ and in particular answer Erd\H{o}s' question of whether $m_n(\log n)> (\log n)^3$. In fact, the authors suspect that $m_n(\log n)$ may be bigger than any fixed power of $\log n$. Our initial examples in Section \ref{sec:prelim} are essentially the classical examples of explicit Ramsey graphs due to Naor \cite{Naor-lex}, following-up on the idea of using lexicographic products to build Ramsey graphs due to Abbott \cite{abbott-explicit-ramsey}. There has recently been some major progress on finding better explicit Ramsey examples \cite{explicit-ramsey-1,explicit-ramsey-2,explicit-ramsey-3}. It would be interesting to see if one can combine these graphs with our ideas to improve our upper bound.

Another possibly interesting perspective arises if we consider a colouring restatement of our problem. Note that $m_n(r)-1$ may be defined as the largest number $m$ such that in any $2$-colouring of $K_n$ we can find $m$-vertices not containing a monochromatic $K_r$ in one of the colours. With this in mind one can define the $m$-local Ramsey number $LR_m(G)$ of a graph $G$ as the smallest $n$ for which in any $2$-colouring of $K_n$ there are $m$ vertices not containing a monochromatic copy of $G$ in one of the colours. For example if $m$ is sufficiently larger than $r$ \Cref{prop:small-r} implies that $LR_m(K_r)=2^{\Theta(m/r)}$. Natural generalisations to more colours or asymmetric graphs might hold some interest as well.

\vspace{0.2cm}
\textbf{Acknowledgements.} We would like to thank David Conlon for useful conversations and remarks and anonymous referees for their comments.
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\href}[2]{#2}

redpajama_set_name: RedPajamaArXiv


text:
Home/world/NATO countries bought 14 Su-25 aircraft from Bulgaria and gave them to Ukraine
NATO countries bought 14 Su-25 aircraft from Bulgaria and gave them to Ukraine
NATO countries bought 14 Soviet-made Su-25 attack aircraft in Bulgaria and handed them over to Ukraine. This is reported by "Bloomberg", referring to the Dutch analytical project Oryx, led by a former military officer, and also to statements by supporters of Ukraine.
In the publication of "Bloomberg" and one of the cited sources, there is no mention of direct supplies of weapons from Bulgaria to Ukraine, but of those sold to other countries, which were subsequently handed over to Kyiv.
Already in May 2022, a publication appeared in the Polish publication Defense.pl, which cited the NEXTA agency, according to which 14 Su-25 attack aircraft from "one of the Eastern European partners" were delivered to Ukraine. The text then noted that the machines had arrived "in parts" by land. And the agency claimed that the number of delivered aircraft corresponded to the size of the Bulgarian Su-25 fleet and recalled that eight of them were recently modernized by the Belarusian aviation plant in Baranovichi.
Immediately after that, "24 Chasa" sought the Ministry of Defense for a comment, but they denied the information: "The SU-25 attack aircraft, which Ukraine allegedly received, are not from Bulgaria."
It is likely that the same SU-25s are now being referred to in the Bloomberg publication, as their numbers match.
The Bulgarian parliament voted at the end of last year to send weapons to Kyiv, and in the previous National Assembly this was the subject of fierce disputes and was rejected by the then coalition partner BSP. At the same time, in the spring of 2022, allegations appeared that the shipment of military equipment to Ukraine was taking place through Poland.
By 2022, it was known that Bulgaria had 14 Su-25s. In October, flights with them stopped after one of them crashed at the Bezmer air base, and the pilot ejected, which saved his life.
The authors also refer to data from Oryx, some of which have been collected in a special publication since the first months of the war. The date is from April 11, but from the arsenal described and the transmission times indicated in the list, it follows that the publication is constantly updated.
The Oryx website mentions 14 Su-25 aircraft "purchased by Bulgaria from NATO member states and delivered to Ukraine". April 2022 is listed as the time of the deal. Another Balkan country has given Su-25, but in August – North Macedonia (other cases of aircraft are not mentioned).
Bulgaria, according to the same publication, bought from the Czech Republic an unclear number of 152 mm D-20 howitzers (in April 2022), BM-21 Grad rocket launchers (May or June 2022), also from the Czech Republic, and an unknown number of T-72M1 tanks in May or June.
The same platform documented the destruction of a total of 40 T-72 tanks used by the Ukrainian army during the war. Flags indicate Poland, Bulgaria and the Czech Republic as origin.
Oryx's creator, Steen Mitzer, is a military history book author who follows conflicts in the Middle East and Africa, the Caucasus and elsewhere, and who also provides data on the war in Ukraine.
These are not the first publications that mention Bulgarian weapons in Ukraine, but usually do not specify when they were sold. At least twice – in the summer and in December – a Ukrainian brigade boasted of its weapons, which, according to reports from Ukrainian sources, were delivered precisely from Bulgaria, writes "Diary".
They discovered why we get sick more in winter. And the reason is unexpected
Putin is grooming his daughter to be his successor in the Kremlin
The military will now look after themselves, giving up private security
they fined the prime minister for driving without a seat belt
I am related to Count Dracula


text:
Regimes of nonlinear depletion and regularity in the 3D Navier-Stokes equations Academic Article
2014 IOP Publishing Ltd & London Mathematical Society. The periodic 3D Navier-Stokes equations are analyzed in terms of dimensionless, scaled, L2m-norms of vorticity Dm(1m<). The first in this hierarchy, D1, is the global enstrophy. Three regimes naturally occur in the D1-Dmplane. Solutions in the first regime, which lie between two concave curves, are shown to be regular, owing to strong nonlinear depletion. Moreover, numerical experiments have suggested, so far, that all dynamics lie in this heavily depleted regime [1]; new numerical evidence for this is presented. Estimates for the dimension of a global attractor and a corresponding inertial range are given for this regime. However, two more regimes can theoretically exist. In the second, which lies between the upper concave curve and a line, the depletion is insufficient to regularize solutions, so no more than Leray's weak solutions exist. In the third, which lies above this line, solutions are regular, but correspond to extreme initial conditions. The paper ends with a discussion on the possibility of transition between these regimes.
Donzis, Diego
NONLINEARITY
Gibbon, J. D., Donzis, D. A., Gupta, A., Kerr, R. M., Pandit, R., & Vincenzi, D.
Gibbon, John D||Donzis, Diego A||Gupta, Anupam||Kerr, Robert M||Pandit, Rahul||Vincenzi, Dario
IOP Publishing Publisher
NONLINEARITY Journal
Navier-stokes
10.1088/0951-7715/27/10/2605


The Mosaic Company (NYSE:MOS) recorded a trading volume of 3113140 shares, below its 5-day volume average of 3730720 shares. The stock kicked off the session at $28.09 and later approached $28.3 with a change of 0.43%.
The Mosaic Company is currently rated as outperform by 19 stock analysts, with the company still having around 2% shares outstanding that can be sold short legally. The value of their shares went down by -15.47% last month. Over the course of a year, the stock has fallen by -3.12%. Financial analysts are becoming more bullish than ever, with the 9 analysts who cover the activities of MOS advising investors to buy. Experts on Wall Street have set a price target of $37.74 for the stock. The decision was arrived after looking at their 33.36% gain from current levels. At the moment, The Mosaic Company (MOS) trading price is around -24.27% below its 52-week high.
The Mosaic Company (MOS) was successful in overcoming the expected $0.57 as the stock recorded an earnings of $0.77 a piece during their last financial report presentation. Revenue, meanwhile, had a distance of -19.02% from the previous quarter, with the company recording $2.37 billion in revenue.
The shares of The Mosaic Company (NYSE:MOS) have struggled and recovered 23.58% ever since it recorded its 52 week low. During the course of 5 days, the stock price volatility has remained at 1.92%, increasing the time frame to a month, the volatility recorded was 2.86%. The price of their shares still hasn't been successful in staying above its 20 day-moving average, staying at a distance of -5.44% and while its 50-day moving average is around -8.83%. In a look at the previous five trading session MOS stock has returned -0.28% and sank -8.94% when compared to its 200-day moving average that is $31.96. The Mosaic Company (MOS) will look to keep its 13.06% increase that it witnessed in the past year.
As the normal trading session ended, the stock of Boston Scientific Corporation (NYSE:BSX), fetched in a $0.29 spike to $38.15. The session began with the stock price at $37.71, hitting a high of $38.26 before down again. For the week, analysts seem to be okay with their bullish with the consensus call staying at 1.5. Boston Scientific Corporation gets 16 buy analyst recommendations while 2 recommend a hold and 0 rated it as a sell. The shares of the company tumbled -6.95% from its high $41, with their overall market value now roughly $52.8 billion.
The mean recommendation of BSX on Reuter's scale remained unchanged from 1.42 to 1.42 within 30 days. This figure implies that the majority of analysts rate it as a buy. The price target for Boston Scientific Corporation (NYSE:BSX) is now around $44.07 a share, implying the stock has what it takes to grow by 15.52%. The recent trading session also indicated the stock needs to add 25.82% more meet analysts' high consensus price target.


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Signet (SIG) Q3 Earnings & Revenues Beat, FY22 View Higher
Zacks Equity Research December 02, 2021
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Signet Jewelers Limited (
SIG - Free Report) reported stellar third-quarter fiscal 2022 results. Both the top and the bottom line beat the Zacks Consensus Estimate and improved year over year.
Signet's e-commerce business performed impressively during the quarter. Consumers are responding positively to SIG's refreshed merchandise assortment as well as the advanced capabilities to the Connected Commerce and fulfillment capabilities. SIG's Inspiring Brilliance strategy is also well on track.
Over the past three months, shares of this presently Zacks Rank #1 (Strong Buy) player have gained 35.8% in the past six months, outperforming the industry's rise of 12.4%. You can see the complete list of today's Zacks #1 Rank stocks here.
Q3 Details
Signet reported adjusted earnings of $1.43 per share that beat the Zacks Consensus Estimate of 67 cents. The bottom-line figure rose significantly from 11 cents earned in the year-ago quarter.
This jewelry retailer generated total sales of $1,537.8 million, surpassing the Zacks Consensus Estimate of $1,452 million. The top line also increased 18.3% year over year and 29.5% from the third-quarter fiscal 2020 actuals. Same-store sales rose 18.9% year over year and 37.2% from the third-quarter fiscal 2020 reading.
Signet Jewelers Limited Price, Consensus and EPS Surprise
Signet Jewelers Limited price-consensus-eps-surprise-chart | Signet Jewelers Limited Quote
E-commerce sales jumped 14.4% from the prior-year quarter's level to $273.1 million. Brick-and-mortar same-store sales grew 20.3% year over year.
A Sneak Peek Into Margins
Gross profit in the reported quarter amounted to $575.6 million, up from $436.5 million in the year-ago fiscal's comparable quarter and $369.1 million in third-quarter fiscal 2020. Gross margin came in at 37.4%, which expanded 400 basis points (bps) year over year.
Selling, general & administrative expenses (SG&A) came in at $470.5 million, up from $389.3 million in the prior fiscal year's comparable quarter. Signet reported an adjusted operating income of $105.2 million, significantly up from $46.8 million recorded in the year-ago fiscal's comparable quarter and against an adjusted operating loss of $29.3 million.
Segment Discussion
Sales in the North America segment increased 17.9% on a reported basis to $1,394.2 million. Same-store sales jumped 19.8% from the year-ago quarter's levels, buoyed by broad-based category strength. Average transaction value (ATV) rose 15.2%, while the number of transactions climbed 3.5%
The segment's e-commerce sales grew 14.9%, while brick-and-mortar same-store sales jumped 21.5%.
Sales in the International segment rose 13.1% on a reported basis to $120.9 million. Same-store sales at the segment increased 8.8% year over year. ATV dipped 4%, while the number of transactions increased 12.8%. E-commerce sales improved 8.6%, while brick-and-mortar same-store sales rose 8.9%.
Signet ended the quarter with cash and cash equivalents of $1,516.9 million, accounts receivable, net of $19.3 million and inventories of $2,148.3 million. Long-term debt was $147 million at the end of the reported quarter, down from $1 billion million at the end of the year-earlier quarter. Total shareholders' equity was $1,534.6 million at the end of the quarter.
For the 39-week period ended Oct 30, 2021, Signet generated net cash of $483.9 million from operating activities. It had adjusted free cash flow of $352.1 million.
During fiscal 2022 to date, Signet repurchased roughly 467,000 shares for $41.1 million. As of Oct 30, 2021, SIG had $183.9 million remaining under the share repurchase authorization.
As of Oct 30, 2021, Signet had 2,851 stores. SIG projects closing about 75 stores in fiscal 2022 while opening around 85.
Signet raised expectations for fiscal 2022. For fiscal 2022, management expects total revenues of $7.41-$7.49 billion, up from $7.04-$7.19 billion predicted earlier. Signet now anticipates same-store sales in the range of 41-43%, up from the earlier view of 35-38%.
Adjusted operating income is anticipated in the range of $777-$814 million, suggesting a rise from the prior projection of $680-$735 million. Capital expenditures are likely to come in the $190-$200 million bracket.
For fourth-quarter fiscal 2022, management expects revenues of $2.40-$2.48 billion. Signet expects same-store sales in the band of 6-9%. Adjusted operating income is expected to be $280-$317 million.
Signet purchased Diamonds Direct, which is included in the fourth quarter and fiscal 2022 outlook from the date of acquisition on Nov 17, 2021. Management anticipates a shift of consumer discretionary spending from the jewelry category to experience-oriented categories in the fiscal fourth quarter.
Cost savings are likely to benefit SG&A and the gross margin. SIG continues focusing on mitigating the impacts of supply-chain disruptions stemming from the pandemic.
However, management is cautious about the rest of the fiscal year due to the uncertainties triggered by the pandemic and the Omicron variant.
More Key Picks
Some other top-ranked stocks include Boot Barn Holdings (
BOOT - Free Report) , Tractor Supply Co. (
TSCO - Free Report) and Target (
TGT - Free Report) .
Boot Barn Holdings, a lifestyle retailer of western and work-related footwear, apparel and accessories, sports a Zacks Rank #1 at present. The stock has jumped 69.2% in the past six months.
The Zacks Consensus Estimate for Boot Barn Holdings' current financial-year sales and earnings per share (EPS) suggests growth of 54.4% and 183.3%, respectively, from the year-ago period's corresponding figures. BOOT has a trailing four-quarter earnings surprise of 35.3%, on average.
Tractor Supply, a rural lifestyle retailer in the United States, flaunts a Zacks Rank of 1, currently. TSCO has a trailing four-quarter earnings surprise of 22.8%, on average. Shares of TSCO have surged 32.6% in the past six months.
The Zacks Consensus Estimate for Tractor Supply's current-year sales and EPS suggests growth of 19% and 23.9%, respectively, from the year-ago period's corresponding readings. TSCO has an expected EPS growth rate of 9.6% for three-five years.
Target, a renowned omnichannel retailer, presently carries a Zacks Rank #2 (Buy). TGT has a trailing four-quarter earnings surprise of 19.7%, on average. The stock has rallied 22% in the past six months.
The Zacks Consensus Estimate for Target's current-year sales and EPS suggests growth of 14% and 39.6%, respectively, from the corresponding year-ago period's levels. TGT has an expected EPS growth rate of 14.4% for three-five years.
In-Depth Zacks Research for the Tickers Above
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Boot Barn Holdings, Inc. (BOOT) - Free Report >>
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