In the direct-to-video "MagiRanger- The Movie: The Bride of Infershia", Kai Oz's soccer team manager and [not-so-secret] crush, Yamazaki Yuki, is kidnapped by King Glum. This Infershia general plans to marry her in order to release and awaken Emperor N. Ma. Though Yamazaki resists, the Sky Saint traitor Memmei casts a spell over her using a diamond ring. To get her back, the five Oz siblings climb a huge Mandoru-boy to reach the sky-bound lands of Magitopia, and its leader the great Sky Saint Magiel [played by the late legendary Super Sentai actress Machiko Soga in her last on-screen role]. When Magiel is convinced of their intentions, she gives to Kai a Majuu Unigolon to help him travel into the Hades underground. As a point of interest to the Oz siblings, their father, the fallen champion Sky Saint Blagel, had used this Unigolon as his steed in battle. The two arrive just in time to prevent the Infershia wedding between King Glum and the Bride of Infershia, and MagiRed rescues Yamazaki. After returning her safely to the surface, King Glum engages the five MagiRanger and is defeated. But the madoushi Memmei uses his dark magic to grow Glum and rebuild him into the giant Sword of Glum- a monster with metal blades and spikes, shaped like a scorpion. To combat the Sword of Glum, MagiRed changes into Saint MagiPhoenix and then merges with the Unigolon to form Tenkuu Gattai Saint Kaiser. Together with MaRyuu Gattai MagiDragon and Mahou Tetsujin Travelion, the Sword of Glum is destroyed with a series of magical attacks.
In Stage 34- "Bond of Courage" in the series, Wolzard is unexpectedly & shockingly revealed to be the lost Magitopia champion Blagel. Not only that, but this also makes him father of the five Oz siblings, whom they thought had died in the war fifteen years earlier! He changes into his human form, Izamu, and teleports them all to the surface of Infershia. There, he explains that he had been trying to seal the Infershia from the inside, but because of damage done in the final battle by the traitorous Sky Saint Raigel, it left open the opportunity for Emperor N.Ma to transform him into the loyal Madou Knight Wolzard, and blocked out his real memories. But before the five siblings have a chance to absorb all of this, Memmei and Vancuria take Blagel back into Infershia, and unleash the meshed Chimera creature once more on the MagiRanger. Just in the nick of time, MagiShine and Lunagel arrive riding a Majuu Unigolon- a legendary horned horse that can freely travel between Marudeonna universes, Magitopia, and the real world. The Unigolon takes the MagiRanger back to the real world; the two Sky Saints, however, chose to remain behind to settle the score against the traitor Memmei. Chimera follows the MagiRanger and grows to gigantic proportions. While the others are still weakened with grief and shock, MagiRed impatiently and aggressively transforms himself and Unigolon into their giant forms and then into Tenkuu Gattai Saint Kaiser. MagiDragon soon follows and is overpowered by Chimera, but Saint Kaiser, in a rage, is able to defeat the monster. Sadly, no sooner does he do that when Mahou Gattai WolKaiser appears, and the two engage in a heated battle... avenging youngest son against cursed trapped father.
Since Saint Kaiser appeared in the OVA movie as well, which had no real continuity with the series, there is no way to tell which version is the 'correct' version. However, for the sake of less confusion, I will use the series as basis for this review, since it was more important to the plot that way than it was in the movie.
Saint MagiPhoenix (back) is nothing but a repaint of the MagiPhoenix figure from the DX Majin Gattai MagiKing (sold separately). The silver on its legs, metallic blue for its belt, and the artistic 'M's on both its chest and head have been repainted gold. There is also a small diamond on his forehead that was originally painted gold that has been left the original red color of the hard PVC plastic. Aside from these few paint alterations, MagiPhoenix and Saint MagiPhoenix are exactly the same. Saint MagiPhoenix forms half of the chest of Saint Kaiser.
The Majuu Unigolon (back) is fairly close to the dark Majuu Barikion, but there are differences other than coloring. The most obvious one is that Unigolon's head has been redesigned to incorporate a long unicorn's horn with an 'M' molded on either side, and a screw-like pattern running across it's length. Its mane has also been changed, with the screw pattern running its length as well. The tail has been re-sculpted to make it curve in a different angle. While the color of the plastic has been changed from black to white, some very minor paint apps were changed as well, such as removing silver paint from all four lower legs and leaving them the natural white of the plastic. Unigolon forms the head, torso, part of the chest, arms, legs, and weapon of Saint Kaiser.
As MagiPhoenix did with Barikion, Saint MagiPhoenix can ride atop Unigolon in an unnamed configuration; there are no reins for the toy to grasp like there are in the both the movie and series. Because MagiPhoenix wielded the Phoenix Sword, Unigolon's horn was designed to detach so that Saint MagiPhoenix would be provided with a sword as well. However, the unnamed sword was not used or mentioned in the show or movie.
Saint Kaiser, in channeling Kai Oz's rage and raw determination, proves to be a powerful opponent against the multi-faceted and multi-faced Chimera. While he may not have had absolute control over the Unigolon's power, it was enough to destroy the monster. Then when WolKaiser appeared, Saint Kaiser dueled against him to an unprecedented but much-anticipated stalemate. Like WolKaiser, he wields a double-blade staff composed of Unigolon's mane, neck and head & horn. With it, Saint Kaiser can perform the Saint Horn Finish- where he jumps high into the air, and impales the unicorn horn into his enemy.
Tenkuu Gattai Saint Kaiser (back) now more closely resembles Yuujou Gattai FireKaiser because of the similarity between Unigolon & Barikion, and Saint MagiPhoenix & MagiPhoenix. There are still differences... Saint Kaiser has a golden ponytail with a smaller unicorn horn on his forehead, and a red-&-gold phoenix-like emblem over his face around his green eyes. There are also two bronze stallions emblazed on his shoulders. The unnamed double-sided staff is also nearly identical to FireKaiser's, except for Unigolon's head and golden PVC horn. However, Saint Kaiser's long ponytail-like queue is shaped differently, adding the smaller unicorn horn which unfolds from his forehead, and a new face design. The mane, also, was changed and has the screw-like pattern woven around it instead of the layered armor blade FireKaiser has on his.
It is this last point that needs clarification- Wolzard merging with Unigolon. Because Wolzard also shares a similar shape with Saint MagiPhoenix, and Unigolon is nearly identical to Barikion, an alternate white & gold version of WolKaiser can be formed. Bandai did not intend for there to be a rotating face under Unigolon's jaw like there is on Barikion. So instead, they designed a single, non-moving piece to represent a 'face'. This is because MagiPhoenix and Wolzard have different neck structures for FireKaiser and WolKaiser, respectively. And so here, since Wolzard requires the horse head for his combination, Bandai placed an alternate 'face' under Unigolon's jaw so that he wouldn't need Barikion's. Also, since Unigolon's head is needed for the alternate WolKaiser, his tail becomes useless, so it is incorporated into the staff, giving it a wicked golden sheppard's hook on one end, and Unigolon's mane on the other. This makes it look near-identical to the Bari-Javelin used by WolKaiser.
There are no differences-! They are perfectly identical in every way except in packaging and instructions! Actually, the Steedergon Fury Megazord is about $30.00 less since it's not imported directly from Japan. All paint apps, materials, and moldings are exactly the same, and there are no decals for either toy. For 2006, you need not have to worry about importing the rare DX Saint Kaiser unless you are concerned about its connection to the "MagiRanger" line and the packaging. (And, if you already paid for an expensive import like I did, I completely sympathize...) Additionally, Bandai America never released the DX WolKaiser in American toy stores due to marketing strategies. They said that the black robot would not sell very well, though they thought that they may have released it later in 2006 as a store-exclusive; alas that never came to pass.
On a more personal note, as this is my last "MagiRanger" review for CDX, I want to give my special thanks to Siu-Wai, owner of Kicks Hobby-Japan (ships across US only), for holding onto this set for me even though he was out of stock. I got all of my DX "MagiRanger" sets from his only physical store (located outside Northgate Mall in northern Seattle, Washington), and I will continue to shop with him.
I can't believe that I had this all typed up and pictures taken two years ago, and then forgot that I'd even covered it! For those of you who were waiting for my take on it, I do apologize for the long delay!
Oh, BTW- what are the chances that I'll cover either the DX Densetsu Gasshin MagiLegend (or Deluxe Manticore Megazord)? Umm... next to none. Someone else will have to cover that disaster. I have all that I want from "MagiRanger", so this truly is my last review on any toys related to it.
There was actually a UK release for the Centaurus-Wolf Megazord which came out in the summer of 2007.
I don't have one (nor a Wolkaiser) so I'm not sure how it fared in translation, but I have a feeling it's untouched.
Thanks for the heads up. If they change even a single paint app, then it's easy to assume that other mods have been made. Typically, however, unless it's die-cast metal they leave materials alone.
It'll be interesting to see what Bandai UK make of the Operation Overdrive releases. They don't seem to have quite made it to our shelves as of yet but I've seen catalogue pictures that look more like the Boukenger toys, of course there's every chance that those photos originated with Bandai Japan and we get the same Drive Max disaster as you do.
Ooh- that's a hard one to predict. As I said, the UK stuff is usually unaltered from Japan; whether you get everything that we get in the US as well, I cannot say. I think I should rely on Bandai UK (or whatever it's called) more than BA to get Super Sentai-accurate stuff. But as for the Drive-Max Megazord...? Man, I just don't know. But fortunately, BA has learned its lesson for "PR Jungle Fury", and has released the new Megazord (called?) just like the original GekiTorja.
But, they also seem to be introducing a new line of characters and Zords that were not present in "GekiRanger"- a brand new DX Megazord has surfaced with a white tiger, panther, an elephant, a bat, and a shark which can separate/combine only with itself, and a 3" Ranger figure can be placed inside of it! My thinking here is that BA somehow has decided to make a new Megazord based on an old-and-unused "GaoRanger" combo to liven up the toy line. At first glance without taking it out of the box it looks interesting, but you can easily tell it is both simple and cheap.
I just got my hand on that guy, and let me tell you something : It's AMAZING! Wolkaiser was one of my favorite but this, this is even beyond. Probably one of the best mecha ever created in the super sentai/ power rangers line.
Yeah, the MagiRanger mecha are pretty awesome... with the exception of the MagiLegend which absolutely no one liked [on either side of the pond]. They really played up the whole magic armor style in a way that fit the Super Sentai format. I mean, the frickin' MagiKing had a tall wizard's hat and pointed-toe boots!
I had a chance to examine that special-edition MagiKing when I was at Kicks one time. Though it was double the price of the regular DX ver (which I think is an awesome mecha), that thing was just beyond-words-type awesome- every single surface was painted, be it black or gray.
That thing is amazing. I'm getting my hand on it somewhere this summer (thanks ebay).
Any chance anyone would be willing to show us what they look like?
"Ok, this is my first post on this site..."
Well, then- welcome to the club, stranger!
The thing about the "MagiRanger" mecha is that it avoids limb-swapping. Instead, it is chest-swapping- the central chest unit of each of the five mecha- MagiKing, WolKaiser, Travelion, MagiLegend, and Saint Kaiser- can be removed, and then attached to any of the others. Now while they don't do this in the series on a regular basis, MagiPhoenix (chest for the MagiKing) swapped at least twice with the evil Majuu Barikion to become FireKaiser (which is a swap using WolKaiser's body). So, since Saint Kaiser is a repaint, you pretty much see how that works out. And we never saw a swap with any of the others.
With the exception of the awful MagiLegend, I've written the reviews on all of the DX "MagiRanger" mecha, and there's no real point to creating an entire review based on this one element. Simply mentioning it is sufficient in my eyes. So, if someone else wants to step up and do that, that's just fine; I've done my piece with it.
Oh, BTW, go to Veoh.com to see both episodes of Power Rangers and TV-Nihon-subbed Super Sentai eps while you're away!
Ok, thank you for replying...I know it's just chest swapping but it still looks pretty cool...lol!
I didn't think you'd do a whole review on it, as I think you did cover it sufficiently as well...I just wanted some pics along with it...a guy can dream, right?
No no no no no, you misunderstand- the DX MagiLegend does also chest-swap with the rest of the line; the MagiLion's head pops off top reveal a big ol' gap which then the Front Engine (Travelion), Wolzard (WolKaiser), or MagiPhoenix (MagiKing) can fit into. MagiLion's head has a panel folded up inside that then extends to cover the gap when it attaches to the other mecha.
The problem with chest-swapping is, even though the DX toys can do it, in the series this breaks continuity on how a majin gattai works. Presumably, if MagiLion lost it's head, then the four Legend MagiRanger which compose it would- in theory- die because their heads would be taken off. (Remember, with the exception of Travelion, all mecha in "MagiRanger" are actually the human-sized characters who have been merged/transformed into larger forms using magic.) That is also why the MagiLegend itself breaks continuity in-series- previously one ranger would form one MagiMajin; whereas MagiLion is composed of four Legend MagiRanger.
Man, I never thought about it like that...man, my brain hurts...all these rules...it makes it all so complicated...ugh!!
Look up the electronic MagiPhone on CDX to see what I mean... or just watch the show.
Ant chance of a video reveiw of the Magi Ranger set?


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Taming myths and legends for brand power
What makes Asda a 'legend' and Tesco a 'myth'? According to research from Lippincott, it's all to do with brand power and the relationship between your marketing and the actual consumer experience.
By MaryLou Costa	 15 Apr 2010 12:00 am
Myths and legends are at the core of all cultures, and in modern times, the line between them has blurred. But in the marketing world, it seems brands can't afford for the line between legend and myth to become blurred.
According to research from agency Lippincott, companies that can be defined as "brand legends" are those which have succeeded at matching the strength of their communications with the delivery of a distinctive, positive customer experience. Meanwhile, "brand myths" are those companies which have rested on the laurels of their reputations for too long and have let it override what a customer takes from the physical product or retail outlet.
Through a survey of 5,000 consumers, Lippincott rated 100 brands on four classifications (see Methodology, p28). As well as the "legends" and "myths", those brands that haven't yet reached legend status but have a loyal and valuable following are termed "tribal". Those companies that have failed to create engagement with customers have been classed as "unattached".
Lippincott senior partner Simon Glynn says that in the age of social media, consumers are openly sharing feedback on both positive and negative brand experiences. As a result, he believes that companies need to help encourage their own legend status by providing the stimulus to generate positive advocacy.
"You need to give people the experiences, the surprises – things to talk about so that they can help build your brand on your behalf," he says.
Virgin Atlantic is rated by consumers as a "legend", with its reputation defined more by service quality than advertising alone. Glynn points out that it achieves 50% advocacy from consumers on their experience, with 37% basing their advocacy on the story power (or marketing) of the brand.
When a brand is seen as having a better experience rating than story power, this means that it exceeds the expectations of consumers. The reality is better than the marketing. When the story power is greater than the experience, it suggests that the company has not yet managed to match the reality to the promise.
In the case of Virgin Atlantic, Glynn notes: "The brand is driven by the experience of what it's actually like to fly. There is some story telling around it, but fundamentally its legend status is driven by the investment and effort it has put into creating the distinctive experience."
Budget brand Ryanair falls into the territory of an unattached brand. It scores less than 5% on experience advocacy and just 13% on story power advocacy. Its strategy of charging fees for what seems like every possible part of the flying process while advertising its flights heavily does not appear to sit well with consumers.
But is there enough of a gap between Ryanair and its budget competitor easyJet for easyJet to have any advantage? EasyJet scores 15% on experience advocacy and 18% on story power advocacy, keeping it in the unattached quadrant.
It seems easyJet as well as Ryanair is disadvantaged by such a strong focus on price rather than service. Lippincott's Glynn suggests that enhancing future service levels and increasing direct engagement through social media could help overcome the limits of being a budget brand.
Supermarket Asda, however, manages to avoid the problems of being seen by consumers as budget. It demonstrates that even lower-cost brands can become legends if they offer a mix of strong branding, customer engagement and experience. It scores 43% on experience power advocacy and 50% on story power advocacy, ahead of fellow legend Marks & Spencer's 38% experience power and 48% story power.
Retail giants Sainsbury's and Tesco might be household names, but they have entered the brand myth sector, demonstrating a high story power (46% and 55% respectively) against low experience power (22% and 23% respectively).
As might be expected of a premium brand, Waitrose is classed as tribal, scoring 33% on both experience power and story power. It is on the threshold of becoming an unattached brand, but this could be combatted through its latest marketing drive featuring celebrity chefs Heston Blumenthal and Delia Smith.
Utilities providers, by their unsexy nature, also struggle to reach the brand legend quadrant. Vodafone and O2 are the only mobile providers with tribal followings. T-Mobile, Orange, 3, Virgin Media and BT are all classed as unattached.
Arguably, telecoms is the kind of market where achieving a tribal following is almost as valuable as becoming a legend. This may also be the case for energy utilities and the finance sector, as it means reaping the benefits of a longstanding, loyal customer base while retaining enough freedom to move into new areas.
BlackBerry may also see the value in its rating as a tribal brand but its increased marketing pushes to new market sectors such as youth and women, indicating it wants to break into the world of legends. This is where its rivals Nokia, Samsung and Sony Ericsson sit, as well as Google, Sony and Apple. Apple has only just entered legend territory, demonstrating that while there are a lot of Apple lovers out there, there are also many who are indifferent.
In the computer market, Apple stands out a mile from its PC manufacturer rivals Dell and HP, and, on an operating system level, Microsoft. These businesses have all seen their reputations dwindle into brand myths. Although Microsoft has a strong story power rating of 62%, indicating good brand messaging, it appears that consumers feel that the marketing doesn't match the experience when they turn on their PCs.
The best chance brands have at achieving legend status is making sure the brand experience permeates across every point where the customer interacts with the business. Glynn says this comes back to the company's structure and how closely the brand and customer service departments work; and how much influence a head of a brand can have over customer experience and operations.
"In most companies, the brand team's origins within the organisation tends to be more focused around the story side than the experience side," he says. "You get a lot of people who understand this but struggle to make it work differently because their organisation doesn't allow for it."
As brands are increasingly being built on the social sharing of customer experiences, Glynn concludes that the role of a chief marketing officer must transcend traditional marketing activities and work alongside operations that contribute to the brand experience. This, he notes, is how brand legends are created and stay alive.
Lippincott interviewed 5,000 people in August last year to rate their perceptions of 100 brands, based on their "story power" versus their "experience power". The story is what the company tells people about its brand (through marketing, for example); the experience is the reality of the product or service. The results of the two measures can be seen in the chart below.
Legend brands
British Gas, Bupa, KFC, Pizza Hut, Asda, Marks & Spencer, Amazon, Virgin Atlantic and Nationwide
Tribal brands
Npower, Starbucks, Waitrose, John Lewis and First Direct
Myth brands
BP, McDonald's, Sainsbury's, Tesco, Barclays and Aviva
Unattached brands
Burger King, Costa, Ryanair, easyJet, Gap and Morrisons
The frontline
Kevin Peake, director of marketing, Npower
Our place in the tribal category reinforces our position as a challenger to the colossus that is British Gas. It supports our belief that the people who have any experience of our brand like us and therefore become loyal. Our continuing challenge is to expand our user base by taking our product to a wider audience by aiming to puncture the British Gas myth, which we are endeavouring to do through new offers to directly rival our competition.
David Kisilevski, EMEA vice-president of marketing, Burger King
I'm disappointed to see Burger King where it is in the table. McDonald's scored higher because of its investment in its restaurant refit. Such a move makes a big difference because it gets people talking about the brand. Improving our store experience and restaurant image is one of the things we have started to invest in, and seeing these results confirms that this is the right thing to do. It will be very interesting to track these trends over the next few years to see whether this has an impact.
Brian Waring, vice-president of marketing and category, Starbucks UK and Ireland
We've grown our tribal following through a strong emotional connection with our customers. We've used the digital space to enhance this connection and enable us to share our story in a relevant way. Examples of this include the number of people using the myStarbucks iPhone app and our Facebook page, which has more than 307,000 UK fans.
Matt Coombe, general marketing manager, Sony UK
What's most important for brands is not just producing a nice bit of TV work, but carrying those same marketing elements through the consumer
journey to purchasing and beyond. The brand needs to come to life in every aspect of this.
At the start of the journey, you make clear what the brand means. As consumers progress towards purchasing, you give them a bit more at each step of the journey to build the relationship. And make sure that at the end of the journey that engagement with store staff is seamless, as this is where people can get disappointed.
News Trends Consumer Goods Marketing & Consumer Trends Retail Travel & Leisure Utilities

redpajama_set_name: RedPajamaCommonCrawl


noun, plural tri·fo·ri·a [trahy-fawr-ee-uh, -fohr-] /traɪˈfɔr i ə, -ˈfoʊr-/. Architecture.
(in a church) the wall at the side of the nave, choir, or transept, corresponding to the space between the vaulting or ceiling and the roof of an aisle, often having a blind arcade or an opening in a gallery.
It will also be noticed that the outer arches of the triforium are not concentric with the sub-arches.
These windows rise as high as the top of those of the triforium.
Above is the triforium, consisting of two arches, each divided into three sub-arches.


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China and Greece
FT on Chinese investments in Greece
On 6 October 2014 FT published an article on Chinese investments in Greece. According to correspondent Kerin Hope, China sees Greece as a pivotal commercial gateway to southern Europe.
COSCO's positive economic results in Piraeus
Piraeus Container Terminal (PCT SA) is continuously playing a critical role for the increase of COSCO's revenue. COSCO's economic results for the first half of 2014 are characteristic.
'2014: The year of China in Greece and of Greece in China'
Mr Vassilis Korkidis is one of the most important representatives of the Greek commercial sector. Having a long-lasting personal experience in leading his family business company, which trades electrical and marine equipment, he currently concentrates on a different, perhaps more difficult mission
Names of Chinese companies participating in the Thessaloniki International Fair
The 79th International Fair of Thessaloniki came to an end. The speeches of Greek Prime Minister Mr Antonis Samaras and the leader of the main opposition SY.RIZ.A party, Mr Alexis Tsipras, did naturally attract the media attention.
Looking for an economic partner in Greece
Ring Ye was one of the representatives of Chinese companies who came to Greece on the occasion of the Thessaloniki International Fair. His ambition was to introduce heating and lighting RYCO Company to Greek clients and create a potential co-operation network in Greece.
The new era of Sino-Greek relations
Sino-Greek relations have entered a new era since the outbreak of the European debt crisis. Greece constitutes a special case for China for two main reasons. First, this country is the epicentre of the European debt crisis.
About UsArticles/InterviewsCompaniesFeaturesIn the MediaNews
Copyright © 2022 China and Greece | About us | Terms and conditions
Development Fotis Routsis


Figure 5.16. Open-book—B1—pelvic fracture. a: Plain x-ray. The bladder neck in this case had been cut by the fractured bone. b: Ligaments disrupted in an open-book fracture.
FIGURE 5.17. Closed-book pelvic fracture B2. a: These are the images obtained at the time of retrograde urethrography, which shows complete disruption of the posterior urethra (contrast does not progress beyond the bulbar urethra). During the process of fracturing, the overlapping bones of the fractured pubis have sheared through the urethra. b: Ligaments and bones disrupted in a closed-book fracture.
fracture being more likely to be associated with a bladder injury and the closed-book fracture with a urethral injury.
In the vertically unstable pelvic fracture (C in the Tile classification), if a sacrotuberous ligament and a posterior sacroiliac ligament are torn, the affected hemipelvis can move upward and posteriorly with respect to the sacrum (vertical instability). A fracture of the transverse process of L5 vertebra is a sign that such a fracture has occurred (i.e., it is a sign of vertical pelvic instability). Again, vessels and nerves can be damaged.
This is based on inlet and outlet views of the pelvis, the x-ray beam being angled accordingly. These views demonstrate antero-posterior (inlet view) and vertical (outlet) displacement of the pelvic ring. CT can provide better definition of sacral, sacroiliac, and acetabular fractures and dislocations.


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Links to Our Favorite
Music Resources
Resources for keeping a song in the heart of your children. Our mission is to nurture the love of music in children and develop musical literacy through instruction, participation and performances, enhancing academic education with musical arts. There are other friends of MFMII that have a similar mission that are listed below in the following categories.
Music Children Love
This is the mother organization that Music for Minors II was modeled after in its formation. Their service area covers San Mateo and Santa Clara Counties.
Providing preschool music programs in the Bay Area for more than 20 years. This program is highly recommended by MFMII.
Educating our children. Inspiring our future. MFMII networks with this wonderful program that nurtures children's development.
EBMT provides a true family music experience.
The leading guide to the best in kid's websites.
Children's Recordings and Books. MFMII greatly appreciates the wonderful partnership established with Linden Tree over the years. Their generosity and support of MFMII's many concerts and workshops is commendable. We are grateful for their quality merchandise that has enriched children and families for many years.
Lots of resources for learning to play the recorder.
2007 Parents' Choice Approved Award Winner among many other awards. Charlotte Diamond's music has won the hearts of children and families everywhere. It has also been instrumental in the establishment of MFMII. Charlotte, a former teacher, continues to teach volumes in her many CDs and teacher workbooks. She knows what children need and how to creatively engage them in song. Her music definitely makes this a better world.
One of the premier entertainers of children and families in America. Red Grammer is an outstanding children's recording artist and Grammy nominee whose music inspires, captivates, and touches the hearts and souls of children everywhere. MFMII incorporates Red's music in our training program for community volunteers and sponsors him in our community concerts as he reaches young and old alike with his unique style and message. He is a great advocate for children and brings out the best in them.
Community Resources & Partners
As a long time member of the Council, MFMII shares its mission of furthering support of the fine arts in the local community. MFMII has worked in partnership with the FCAC to promote the establishment of a Fremont Cultural Arts Center as well as networked with organizations within this clearinghouse.
Supports all art forms – dance, literature, media arts, music, theater, visual arts, traditional arts and crafts, and multidisciplinary programs.
Conveniently located in the heart of San Francisco's East Bay Area, Big Toe Audio is the area's premiere audio post-production facility offering superior service for a wide variety of both creative and commercial audio needs.
I make copies of the songs for their poetry and songbooks which enhances their reading skills. - 1st Grade, Patterson School, Fremont


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ldoc

Index

Contents

Class CommandManager

Modules

command.command

command.command_manager

command.compound_command

command.set_multiple_command_mode_command

model.object.object_bridge

state.abstract_state

state.state_manager

view.editor

view.fields.array_field

view.fields.asset_field

view.fields.boolean_field

view.fields.choice_field

view.fields.color_field

view.fields.field

view.fields.group_field

view.fields.material_field

view.fields.numeric_field

view.fields.object_field

view.fields.object_type_field

view.fields.string_field

view.fields.team_field

Module command.command_manager

CommandManager module.

 Available globally as SB.commandManager.

Class CommandManager

	CommandManager:execute (Instance[, widget=false])
	Execute a command
 If the command is to be executed in a different Lua state than where this method is invoked from, it will be sent as a message to the proper state using the message mechanism.

	CommandManager.CommandManagerListener
	Listener definition

Class CommandManager

        CommandManager class

    CommandManager:execute (Instance[, widget=false])

    Execute a command
 If the command is to be executed in a different Lua state than where this method is invoked from, it will be sent as a message to the proper state using the message mechanism.

Parameters:

Instance
            command.Command
         of a class that implements the Command interface.

widget
            boolean
         Specifies whether the command should be executed in LuaUI(true) or LuaRules(false).
         (default false)

Usage:

SB.commandManager:execute(PrintMessageCommand("hello"))

    CommandManager.CommandManagerListener

    Listener definition

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Last updated 2017-09-23 19:07:28

redpajama_set_name: RedPajamaGithub


text:
Korkoya Admonishes Accusers To Seek Legal Redress On U.S. Citizenship Allegation
D. S. Menjor
The challenge from the Chairman of the National Elections Commission (NEC), Jerome Korkoya and the Executive Mansion to anyone claiming that he holds a U.S. passport and is an American citizen should take the NEC boss to court is at the verge of being granted. Two prominent lawyers and other citizens have galvanized to do just that. According to renowned Liberian musician and women's rights advocate, Miatta Fahnbulleh, she and other prominent Liberians are leading a well organized campaign to ensure that the Liberian constitution is respected in this case.
She disclosed after a peaceful protest at the headquarters of the NEC in Monrovia that Cllr. Laveli Supuwood and Cllr. Tiawon Gongloe are preparing a lawsuit to take the NEC boss to task, and prove in court that he is indeed a US citizen.
Ms. Fahnbulleh, who is popularly known as Aunty Miatta, is leading the advocacy aspect of the campaign to remove Cllr. Korkoya. She and several others gathered at the NEC yesterday demanding that the NEC boss should do the honorable thing by resigning.
The campaign is under the auspices of the Concerned Citizens to Protect the Constitution (CCPC), on which Aunty Miatta serves as Interim Coordinator.
She said, unlike other parts of the world, public officials in Liberia lack integrity and that is why they are not deterred by scandals. "But the power of national governance is inherent in us so we must use that power to bring them to book and this is what we have resolved to do in this case," she said.
She indicated that the facts about Cllr. Korkoya's US citizenship are clear, tangible and realistic, "but since he wants people to take him to court, we will do that and thank God my brothers, Gongloe and Supuwood are working on the legal aspect of this campaign."
Making reference to the provision of the Constitution that says only Liberian citizens with impeccable character are eligible to serve as commissioners of the NEC, she said Cllr. Korkoya is legally unfit, legally improper and legally incompetent to superintend the October 10 elections.
"His successful confirmation by the Liberian Senate in 2012 as chairman of the National Elections Commission strongly and clearly suggests to us that Chairman Korkoya must have perjured himself by lying under oath about his citizenship status," Fahnbulleh said.
She said "In our opinion Chairman Korkoya cannot and should no longer continue to serve in this stead and he should do the most honorable thing to step aside and give room to a legally competent individual with proven integrity to serve as chairman of the NEC."
She indicated that the group is doing everything legally required to ensure the removal of the NEC Chairman. "We started with today's (yesterday's) peaceful protest which we had earlier written the Justice Ministry about," she said.
The group, in the letter to the MOJ, denounced the ineligibility of the NEC boss, warning that something must be done to avert the level of embarrassment the country is at the verge of facing.
"Our constitution forbids foreigners from serving as commissioners of the NEC simply because it has to do with trust and credibility and Cllr. Korkoya's presence at the NEC as chairman is a complete violation of the nation's law on elections," the letter, a copy of which is in the possession of the Daily Observer, said.
The letter, dated May 1, 2017, informed the Ministry of her organization's plan to peacefully assemble yesterday at the headquarters of the NEC in Sinkor, Monrovia to protest against Korkoya's continuance in office as the NEC boss.
She received a response from the Ministry asking the group to postpone the protest until consultations into the matter were completed.
"In recognition of your constitutional rights to peaceful assembly and protest vis-à-vis the need to ensure the safety and security of the nation and people, and considering the current security environment, we respectfully urge that the planned peaceful protest be postponed until further consultations are held with the security agencies," the MOJ replied.
Meanwhile, under relentless pressure from the public and the media, Cllr. Korkoya, while not denying his United States citizenship, has instead repeatedly maintained that he is a Liberian and anyone challenging his citizenship should seek redress through the court.
Fahnbulleh added that the Concerned Citizens to Protect the Constitution will not fall to such a ploy when the facts are clear to all and sundry that he, Korkoya, is a citizen of the United States.
She said while taking the matter to the court is vital, her organization is also calling on ECOWAS, EU, the U. S. Government and all other friends of Liberia to prevail on the Liberian Government to consider with exigency the immediate resolution of this matter which has the potential to adversely affect the ongoing electoral process and undermine national peace and stability.
"We also call on the U.S. Government to reign in its rogue citizens who are deliberately impersonating as Liberians, wantonly violating the Constitution of our country and provoking a crisis with unforeseen consequences to take back home their citizen, Korkoya," she said.
It may be recalled that the National Democratic Coalition (NDC) told a news conference on April 21 in Monrovia that it had 'prima facie' evidence that the NEC Chairman is a naturalized citizen of the United States of America, and threatened legal action to remove him from the position.
The NDC's political leader Nyanquoi K. Kargbo accused Cllr. Korkoya for lying under oath and said the Party had written President Ellen Johnson Sirleaf, urging her to remove Cllr. Korkoya from office; else its lawyers both in Liberia and in the United States would pursue legal action to remove him as chairman of the NEC.


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HomeNEWSBYO containers policy live from June 1 at Foodstuffs stores
BYO containers policy live from June 1 at Foodstuffs stores
Customers at Foodstuffs supermarkets' instore butchery, seafood counter, delicatessen and bakery will be able to have food packed into their own clean, leak-proof containers from June 1.
Auckland's New World Long Bay launches the new policy slightly earlier on May 28. New World, Pak'n Save and Four Square stores in the North Island will offer it from June 1.
The service is only available at counter departments in store for the time being, and does not apply to bulk bins. This is so the store can inspect containers to make sure they're clean and fit for purpose, and so the container's weight can be subtracted from the weight of the product being purchased.
Mark Casey, group manager regulatory services at Foodstuffs North Island, says introducing a BYO container policy was not as easy as might be expected.
"We've run successful trials at a number of our stores where we worked out what we had to do for customers to bring their own containers. It's not as simple as it looks. We take our responsibilities as a retailer of food very seriously – food safety is a top priority, so making sure our customers' groceries aren't compromised through poor hygiene is very important."
"Many people don't realise that products must be sold minus the cost packaging might add to a product. That's why we restrict BYO to counters where we can subtract the weight of the container and produce a price label for the cost of the product only," says Casey.
"We have to make quite significant changes to the way we operate to take the BYO option storewide, but this new policy in our counter areas is a major step towards zero waste."
The new policy follows the phasing out of single-use shopping bags. Many retailers like Foodstuffs have voluntarily adopted changes ahead of new regulations against the bags, which will come into force from 1 July 2019.
Global ice cream giant buys Tip Top for $380m
Auckland design agency wins gold for retail packaging


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Q: Right-angled triangle on a cylinder I have a question about trigonometry on the surface of a cylinder, and whether it's different from regular trigonometry on a plane.
Suppose I have a straight line of length $L$, and I tilt it by an angle $\alpha$. Then the horizontal and vertical projections of this tilted line are given by $L \cos \alpha$ and $L \sin \alpha$ resp.
Now, consider the equivalent problem on the surface of a right circular cylinder of radius $R$. Then the straight line is an arc of radius $R$. If I now tilt this arc by an angle $\alpha$, so that the arc still lies on the cylinder, what are the horizontal and vertical projections in the cylinder?
I tried doing this by unwrapping the cylinder onto a plane. But I got the same values as before: $L \cos \alpha$ and $L \sin \alpha$. Does this make sense? I would expect it to be different.

A: I realised in hindsight that I should not expect the two answers to be different. The 'rolling' map that takes the plane to the cylinder, will also roll up the flat line and the tilted line. Moreover, the map is isometric and conformal, thus the angle between the two lines will remain the same.
I hope this question is useful to someone else in the future.

redpajama_set_name: RedPajamaStackExchange


I know that we are beginning a new season of wedding dresses and more than one and more than two you are scrutinizing this spring/summer 2010 trends to be radiant. Today I want to introduce you to this beauty of a dress designed by Cosmina Englezian, sewn by hand with more than 55,000 pearls and Swarovski crystals, which have been used about 40 meters of fabric.
It is inspired by the fashion of the twenties and has been called "Crystal Dream" (Dream of crystals) due to the raw material with which it is made. I like your design, tailored and ruffled bottom, with a certain air of Spanish. That Yes, who put it should know that it weighs about 12 pounds, so it won't have it easy to move.
This work of art has been sold recently for around 35,000 euros, including veiled (although my feathers and tulle veil me not just convince).


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800 - General Government (16)
(-) 950 - Undistributed offset Receipts (1)
(-) House Committee on Oversight and Government Reform (1)
(-) House Committee on Transportation and Infrastructure (2)
H.R. 3972, A bill to amend the Outer Continental Shelf Lands Act to prohibit the Secretary of the Interior from charging state and local government agencies for certain uses of the sand, gravel, and shell resources of the outer continental shelf
Cost estimate for the bill as ordered reported by the House Committee on Resources on August 5, 1998
H.R. 4275, Economic Development Partnership Act of 1998
Cost estimate for the bill as ordered reported by the House Committee on Transportation and Infrastructure on July 23, 1998
H.R. 3869, Disaster Mitigation Act of 1998
Cost estimate for the bill as ordered reported by the House Committee on Transportation and Infrastructure on June 25, 1998
H.R. 4259, Haskell Indian Nations University and Southwestern Indian Polytechnic Institute Administrative Systems Act of 1998
Cost estimate for the bill as ordered reported by the House Committee on Government Reform and Oversight on July 23, 1998
H.R. 1983, Narragansett Justice Act
H.R. 3797, Wyandotte Tribal Settlement Act of 1998
Cost estimate for the bill as ordered reported by the House Committee on Resources on May 20, 1998
H.R. 1154, Indian Federal Recognition Administrative Procedures Act of 1998
Cost estimate for the bill as ordered reported by the House Committee on Resources on May 20, 1998, with a subsequent technical amendment
H.R. 2742, California Indian Land Transfer Act
Cost Estimate June 8, 1998
H.R. 3069, Advisory Council on California Indian Policy Extension Act of 1997
Cost estimate for the bill as ordered reported by the House Committee on Resources on March 25, 1998


text:
Readers ask: Where Were The Blues Born?
1 What city is the birthplace of the blues?
2 What is the birthplace of blues music?
3 What city is known for blues?
4 Who invented blues?
5 Is Blues and Jazz the same?
6 Who is known as the father of the blues?
7 Why is blues called devil's music?
8 What is the blues capital of the world?
9 Which city is the music capital of the world?
10 Which city is known for jazz?
11 Who is most famous blues artist?
12 Who are the famous blues musicians?
13 What blues means?
What city is the birthplace of the blues?
Yes, Mississippi is the birthplace of the blues Mr.
What is the birthplace of blues music?
The men from Clarksdale. Located in the heart of the Delta, Clarksdale has been dubbed the birthplace the Blues, with the exact spot attributed to the Dockery Plantation. Many sharecropping families and field workers occupied this space, including one Henry Sloan.
What city is known for blues?
Chicago. The great city of Chicago has music venues for every taste, but it is particularly known for the Blues. The "Chicago Blues " style came about in the post-World War II-era when many African-American southerners moved to the industrial north in search of jobs.
Who invented blues?
Blues is a music genre and musical form which was originated in the Deep South of the United States around the 1860s by African-Americans from roots in African-American work songs, and spirituals.
Is Blues and Jazz the same?
Blues is derived from Bluegrass, Jazz, R&B, and Rock. Jazz comes from Calypso, Funk, Soul, and Swing. It's important to know that Blues was around before Jazz; thus, Blues can be considered an element of Jazz music. Jazz is from New Orleans, while Blues is from Mississippi.
You might be interested: FAQ: How To Play Blues Scales For Beginners?
Who is known as the father of the blues?
Today's blog celebrates the career of W.C. Handy. Born in Florence, Alabama on November 16, 1873, William Christopher Handy became interested in music at an early age.
Why is blues called devil's music?
Not sure much the Blues but it's reinvention through early rock and roll Like Elvis and Chuck Barry etc… People referred to this as the Devil's Music because it was widely believed, at the time, that it caused teens of the day to indulge in sinful activities like premarital sex and drugs.
What is the blues capital of the world?
I'll tell you in no uncertain terms what is and has been for the past 30 plus years the capital of the modern blues world Long Beach, California. That's right Long Beach. For anyone who has lived in California for any length of time, they are now uttering under their breath or hopefully out loud, "DUH."
Which city is the music capital of the world?
HOUSTON – Not only is Austin the capital of Texas, but is also dubbed the live music capital of the world.
Which city is known for jazz?
New Orleans: Birthplace of jazz.
Who is most famous blues artist?
The main source being The All Music Guide To The Blues (Miller Freeman Books).
The Top Ten.
Charley Patton (1887 – 1934)
Blind Blake (early 1890s – 1933)
Blind Lemon Jefferson (1897 – 1929)
Lonnie Johnson (1899 – 1970)
Louis Jordan (1908 -1975)
T-Bone Walker (1910 – 1975)
John Lee "Sonny Boy" Williamson (1914 -1948)
You might be interested: Quick Answer: Where Did Rhythm And Blues Originate?
Who are the famous blues musicians?
BIG BILL BROONZY (Lee Conley Bradley; 1903 – 1958)
SON HOUSE (Eddie James "Son" House, Jr.; 1902 –1988)
BLIND WILLIE McTELL (William Samuel McTier; 1898 –1959)
MAMIE SMITH (née Robinson; 1891 – 1946)
ARTHUR "BIG BOY" CRUDUP (Arthur William Crudup; 1905 –1974)
SKIP JAMES (Nehemiah Curtis "Skip" James; 1902 –1969)
What blues means?
Since the fourteenth century, the word blue has been used to mean "sad." The noun blues came into use in the 1700s to describe a state of sadness or melancholy. Lyrics in blues music often describe hard times, struggles, and a particular kind of melancholy mood.


In my last column, we explored the interaction of the economy and real estate cycles and how they are not the same. In that column, we used the graph below to look at the two cycles.
As you can see in the above chart, the ongoing economic cycle, shown in black, does not run simultaneously to the one occurring in the real estate industry (shown in blue), but usually precedes it by as much as a year. Contrary to popular belief, the real estate cycle is not fed directly by the local or national economic cycle, but instead by what we will call the "investor demand" cycle. Likewise, the investor demand cycle is also not fed by the real economy, but instead fueled primarily by what investors are reading or hearing from the media. This information shapes their decisions.
It is through the investor demand cycle that money flows, in the form of financing, into the real estate industry. These investors, both institutions and individuals, are the savings and loans, banks, insurance companies, as well as, pension plans, public and private syndications, and wealthy individuals.
The reason the real estate cycle lags behind the economic cycle is that the media can only report data and statistics concerning the economic cycle after they have been accumulated. In addition, their true job is to report what they feel we are most interested in hearing. If they miss the public's interests too consistently, they are not going to grow.
By looking at our original economic example and tracking what the media are reporting and how the investors are reacting to it, we can see how these two cycles interact.
Using the chart above, starting at the bottom of the economic cycle, we are at the pit of the recession. During this period, the media would be reporting what I will call "Bad Press." Top stories, because it has been so bad for so long, would probably be exploring the more tragic, human side of problems caused by the bad economy. Unemployment and the pain of failed, small businesses, etc. are the primary economic news.
Since the news has been bad for some time before this bottom actually occurs, the investor and his funds have long since fled this marketplace. Because of the investor's alarm, he will have sought new places or areas for his capital to be invested. The real economic recession, therefore, will become paralleled with a subsequent, complete halt of development in the real estate industry.
As the true economy starts to climb out of the recession, a real demand/growth cycle will be in effect for 6-12 months before any new improved statistics can become available and subsequently reported. During this economic startup period, since no new positive statistics are yet available, the news from the media will be flat. Since flat news is uninteresting, the local media will choose to report on non-economic issues. The national media will be diverting investor awareness to areas of the economy and country that are currently booming, such as maybe the stock market or what is happening in China, etc. Since investor awareness is diverted to where the "Good Press" is, no new construction money will be available to this real estate market from investors.
Because such little money is available, very little new construction is contemplated. As the current supply of realty inventory is absorbed, the newly developed projects of the inside investor will gradually start to appear. As this occurs, some members of the media will see the turnaround more quickly than others and release some brief, positive stories. However, once scared away, the investors will require overwhelming proof that the market is strong and very safe before they will choose to re-enter it.
During the oncoming, stronger economic upswing, the press will start to give more and more positive signals. The first substantial proof of the recovery, in the form of statistical data, will now finally become available and be published.
The press, because of these confirming statistics, now will swing into a full "Good Press" mode. Reading the repetitive "Good Press," investors finally start to appear. Unfortunately, however, because of the lack of development activity, real shortages are already present and are now getting worse. The reality of the two-year time lag from project contemplation to project completion will over-exaggerate this shortage problem, causing even higher increases in rent and sales prices. Total consumption of the available material and labor force will drive the prices upward and thus creates the long awaited boom period for the real estate industry. The "Good Press" keeps getting better and stronger, and the money that had just previously started to trickle into the market now comes flooding in.
The institutional investor, finally feeling safe because it can no longer be criticized, opens its money gates. As new businesses are formed and economic expansion continues, everyone and his brother begin to enter the real estate industry to capitalize on the boom.
The problem, at this point is that the real economy has just begun to peak and start its normal slowdown. However, it will be a year before the statistics finally confirm this slowdown, during which the period the "Good Press" will still run at full steam. Meanwhile, every small investor in the country is now aware of the "Good Times" in real estate and wants a piece of the action. Billions of dollars demanding placement through the private syndicators will become available. The financial institutions, having a difficult time competing, will become even more aggressive in lending.
By this point, when the building should have stopped, it just now shifts into high gear. Two years of oversupply is built-in approximately a six-month period. In addition, as we discussed in my last column, builders will build as long as they can get money, and the investors, following the positive press, are only too happy to comply.
Then, unfortunately, the historical bad statistics finally become available and confirm the reality of a recession, which has now really been underway for the last 6-12 months, and the "Bad Press" begins. The non-sophisticated investor remains for a while longer, but the institutional investors, coming face to face with reality on the front page of their newspapers, slam on the brakes. People, who thought they had strong banking relationships find out that they have no banking relationship. Many businesses that could have survived with a prudent working banking relationship will now go bankrupt, furthering the economic collapse.
Within a short period, the reporting becomes "Very Bad Press." All remaining investors flee the market, which is just prior to the bottom of the real economy's recession cycle.
For the next period, "Very Bad Press" becomes the word for the day, and big headlines appear regularly. The slide into total recession occurs almost immediately as soon as the money vanishes. Only projects that have been pre-funded will start during the period that follows. Soon, we are at the point, which this example started.
In my next column, we will explore a method I developed to predict where we are in the real estate cycle and what will happen next.
Read more about Jody Tallal, a pioneer in the financial-advice industry, in the WND story announcing his column.
This entry was posted in Uncategorized on February 17, 2018 by admin.


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\section{Introduction}

Given $\epsilon>0$ and $d \ge 1$, the Borsuk Graph $\text{Bor}^d(\epsilon)$ is the graph with vertex set corresponding to points on the $d-$dimensional unit sphere $S^d\subset\mathbb{R}^{d+1}$ and edges $\{x,y\}$ if and only if $\d{x}{y}>2-\epsilon$, that is, if the two points are $\epsilon$-near to antipodal. Here distance is measured in the ambient Euclidean space $\mathbb{R}^{d+1}$. It is well known that when $\epsilon$ is sufficiently small, its chromatic number is $d+2$, in fact this is equivalent to the Borsuk--Ulam theorem.

The Borsuk graph was part of Lov\'asz's inspiration for his proof of the Kneser conjecture \cite{Lovasz1978}. Among other properties, this graph constitutes a nice example of a graph with large chromatic number and odd girth. See for example \cite{Gabor2011,Prosanov-Raigorodskii-Sagdeev2017,Erdos-Hajnal1966,Erdos-Hajnal1967}. It has also been studied because of its relation with Borsuk's conjecture and distance graphs \cite{Raigorodskii2012, Barg2014,Prosanov2018,Sagdeev18}.

We are interested in the chromatic number of random induced $n$-vertex subgraphs of the Borsuk graph. Our main point is that if $\epsilon \to 0$ slowly enough as $n \to \infty$, then topological lower bounds on chromatic number are tight. This contrasts with the situation studied by Kahle in \cite{Kahle2007}, where topological lower bounds are not efficient for the chromatic number of Erd\H{o}s--R\'enyi random graphs. Similar problems have also been studied for random Kneser graphs in \cite{Kupavskii2018} and \cite{Kiselev-Kupavskii2018}.

The rest of the paper is organized as follows. We finish this section with some definitions and notation. In section \ref{section_2} we prove Theorem \ref{thm_eps_constant} when $\epsilon$ is fixed.

\begin{theorem}\label{thm_eps_constant}
	Let $d\geq 1$ and $0<\epsilon<2-\lambda_d$ be fixed. Then a.a.s. $$\chi\left(\Bord{\epsilon}{n}\right)=d+2.$$
\end{theorem}

In section \ref{section_3} we prove Theorem \ref{thm_lower_bound}, stating that the chromatic number is still the same when $\epsilon\to0$ slowly.

\begin{theorem}\label{thm_lower_bound}
	Let $\epsilon(n)=C\left(\dfrac{\log{n}}{n}\right)^{2/d}$, where $$C\geq \dfrac{64}{3} \left( \dfrac{3\pi^2}{4} \right)^{1/d}.$$ Then a.a.s. $\chi\left(\Bord{\epsilon(n)}{n}\right)=d+2.$
\end{theorem}

Finally in section \ref{section_4} we prove Theorem \ref{thm_upper_bound}, showing that this rate is tight, up to a constant, in the sense that if $\epsilon\to 0$ faster, then the random Borsuk graph is $(d+1)$-colorable, a.a.s.

\begin{theorem}\label{thm_upper_bound}
Let $\epsilon(n) = C (\log n / n)^{2/d}$,  where $$C< \frac{3(4-\lambda_d^2)}{64}\sqrt[d]{\frac{9}{4 d^2}}.$$
Then a.a.s. $\chi\left(\Bord{\epsilon(n)}{n}\right)\leq d+1$.
\end{theorem}

\begin{defi}[Random Borsuk graph]
	Given $n \ge 1$, $d\geq 1$, and $\epsilon>0$, we define a \textbf{random Borsuk graph}, $\Bord{\epsilon}{n}$, as follows.

\begin{itemize}
		\item Its vertices are $X_1, X_2,\cdots,X_n$, $n$ independent and identically distributed uniform random variables over the $d$-dimensional sphere $S^d\subset \mathbb{R}^{d+1}$ of radius 1.
		\item $X_i$ and $X_j$ for $i\neq j$ are connected by an edge, if and only if $\d{X_i}{X_j}>2-\epsilon$, where $\lVert\cdot\rVert$ is the Euclidean distance.
	\end{itemize}

\end{defi}

Throughout this paper we will think of random Borsuk graphs on $S^d$ for a fixed dimension $d$. However we will explicitly point out the constants that depend on $d$ in the statement of the results. We will denote the closed ball with center $x$ and radius $r$ by $B(x,r)=\left\{y\in\mathbb{R}^{d+1}: \norm{x-y}\leq r \right\}$. Similarly, we denote intersections of closed balls with the $d$-sphere by $\B{x}{r}$, and we call them spherical caps, so $$\B{x}{r}:=B(x,r)\cap S^d=\left\{y\in S^{d}: \norm{x-y}\leq r\right\}$$
Given a Borel set $F\subset \mathbb{R}^{d+1}$, we denote its volume, i.e. Lebesgue measure, as $\vol{F}$. Similarly, for a Borel set $F\subset S^d$, we denote its area on the surface of the sphere, by $\area{F}$. Also, we denote $\omega_d=\vol{S^d}$ and $\alpha_d=\area{S^d}$. Given a graph $G$, we denote its chromatic number by $\chi(G)$.

We say that an event happens \emph{asymptotically almost surely (a.a.s)} if the probability approaches $1$ as $n \to \infty$.

\section{Random Borsuk Graph with $\epsilon$ constant}\label{section_2}

We start by proving that when $\epsilon>0$ is constant and small, $\chi(\Bord{\epsilon}{n})=d+2$ a.a.s.

\begin{lemma}\label{lemma_min_distance}
	For $x,y\in S^d$, $\norm{x-y}>2-\epsilon$ if and only if $\norm{x+y}<2\sqrt{\epsilon-\frac{\epsilon^2}{4}}$
\end{lemma}
\begin{proof}
	Since $\overline{(-x)x}$ is a diameter, $\overline{(-x)y}\perp \overline{xy}$. Thus $\norm{x+y}^2=4-\norm{x-y}^2$, so the claim follows.
\end{proof}

Before getting into the analysis of the chromatic number, let us point out the fact that the odd girth of the Borsuk graph is $> 1/\sqrt{\epsilon}$. While this has been observed before (see \cite{Erdos-Hajnal1966,Erdos-Hajnal1967,Gabor2011}), we include a proof for completeness.

\begin{lemma}\label{lemma_odd_girth}
    Let $\epsilon>0$ and $x_0\in S^d$. If $x_0y_1x_1y_2\cdots x_ny_{n+1}=x_0$ is an odd cycle in the Borsuk graph $\Bor{\epsilon}$, then $2n+1\geq 1/\sqrt{\epsilon}$. In other words, all odd cycles in $\Bor{\epsilon}$ have length greater than $1/\sqrt{\epsilon}$.
\end{lemma}

\begin{proof}
    Since $\norm{x_i-y_i},\norm{y_i-x_{i-1}}>2-\epsilon$, for any $i$, by applying Lemma \ref{lemma_min_distance}, we get

    \begin{align*}
    \norm{x_i-x_{i-1}} &\leq \norm{x_i+y_i}+\norm{-y_i-x_{i-1}} \\
    & =\norm{x_i+y_i} +\norm{y_i+x_{i-1}} \\
    & \leq 4 \sqrt{\epsilon - {\epsilon^2}/{4}} \\
    & <4\sqrt{\epsilon}.
    \end{align*}

    Thus $$\norm{x_n-x_0}\leq\norm{x_n-x_{n-1}}+\norm{x_{n-1}-x_{n-2}}+\cdots+
    \norm{x_1-x_0}\leq 4n\sqrt{\epsilon}.$$

    Finally, $$2=\norm{2x_0}=\norm{x_0+y_{n+1}}\leq \norm{x_0-x_n}+\norm{x_n+y_{n+1}}<2(2n+1)\sqrt{\epsilon}.$$

    Therefore $$\frac{1}{\sqrt{\epsilon}}<2n+1.$$

\end{proof}

\begin{lemma}\label{lemma_color_d+2}
	For each $d\geq 1$, there exist a constant $\lambda_d<2$ such that, whenever $0<r<2-\lambda_d$, the Borsuk graph $\text{Bor}^d(r)$ has a proper coloring with $d+2$ colors.
\end{lemma}
\begin{proof}
	Let $\Delta$ be the regular $(d+1)$-simplex inscribed in the unit $d$-sphere $S^d$. Consider the map $\Phi:\partial\Delta\to S^d$ from the boundary of $\Delta$ to $S^d$ given by $\Phi(x)=x/\norm{x}$. Note then that $\Phi$ is a homeomorphism. Let $\tau\in\partial\Delta$ be a maximal face, and let $\lambda_d=\diam{\Phi(\tau)}$. Since $\Delta$ is regular, the value of $\lambda_d$ does not depend on the face $\tau$.

	Note now that $\lambda_d<2$. To see this, suppose that $\lambda_d=2$. Since $\tau$ is closed, so is $\Phi(\tau)$, so there exist $x,y\in\tau$ such that $\norm{\Phi(x)-\Phi(y)}=2$. This means $\Phi(x)$ and $\Phi(y)$ are antipodal, and so $y=-\frac{\norm{y}}{\norm{x}}x$. Since $\tau$ is convex, $0=\sfrac{\norm{y}}{(\norm{x}+\norm{y})}x+\sfrac{\norm{y}}{(\norm{x}+\norm{y})}y$ must also be in $\tau$, but this is a contradiction, since $\tau\subset\partial\Delta$, proving the claim.

	We now give a coloring for $S^d$ as follows. We start by coloring $\partial\Delta$: give a different color to each of the $d+2$ facets, and for the lower dimensional faces, assign an arbitrary color among the facets that contain them. Finally, color $\Phi(x)\in S^d$, with the color of $x$.

	Note this is indeed a proper coloring for $\Bor{r}$, since all points in $S^d$ of the same color lie on the image of a facet $\Phi(\tau)$, of diameter $\lambda_d$; so if $x$ and $y$ have the same color, $\norm{x-y}\leq\lambda_d< 2-r$ so they are not connected by an edge in the Borsuk graph.
\end{proof}

The upper bound for the chromatic number follows immediately from Lemma \ref{lemma_color_d+2}. The proof we give below for the lower bound, is a direct application of the Lyusternik--Shnirelman--Borsuk Theorem \cite{LS30,Borsuk33}. We state this well-known theorem without proof; for more details and a self-contained proof see, for example, Chapter 2 of Matousek's book \cite{Matousek2008}.

\begin{theorem*}[Lyusternik--Shnirelman--Borsuk] For any cover $U_1, \dots , U_{d+1}$ of the sphere $S^d$ by $d+1$ open (or closed) sets, there is at least one set containing a pair of antipodal points.
\end{theorem*}

We note that B\'ar\'any gave a short proof of Kneser's conjecture using this theorem \cite{Barany78}. See also Greene's proof \cite{Greene02}. For the rest of the paper, we refer to this theorem as the LSB Theorem.

\begin{proof}[Proof of Theorem \ref{thm_eps_constant}]
	By Lemma \ref{lemma_color_d+2}, since $\Bord{\epsilon}{n}\subset \text{Bor}^d(\epsilon)$, $d+2$ is an upper bound for the chromatic number of the random Borsuk graph.\\

	Let $F_1,F_2,\dots, F_N$ be a cover of $S^d$ by Borel sets, such that $\text{diam}(F_i)\leq\frac{\sqrt{\epsilon}}{2}$ and $\area{F_i}>0$ for all $i$. Note we can construct such a family of sets in many ways, for instance, as we do in the next section, we can consider a $\delta$-net of $S^d$ and let the sets $F_i$ to be spherical caps centered on the $\delta$-net of radius $\sqrt{\epsilon}/4$ where $\delta\leq\sqrt{\epsilon}/4$. Note here that the sets $F_i$ and $N$ depend only on $\epsilon$, which is fixed. \\

	Let $c=\min_{i}\frac{\area{F_i}}{\area{S^d}}$ and $G=\Bord{\epsilon}{n}$. The following computation shows that, a.a.s., $G$ contains at least one vertex in each of the sets $F_i$.

	\begin{align*}\label{eqn-prob-Fi-has-vertices}
	\P{\bigwedge_{i=1}^N \left(V(G)\cap F_i \neq \emptyset\right)} & =
	1-\P{\bigvee_{i=1}^N\left(V(G)\cap F_i=\emptyset\right)}\\
	&\geq 1 - \sum_{i=1}^N\P{V(G)\cap F_i=\emptyset}\nonumber\\
	& = 1 - \sum_{i=1}^N \left( 1-\frac{\area{F_i}}{\area{S^d}}\right)^n  \geq 1 -N(1-c)^n
	\end{align*}
	since $N$ and $c$ are constant, $1-N(1-c)^n\to 1$ as $n\to\infty$, proving the claim. \\

	We may assume then, $G$ has a vertex $y_i\in F_i$ for $i=1,\dots, N$. Proceeding by way of contradiction, suppose there exists a proper coloring of $G$ with $d+1$ colors. For each $j=1,\dots, d+1$ define
	$$U_j=\bigcup \B{y_k}{\frac{\sqrt{\epsilon}}{2}}$$
	where the union is taken over all the $y_k$'s of color $j$.\\

	Since $F_i\subset \B{y_i}{\frac{\sqrt{\epsilon}}{2}}$, the sets $U_1,\dots, U_{d+1}$ are a closed cover of $S^d$. Thus, by the LSB Theorem, there exists an antipodal pair in one of the closed sets. Without lost of generality, say $x,(-x)\in U_1$, so $x\in \B{y_1}{\frac{\sqrt{\epsilon}}{2}}$ and $(-x)\in \B{y_2}{\frac{\sqrt{\epsilon}}{2}}$, with both $y_1$ and $y_2$ having color 1. Then,
	$$\norm{y_1+y_2}\leq \norm{y_1-x}+\norm{x+y_2}\leq \sqrt{\epsilon}\leq 2\sqrt{\epsilon-\frac{\epsilon^2}{4}},$$
	where the last inequality holds because $\epsilon<2$. Lemma \ref{lemma_min_distance} then implies $\norm{y_1-y_2}>2-\epsilon$, so $y_1$ and $y_2$ are connected by an edge in $G$, giving the desired contradiction.
\end{proof}

\section{Random Borsuk Graph with $\epsilon\to 0$}\label{section_3}
\subsection{Lower Bound}
The proof we gave for the lower bound in Theorem \ref{thm_eps_constant}, suggests that we should be able to let $\epsilon\to 0$ and still a.a.s.\ get the same chromatic number. Indeed, this will be the case. We just need to control the number of sets $N$ we use to cover the sphere and their area, in such a way that
$$\lim_{n \to \infty} 1 - N (1-c)^n \to 1.$$
In this section we discuss how to do this using $\delta$-nets on $S^d$, and then adapt the proof of Theorem \ref{thm_eps_constant} to get Theorem \ref{thm_lower_bound}.

We start with a technical lemma on spherical caps.

\begin{lemma}\label{lemma_area_spherical_cap}
	Given $x\in S^d$ and $0<r<1$, the following hold for the spherical cap $B=\B{x}{r}$.
	\begin{enumerate}
		\item The boundary $\partial B$, is a $(d-1)$-dimensional sphere with radius $$r'=r\sqrt{1-r^2/4}.$$
		\item $B$ is indeed a cap, i.e. there exist a $d$-hyperplane in $\mathbb{R}^{d+1}$, such that $B$ is the portion of $S^d$ contained in one of the semi-spaces defined by the hyperplane.
		\item The area $\area{B}$ satisfies the inequalities

		$$\frac{1}{\pi}\left(\frac{\sqrt{3}}{2}\right)^{d-1}r^d\leq \frac{\area{\B{x}{r}}}{\area{S^d}}\leq\frac{d}{3} r^d. $$

	\end{enumerate}

\end{lemma}

\medskip

Spherical caps are well studied in the literature. See, for example, Lemmas 2.2. and 2.3 in \cite{Ball97}. We note that Lemma 2.2 in \cite{Ball97} is better than our Lemma \ref{lemma_area_spherical_cap} in the case that $r$ is fixed and $d \to \infty$, but we are interested in the case that $d$ is fixed.

\begin{proof}[Proof of Lemma \ref{lemma_area_spherical_cap}]
Without loss of generality, we may choose $x=N=(0,\dots,0,1)$ to be the north pole by rotating the sphere. Thus $x\in\partial B$ if and only if $\norm{x}=1$ and $\norm{x-N}=r$. Then
	\begin{align*}
	r^2 &=\norm{x-N}^2 \\
	&=x_0^2+\cdots +x_{d-1}^2+(x_d-1)^2 \\
	&=x_0^2+\cdots+x_{d-1}^2+x_d^2-2x_d+1 \\
	&=2-2x_d. \\
	\end{align*}

So $x_d=1-\frac{r^2}{2}$. Therefore, the hyperplane $\{x_d=1-\frac{r^2}{2}\}$ determines the cap $B$, proving (2). To get (1), note $1=\norm{x}^2=x_0^2+\cdots+x_d^2$, so we get $x_0^2+\cdots+x_{d-1}^2=r^2-r^4/4=(r')^2$, for all $x\in\partial B$, proving it is indeed a $(d-1)$-dimensional sphere with the desired radius.\\

For (3), recall we get the area of $B=\B{x}{r}$ by integrating the length $\ell$ of the arc from $x$ to the boundary $\partial B$, over all the possible unit vectors. Thus
	$$\area{\B{x}{r}}=\int_{\hat{u}\in S^{d-1}} \ell (r')^{d-1}d\!\hat{u}=\ell(r')^{d-1}\int_{\hat{u}\in S^{d-1}}d\!\hat{u}=\ell(r')^{d-1}\alpha_{d-1}.$$

	Some planar geometry gives the bounds $$r\leq \ell=2\arcsin(r/2)\leq\frac{\pi}{3}r,$$ and $$\frac{\sqrt{3}}{2}r\leq r'\leq r,$$ since $0<r<1$. This gives
	$$\left(\frac{\sqrt{3}}{2}\right)^{d-1}r^d\alpha_{d-1}\leq \area{B}\leq \frac{\pi}{3}r^d\alpha_{d-1}$$

	Recall the formula for the surface area of the unit $d$-dimensional sphere $$\alpha_d=\dfrac{2\pi^{(d+1)/2}}{\Gamma\left((d+1)/2\right)}.$$
By using the fact that the Gamma function is increasing on $[2,\infty)$ and treating the first few cases separately, we have that for $d\geq 1$,
	$$\frac{1}{\pi}\leq\frac{\alpha_{d-1}}{\alpha_d}=\frac{\Gamma((d+1)/2)}{\sqrt{\pi}\ \Gamma(d/2)}\leq\frac{d}{\pi},$$
from which the desired result follows.

\end{proof}

For the sake of completeness, we include the following discussion on $\delta$-nets. See, for example, \cite[Chapter 13]{Matousek2002} for more details.

\begin{defi}[$\delta$-Nets]
	Given a metric space $X$ with metric $d$, a \textbf{$\boldsymbol{\delta}$-net} is a subset $\mathcal{B}\subset X$ such that for every $x\in X$, there exists $y\in \mathcal{B}$ with $d(x,y)<\delta$.
\end{defi}

We can then construct a $\delta$-net for any compact metric space $M$ inductively. Indeed, choose any point $y_1\in M$.
For each $m\geq 2$, if $B_m=\cup_{i=1}^m \B{y_i}{\delta}\subsetneq M$, choose any $y_{m+1} \in  M \setminus B_m$. Otherwise, stop and let $\mathcal{B}=\{y_1,y_2,\dots\,y_m \}.$

Compactness ensures that the process stops. It's clear that $\mathcal{B}$ is a $\delta$-net and moreover it is also a maximal $\delta$-apart set. That is, $d(y_i,y_j)>\delta$ whenever $i\neq j$, and we can not add any other point to $\mathcal{B}$ without destroying this property. This implies that the balls of radius $\delta$ and center on the points $y_i$'s cover $M$, while the open balls of radius $\delta/2$ with center on the $y_i$'s are all disjoint. We now show that we can control the size of the $\delta$-net in the case that $M=S^d$.

\begin{lemma}\label{lemma-nets}
	For every $d \ge 1$ and $0<\delta<1$ there exists a $\delta$-net $\mathcal{B}\subset S^d$ , such that (1) for every two points $y_i, y_j\in\mathcal{B}$, $\norm{y_i-y_j}>\delta$, and (2) its cardinality $\mathcal{N}=|\mathcal{B}|$ satisfies:
	$$\frac{3}{d\delta^d}\leq \mathcal{N}\leq \frac{2(3^d)(d+1)}{\delta^d}.$$
\end{lemma}

\medskip

In the literature, it seems more common to find an upper bound such as
$$\mathcal{N}\leq(4 / \delta)^{d+1},$$ which is a better bound when $\delta$ is constant and $d\to\infty$ \cite[Lemma 13.1.1]{Matousek2002}. However, the bound we give is more useful for us since we are dealing with $d$ constant and $\delta\to0$.

\begin{proof}[Proof of Lemma \ref{lemma-nets}]
	By letting $\mathcal{B}$ be the $\delta$-net defined above, we already have a $\delta$-net for the sphere $S^d$ that is also a $\delta$-apart maximal set. To prove the inequalities on its cardinality we give a volume and an area arguments. \\

	Since the points $y_i\in\mathcal{B}$ are $\delta$ apart, the open balls $\text{int}\left(B\left(y_i,\frac{\delta}{2}\right)\right)$ are disjoint. So
	$$ \vol{\bigcup_{i=1}^N B\left(y_i,\frac{\delta}{2}\right)}= \mathcal{N}\omega_{d}\left(\frac{\delta}{2}\right)^{d+1},$$
where $\omega_{d}$ denotes the volume of the $d+1$-dimensional unit ball in $\mathbb{R}^{d+1}$.	Morever, all such balls are contained in the set $B\left(0,1+\frac{\delta}{2}\right) \setminus B\left(0,1-\frac{\delta}{2}\right)$. Thus
	\begin{align*}
	\vol{\bigcup_{i=1}^N B\left(y_i,\frac{\delta}{2}\right)} &\leq
		\vol{B\left(0,1+\frac{\delta}{2}\right)} - \vol{B\left(0,1-\frac{\delta}{2}\right)}\\
		& = \omega_{d}\left(\left(1+\frac{\delta}{2}\right)^{d+1} - \left(1-\frac{\delta}{2}\right)^{d+1}\right)\\
		&= \omega_{d}\delta \sum_{r=0}^{d}\left(1+\frac{\delta}{2}\right)^{d-r}\left(1-\frac{\delta}{2}\right)^r \\
		& \leq \omega_{d}\delta \sum_{r=0}^{d}\left(1+\frac{\delta}{2}\right)^{d}\\
		&=\omega_{d}\delta (d+1)\left(1+\frac{\delta}{2}\right)^d \\
		& \leq \omega_d\delta(d+1)\left(\frac{3}{2}\right)^d,
	\end{align*}
	and the upper bound for $\mathcal{N}$ follows.

	For the lower bound, we consider the area of the spherical caps. Since all points in $S^d$ are within distance $\delta$ of the points $y_i\in\mathcal{B}$ we must have
	$$\area{S^d}=\area{\bigcup_{i=1}^{\mathcal{N}}\B{y_i}{\delta}}\leq\sum_{i=1}^{\mathcal{N}}\area{\B{y_i}{\delta}}=\mathcal{N}\area{\B{y_1}{\delta}}$$

	Therefore, Lemma \ref{lemma_area_spherical_cap} yields	$\displaystyle\mathcal{N}\geq \frac{\area{S^d}}{\area{\B{y_0}{\delta}}}\geq\frac{3}{d\delta^d}$.
\end{proof}

We now proceed to prove Theorem \ref{thm_lower_bound}.

\begin{proof}[Proof of Theorem \ref{thm_lower_bound}]
	Let $G=\Bord{\epsilon(n)}{n}$. Since $\epsilon\to 0$, eventually $\epsilon<2-\lambda_d$, so by Lemma \ref{lemma_color_d+2}, $\chi(G)\leq d+2$. We now proceed to prove the lower bound by a modification of the proof of theorem \ref{thm_eps_constant}. \\

	Let $\delta = \sqrt{\epsilon} / 4$. Let $\mathcal{B}$ be the $\delta$-net given by Lemma \ref{lemma-nets}. Say $\mathcal{B}=\{y_1,y_2,\cdots,y_N\}$, where $$N\leq \frac{2(3^d)(d+1)}{\delta^d}=\frac{A_d}{\epsilon^{d/2}},$$ and $A_d$ is a constant which only depends on $d$. For each $i=1,\cdots, N$, define $F_i=\B{y_i}{\delta}$. Note then the $F_i's$ cover the sphere and $\diam{F_i}\leq2\delta=\frac{\sqrt{\epsilon}}{2}$. \\

	Applying Lemma \ref{lemma_area_spherical_cap}, we have
	$$\frac{\area{F_i}}{\area{S^d}}\geq
	\frac{1}{4\pi}\left(\frac{\sqrt{3}}{8}\right)^{d-1}\epsilon^{d/2}=
	B_d \epsilon^{d/2}=:c,$$
	where $B_d$ is constant.\\

	Finally, all that remains to prove is that $1-N(1-c)^n\to 1$ as $n\to\infty$, even when $N$ and $c$ depend on $n$. This is as follows
	\begin{align*}
	N(1-c)^n &\leq \frac{A_d}{\epsilon^{d/2}}\left(1-c\right)^n \\
	&=
	\frac{A_d}{\epsilon^{d/2}}\left(1-B_d\epsilon^{d/2}\right)^n \\
	&=
	\frac{A_dn}{C^{d/2}\log{n}}\left(1-\frac{B_dC^{d/2}\log{n}}{n}\right)^n\\
	&\leq \frac{A_dn}{C^{d/2}\log{n}}\exp\left(-B_dC^{d/2}\log{n}\right)\\
	&= \frac{A_d}{C^{d/2}\log{ n}} n^{1-B_dC^{d/2}}
	\end{align*}
	The last expression goes to zero as $n\to\infty$, since $$C\geq \frac{64}{3}\sqrt[d]{\frac{3\pi^2}{4}},$$ so $B_d C^{d/2}\geq 1$ and this completes the proof.
\end{proof}

\begin{coro}
	If $$\frac{64}{3}\sqrt[d]{\frac{3\pi^2}{4}}\left(\dfrac{\log{n}}{n}\right)^{2/d}\leq \epsilon(n)\leq 2-\lambda_d$$ for all sufficently large $n$, then $\chi\left(\Bord{\epsilon(n)}{n}\right)=d+2$ a.a.s.
\end{coro}
\begin{proof}
	The chromatic number is monotone with respect to $\epsilon$, so this follows directly.
\end{proof}

\subsection{Upper Bound}\label{section_4}
Theorem \ref{thm_lower_bound} and its Corollary shows that if $\epsilon \to 0$ sufficiently slowly then the chromatic number of the random Borsuk graph is a.a.s. $d+2$. In this section we show that the rate obtained is tight, up to a constant factor. That is, we show an upper bound for $\epsilon$ for which the random Borsuk graph is $(d+1)$-colorable.

We start our analysis by constructing a proper coloring of $\Bor{\epsilon}\setminus\B{x}{\delta}$ with exactly $d+1$ colors, for a suitable $\delta$ that depends on $\epsilon$. Lemma \ref{lemma_area_spherical_cap} establishes that the boundary of an spherical cap on $S^d$ is a $S^{d-1}$ with radius $\delta'$, and Lemma \ref{lemma_color_d+2} allows to color it with $d+1$ colors. We will provide the technical details to translate this coloring into a proper coloring of the desired graph.\\

For the following analysis consider the spherical cap $A=\B{N}{r}$, where $N$ is the north pole. For each $x\in S^d\setminus\{N,-N\}$, let $\gamma_x:[0,\pi]\to S^d$ be the great semi-circle going from $N$ to $-N$ and passing through $x$. Define $f: S^d\to\partial A$ by letting $f(x)$ be the intersection of $\gamma_x$ with $\partial A$. Note this is a well defined function, since if $x=(x_0,\dots, x_d)$, we can parametrize $$\gamma_x(t)=\left(\frac{\sin{t}}{\sqrt{1-x_d^2}}x_0,\dots,\frac{\sin{t}}{\sqrt{1-x_d^2}}x_{d-1},\cos{t}\right)$$
so its last coordinate takes all values in $[-1,1]$ exactly once for $0\leq t\leq \pi$, and from Lemma \ref{lemma_area_spherical_cap} we know $\partial A$ consists of all points with last coordinate $a:=1-\frac{r^2}{2}$.\\

The following lemmas construct the desired coloring.

\begin{lemma}\label{lemma_distance_geodesics}
	Let $x, y\in S^d\setminus\{N,-N\}$ such that $\norm{x-y}\leq\delta$. Define $y'=(y'_0,\dots,y'_d)$ to be the point in the geodesic $\gamma_y$, such that $y'_d=x_d$. Then $\norm{x-y'}\leq 2\delta$.
\end{lemma}

\begin{proof}
	Without lost of generality, we may assume $y=(0,0,\dots,0,\sqrt{1-y_d^2},y_d)$, since we can get this by a rotation of the sphere that leaves the last coordinate fixed. This rotation fixes the north and south poles, so it also transforms the geodesic through $y$ into another geodesic through $y$. Thus, the formula for the geodesic simplifies to
	$$\gamma_y=(0,\dots,0,\sin{t},\cos{t}),\text{    for    } 0\leq t\leq\pi.$$
	So, $y'=(0,\dots,0,\sqrt{1-x_d^2},x_d)$. Then
	\begin{align*}
	\norm{x-y}^2
	&=x_0^2+\cdots+x_{d-2}^2+\left(x_{d-1}-\sqrt{1-y_d^2}\right)^2+(x_d-y_d)^2\\
	&= (1-x_d^2)+(1-y_d^2)-2x_{d-1}\sqrt{1-y_d^2}+(x_d-y_d)^2\\
	\end{align*}
	and
	\begin{align*}
	\norm{y-y'}^2
	&=\left(\sqrt{1-y_d^2}-\sqrt{1-x_d^2}\right)^2+(x_d-y_d)^2\\
	&= (1-x_d^2)+(1-y_d^2)-2\sqrt{1-x_d^2}\sqrt{1-y_d^2}+(x_d-y_d)^2
	\end{align*}

Since $x_{d-1}\leq |x_{d-1}|\leq \sqrt{x_0^2+\cdots+x_{d-1}^2}=\sqrt{1-x_d^2}$, we get $\norm{y-y'}\leq\norm{x-y}$, and so $$\norm{x-y'}\leq\norm{x-y}+\norm{y-y'}\leq 2\norm{x-y}\leq 2\delta.$$
\end{proof}

\begin{lemma}\label{lemma_distance_F}
	Let $x, y\in S^d\setminus\{N,-N\}$ such that $x\not\in A\cup (-A)$ and $\norm{x-y}\leq\delta$. Then $$\norm{f(x)-f(y)}\leq 2\delta.$$
\end{lemma}
\begin{proof}
	Let $y'\in S^d$ such that its last coordinate is $y'_d=x_d$. From the parametrization for $\gamma_x$, we see $f(x)=\gamma_x(t_1)$, where $\cos{t_1}=a$ and $\sin{t_1}=\sqrt{1-a^2}=\delta'$, the radius of $\delta A$, hence
	$$f(x)=\left(\frac{\delta'}{\sqrt{1-x_d^2}}x_0,\dots,\frac{\delta'}{\sqrt{1-x_d^2}}x_{d-1},a\right).$$
	A similar expression holds for $f(y')$, with $y'_d=x_d$, so we get
	\begin{align*}
	\norm{f(x)-f(y')}&=\sqrt{\sum_{i=0}^{d-1}\frac{\delta'^2}{1-x_d^2}(x_i-y_i')^2}\\
	&=\frac{\delta'}{\sqrt{1-x_d^2}}\sqrt{\sum_{i=0}^{d-1}(x_i-y'_i)^2}\\
	&\leq \frac{\delta'}{\sqrt{1-x_d^2}} \norm{x-y'}
	\end{align*}

	Moreover, since $x\not\in A\cup (-A)$, $|x_d|<a$, so $\frac{\delta'}{\sqrt{1-x_d^2}}<1$, so $\norm{f(x)-f(y')}\leq \norm{x-y'}$. Finally, if we let $y'$ be the one defined in Lemma \ref{lemma_distance_geodesics}, $f(y')=f(y)$, and therefore $\norm{f(x)-f(y)}=\norm{f(x)-f(y')}\leq \norm{x-y'}\leq 2\delta$.

\end{proof}

\begin{lemma}\label{lemma_color_d+1}
	Let $0<\epsilon<1$, such that $$ r=\frac{8\sqrt{\epsilon}}{\sqrt{3(4-\lambda_{d-1}^2)}}<1,$$ $x\in S^d$, and $A=\B{x}{r}$. Let $H$ be the induced subgraph of $\Bor{\epsilon}$ by the vertex set $S^d\setminus A$. Then $\chi(H)\leq d+1$.
\end{lemma}
\begin{proof}
	Without loss of generality let $x=N$ the north pole, so $A=\B{N}{r}$. Lemma \ref{lemma_area_spherical_cap} says $\partial A$ is a $S^{d-1}$ sphere of radius $r'=r\sqrt{1-\frac{r^2}{4}}\geq\frac{\sqrt{3}}{2}r$. Thus adapting Lemma \ref{lemma_color_d+2}, we can color it in such a way that every two points with the same color are at a distance of at most $\lambda_{d-1}r'$. We then color $H$ by giving each point $y\in S^d\setminus A\setminus\{-N\}$ the color of $f(y)$, and giving the south pole $-N$ any color. We proceed to prove this is a proper coloring of $H$. \\

	From Lemma \ref{lemma_min_distance}, the neighbors of the south pole lie in $\B{N}{\sqrt{\epsilon-\epsilon^2/4}}\subset A$, so $-N$ is isolated in $H$. Let $y,z\in S^d\setminus A\setminus\{-N\}$ such that $\norm{y-z}>2-\epsilon$. Lemma \ref{lemma_min_distance} implies $\norm{y+z}<\delta:=2\sqrt{\epsilon-\frac{\epsilon^2}{4}}$. If we had $(-y),(-z)\in A$, that would mean $y,z\in -A$, but then $\norm{y-z}\leq r\leq 2-\epsilon$ for small $\epsilon$. So we may assume $(-y)\not\in A$, and since $y\not\in A$, $(-y)\not\in -A$. Thus $(-y)\not\in A\cup (-A)$ and $\norm{-y-z}\leq\delta$, thus Lemma \ref{lemma_distance_F} implies
	$$\norm{f(-y)-f(z)}\leq 2\delta=4\sqrt{\epsilon-\frac{\epsilon^2}{4}}<4\sqrt{\epsilon}=\sqrt{4-\lambda_{d-1}^2}\frac{\sqrt{3}}{2}r\leq \sqrt{4-\lambda_{d-1}^2}r'$$
	From the definition of $f$, it is clear that $f(-y)=-f(y)$, thus Lemma \ref{lemma_min_distance} implies $\norm{f(y)-f(z)}>\lambda_{d-1}r'$, and so $f(y)$ and $f(z)$ have different colors, meaning $y$ and $z$ have different colors as well. Therefore $\chi(H)\leq d+1$.

	\end{proof}

As an immediate application, if a random Borsuk graph leaves some spherical cap in $S^d$ of radius bigger than $r$ with no vertices, then it can be colored with $d+1$ colors. We will show that this is indeed the case when $\epsilon\to 0$ at the said rate. We now include some theorems about Poisson Point Processes and Poisson distributions. For their proofs and a complete discussion refer to \cite{Penrose2003} or \cite{Kingman1993}.

\begin{theorem}[Poissonization]\label{thm_poissonization}
	Let $X_1, X_2, \dots$, be uniform random variables on $S^d$. Let $M\sim \text{Pois}(\lambda)$ and let $\eta$ be the random counting measure associated to the point process $P_\lambda$= $\{X_1,X_2,\dots,X_M\}$. Then $P_\lambda$ is a Poisson Point Process and for a Borel $A\subset S^d$, $\eta(A)\sim \pois{\lambda\frac{\area{A}}{\area{S^d}}}$.
\end{theorem}

\begin{lemma}\label{lemma_poisson}
	For $n\geq 0$, $\P{\pois{2n}<n}\leq e^{-0.306n}.$
\end{lemma}

We are now ready to prove the Theorem \ref{thm_upper_bound}.

\begin{proof}[Proof of Theorem \ref{thm_upper_bound}]
	Let $X_1, X_2, \dots$, be uniform random variables on $S^d$. Let $M\sim\pois{2n}$. Let $\eta$ be the random counting measure of the Poisson Point Process $\left\{X_1,\dots,X_M\right\}$. Similarly, let $\eta_1^n$ be the counting measure of the Random points $\left\{X_1,\dots, X_n\right\}$.\\

    Let $$\delta=\frac{16\sqrt{\epsilon}}{\sqrt{3(4-\lambda_{d-1}^2)}}=A_d\sqrt{\epsilon},$$ where $A_d$ is a constant which only depends on $d$. Let $\mathcal{B}=\left\{y_1,\dots,y_N\right\}$ be the $\delta$-net given by Lemma \ref{lemma-nets}, so $$N \geq \frac{3}{d\delta^d}=\frac{B_d}{\epsilon^{d/2}},$$ and $B_d$ is constant. Let $F_i=\B{y_i}{\delta/2}$ for $i=1,\dots,N$ be spherical caps centered at the $\delta$-net. Thus, as in the proof of \ref{lemma-nets}, the $F_i$'s are disjoint.
	Lemma \ref{lemma_area_spherical_cap} gives $$\frac{\area{F_i}}{\area{S^d}}\leq\frac{d}{3}\left(\frac{\delta}{2}\right)^d=D_d\epsilon^{d/2},$$
where $D_d$ is constant.

	Note that these spherical caps have the same radius required by Lemma \ref{lemma_color_d+1}, so if we prove that a.a.s.\ one of these $F_i$'s doesn't contain any vertices of the random Borsuk graph, then it must be contained in $S^d\setminus F_i$, and the Lemma \ref{lemma_color_d+1} gives a proper $(d+1)$ coloring. This is what we do.\\

	Note that
	\begin{equation}
		\P{\min_{1\leq i\leq N}\eta(F_i)=0} \leq \P{\min_{1\leq i\leq N}\eta_1^n(F_i)=0}+\P{M<n}.
		\label{eqn_prob_unif_vs_poisson}
	\end{equation}

We have
	\begin{align*}
	\P{\min_{1\leq i\leq N}\eta(F_i)=0}&=1-\P{\bigwedge_{i=1}^N\eta(F_i)>0}=1-\prod_{i=1}^N\P{\eta(F_i)>0}\\&=1-\left(1-\P{\pois{2n\frac{\area{F_1}}{\alpha_d}}=0}\right)^N\\
	&\geq 1-\exp\left(-\exp\left(-2n\frac{\area{F_1}}{\alpha_d}\right)N\right)\\
	&\geq 1-\exp\left(-\exp(-2nD_d\epsilon^{d/2})\frac{B_d}{\epsilon^{d/2}}\right)\\
	&=1-\exp\left(-\frac{B_d}{C^{d/2}\log{n}}n^{1-2D_dC^{d/2}}\right).\\
	\end{align*}

    This last expression tends to 1 as $n\to\infty$, since $C$ is such that $1-2D_dC^{d/2}>0$.\\

Lemma \ref{lemma_poisson} assures that $$\P{M<n}=\P{\pois{2n}<n}\to 0$$ as $n\to \infty,$ and therefore (\ref{eqn_prob_unif_vs_poisson}) gives $\displaystyle\P{\min_{1\leq i\leq N}\eta_1^n(F_i)=0}\to 1$, as desired.\\
\end{proof}

\section{Further Questions}

\begin{enumerate}

\item It might be possible to find sharper constants in Theorems \ref{thm_lower_bound} and \ref{thm_upper_bound}. For $d=1$, it is certainly possible. The following can be achieved with similar methods to the ones used throughout this paper, so we include the statement without proof.

\begin{theorem}
Let $\epsilon=C \left(\log n / n \right)^2$.
\begin{enumerate}
    \item If $C\geq 9\pi^2 /4$, then a.a.s. $\chi\left(\Borone{\epsilon}{n}\right)=3$.
    \item If $C<\pi^2/4$, then a.a.s. $\chi \left( \Borone{\epsilon}{n} \right) \le 2$.
    \end{enumerate}
\end{theorem}

\item We wonder whether there exist functions $\epsilon=\epsilon(n)$ such that the chromatic number of the random Borsuk graph $\Bord{\epsilon}{n}$ a.a.s.\ equals $i$, for $1 \le i \le d+1$.

\item We only studied here the case that $d$ is fixed and $\epsilon$ is either fixed or tends to zero at some rate. It also seems interesting to let $d \to \infty$ at some rate, or to let $d$ be fixed and $\epsilon \to 2$. See, for example, Raigorodskii's work on coloring high-dimensional spheres \cite{Raigorodskii2012}.
\end{enumerate}

\noindent We thank our anonymous referees for careful reading and helpful comments.

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