Infections of the ear are common and may occur anywhere in the ear including the external canal (otitis external), the middle ear (otitis media), the mastoid portion of the skull, or other areas of bone surrounding these structures. Infections of the middle ear and mastoid are particularly common following upper respiratory infections.
While occasional ear infections are common, chronic infections of the middle ear and mastoid can cause significant damage to the hearing and balance mechanisms, facial nerves, and in some cases it can actually spread beyond the ear to surrounding structures including the brain. One result of chronic infections is cholesteatoma, a condition where dead skin and chronic inflammation invades the eardrum into the middle ear resulting in the damage and erosion of the middle ear bones. If left untreated, the damage from chronic ear infections can extend to the inner ear, causing sensorineural hearing loss and vertigo. Rare but possible is that it can expand into the cranial cavity and spinal fluid spaces leading to meningitis or brain abscess.
Many tumors can affect the external, middle, and inner ear. The acoustic neuromas (vestibular schwannoma) are benign tumors that come from the vestibular nerve often within the internal auditory canal between your ear and brain. They commonly produce a progressive hearing loss and vertigo usually in one ear. Glomus tumors arise in blood vessel structures in your middle ear or adjacent skull base.
Malformation of the external ear, ear canal, middle ear, and inner ear sometimes actually occur in the womb. Many deformities may be surgically reconstructed: particularly disorders of the auricle, external, and middle ear.
Ototoxicity is ear poisoning resulting from exposure to drugs or chemicals that damage the inner ear or the vestibulo-cochlear nerve (the nerve sending balance and hearing information from the inner ear to the brain). Because the inner ear is involved in both hearing and balance, ototoxicity can result in disturbances of either or both of these senses. Currently, available treatments focus on reducing the effects of the damage and rehabilitating function and individuals who have suffered from hearing loss may benefit from cochlear implants.
Your body maintains balance and position through a carefully composed network known as the vestibular system or your body's navigation system. Your inner ear is affected when something goes wrong with balance and your ear however is only one part of the vestibular system that includes vision, muscles, joints and your brain. When any part of the system does not work properly, including your inner ear, the result is vertigo, dizziness, falls, and a sense of spinning. There are only an elite number of otologists (ear doctors) who specialize in vestibular problems and fortunately; Dr. V is experienced in correcting the demands of dizziness and balance.
Meniere's disease is a syndrome that causes episodes of severe vertigo. Vertigo is a sensation of spinning or motion that can frequently causes nausea and vomiting. If a person develops vertigo that lasts several hours, they have a feeling of pressure in one ear, they hear a roaring noise in the same ear, and they experience fluctuating hearing loss, it is likely that they have developed Meniere's disease. It is not known what causes this condition but a patient who has Meniere's disease has dilated inner ear fluid spaces, which suggest that the fluid (endolymph) is under increased pressure. Many medical researchers believe autoimmunity plays a role in Meniere's disease which means that your own body's' immune system attacks an organ system or tissue in your body. Over the last several years, major breakthroughs have become available to treat Meniere's disease and at The Virginia Institute of Surgical Arts, we strive to control vertigo episodes, as this is the most incapacitating part of the syndrome before we determine who is a candidate for surgical procedures.
When a patient has long-term and recurrent problems with ear infections and drainage from the ear, this is termed as chronic ear disease. The draining of pus and continual infections are abnormal and chronic ear disease causes problems that might include a hole in the eardrum, thickening of the mucosal lining of the middle ear space (mucosal hypertrophy), the growth of skin from the outer surface of the eardrum to the middle ear space, and the formation of a skin cyst called a cholesteatoma.
Cholesteatoma is an infected skin cyst that occurs as a result of chronic negative pressure in the middle ear space. The negative pressure usually occurs as a result of the malfunction of the Eustachian tube, which is responsible for allowing air equalization between the upper throat and nose with the middle ear space. If this does not work properly, negative pressure will develop in the middle ear space and eventually this negative pressure will drain in the eardrum where it contacts structures of middle ear or the inside walls of the middle ear. When this happens, the skin will start to grow across the structures or within the middle ear space and eventually form a skin cyst. The inside of this cyst contains dead skin and bacteria and it is usually infected. The leading edges of the cyst secrete enzymes that can destroy bone on the wall on the surface of the middle ear or the small bones of hearing, which transmit sound from the eardrum to the inner ear (ossicles). Cholesteatomas destroy local tissues and if left untreated, they can cause permanent hearing loss, dizziness, and facial paralysis.
Otosclerosis is a condition, which can involve your ossicles (small bones of hearing) creating conductive hearing loss. This condition arises when abnormal and immature bone grows on one of the small ossicles usually the stapes. The stapes is the third bone in the chain of three ossicles and it receives sound from the eardrum while the first two bones transfer the vibration into the inner ear. Once the Otosclerosis bone has grown around the footplate of the stapes, it will not vibrate and therefore does not conduct sound in your inner ear. This is the cause of the conductive hearing loss and it is unknown what causes Otosclerosis. If a person is diagnosed with Otosclerosis, there is approximately a 40% chance that they will develop it in their other ear.
The facial nerve reaches the muscles of your face through the internal auditory canal, the mastoid portion of the skull, and the middle ear. Because the facial nerve exists in you ear, problems that occur in your ear will affect the facial nerve and this may cause weakness and paralysis in the face or abnormal twitching and distortion. After proper diagnosis, there are a variety of medical and surgical treatments that may reverse or at least improve deformity and dysfunction caused by facial nerve disorders.
EAC (External Auditory Canal) Stenosis is a milder form of atresia, which refers to the absence an external ear canal. Rather than absent, it is simply narrower than normal although narrow (stenosed) canals can be associated with the same ossicular malformations as regular atresia. One of the symptoms is conductive hearing loss and The Virginia Institute of Surgical Arts has treatment options available for atresia.
Tinnitus is a ringing or other type of noise that originates in the ear or head. Although sometimes considered a nuisance, Tinnitus can represent a serious health condition. Nearly 36 million Americans suffer from this disorder and Tinnitus can arise in any of the following areas: the outer ear, the middle ear, the inner ear, or by abnormalities in the brain. Fluid, infection, or disease of the middle ear bones or eardrum (tympanic membrane) can cause Tinnitus and one of the most common causes is damage to the microscopic endings of the hearing nerve in the inner ear. When left untreated, Tinnitus results in a certain amount of hearing nerve impairment and consequently chronic Tinnitus.


Who are these people? And what in the world are they thinking?
So, imagine this. Imagine that you have never heard about the game kubb before. Pick a state outside of WI, MN, IA, and IL, and you live there. Say, Oregon or North Carolina. It is last weekend. Beautiful summer day. One of your friends stumbled upon this game and made a set. You and your family head over there and play a little. You like it, you have a good time. You are tossing kubbs with no real thought. Your batons are pretty vertical, and they are spinning like a fan. Lots of fun and lots of laughs.
Monday morning rolls around and you do a quick Internet search on it. Let's say one of the first websites that pops up is www.usakubb.org. You think to yourself, there is a United States National Kubb Championship in a town called Eau Claire, WI. Heck, you might not even know how to say Eau Claire. You think, are you kidding me, a US Championship in kubb, and there are other tournaments all over the Midwest? You do a little more poking around and find some other websites, pictures, videos, and more.
The crazy thing is, this happens all the time. The US Championship was nice enough to do a little screen shot of the website hits they got over last weekend and Monday. Think about it, who are these people, and what in the world do they think when they learn more about kubb.
Yes, for the most part, this is pretty normal with most of the activity being in the upper Midwest. But people from all over are new to the game and finding out more about kubb. What are they looking at? Hmmmm. Let's think about it a little.
They are looking at people kneeling and sitting on the ground thinking how they are going to put kubbs up. And these people look like they are thinking about it for a long time.
They are looking at guys carrying around big kubb kings painted like a royal king....and big kubb tattoos. What are these people in Oregon, NC, and beyond thinking? Really, think about it. What. Are. They. Thinking.
They look at hip young folks that play this game and what they think are total party animals as well.
They tell their friends, people really travel with kubb pieces and kubb shirts while they travel...the world.
They think, people make names for drinks after the game and put the recipe on the back of their team shirt?
And people actually get interviewed when they play?
And two guys (one with a Brazil flag as a cape) doing a THREE HOUR Live Team Draw for the US Championship with the help of dozens of others, not to mention the six people that drove five plus hours and two that drove four hours, two weeks before they will drive the say trip again.
We get so curious who these people are. Just like when we look up in the sky and see an airplane. Who are those people up there? Where are they from? Where are they going? Are they looking down at our city right now, and if so, what are they thinking about? We often wonder to ourselves, who are these people finding out about kubb, and what in the world are they thinking about us? To be honest, we hope they think we are a little strange and/or have tons of passion for this game. You know why? That is who we are.


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Tobacco Plain Packaging Bill 2011
war / 14 July 2011 14 July 2011 / Trade marks
Last month Phillip Morris announced it plans to sue Australia over the proposed plain packaging legislation for tobacco products.
The Tobacco Plain Packaging Bill 2011 was introduced into Parliament on 6 July and, following its 1st and 2nd Readings, has been referred to the House of Representatives' Standing Committee for Health and Ageing. The Bill was also introduced with the Trade Marks Amendment (Tobacco Plain Packaging) Bill 2011.
The Tobacco Plain Packaging Bill and the EM can be downloaded from here.
The Trade Marks Amendment … Bill and its EM from here.
The House Committee has requested submissions on the Bill by 22 July 2011.
The Committee's Media Alert.
8 Comments on Tobacco Plain Packaging Bill 2011/
A business method patent (not yet)
war / 12 July 2011 3 October 2013 / Patents
RPL applied for an innovation patent of a method entitled 'Method and System for Automated Collection of Evidence of Skills and Knowledge'.
The applied for innovation relates to a method for people to obtain recognition for their prior learning. Apparently, there are some 35,000 qualifications and 34,000 units in the vocational educational and training sector in Australia. Hence it can be difficult for people to identify a particular qualification which they have qualified for as there is no single point of access to the system. The applied for innovation claimed methods for using technology, particularly the internet, to automate the process of identifying relevant criteria and applying for the relevant recognition.
Claim 1 claimed:
1. A method of gathering evidence relevant to an assessment of an individual's competency relative to a recognised qualification standard, including the steps of:
a computer retrieving via the Internet from a remotely-located server a plurality of assessable criteria associated with the recognised qualification standard, said criteria including one or more elements of competency, each of which is associated with one or more performance criteria;
the computer processing the plurality of assessable criteria to generate automatically a corresponding plurality of questions relating to the competency of an individual to satisfy each of the elements of competency and performance criteria associated with the recognised qualification standard;
an assessment server presenting the automatically-generated questions via the Internet to a computer of an individual requiring assessment; and
receiving from the individual via said individual's computer a series of responses to the automatically generated questions, the responses including evidence of the individual's skills, knowledge and/or experience in relation to each of the elements of competency and performance criteria,
wherein at least one said response includes the individual specifying one or more files stored on the individual's computer, which are transferred to the assessment server.
Myall opposed alleging lack of novelty based on prior use. The applicant didn't even need to file evidence to defend this allegation which failed on the evidence submitted by the opponent.
However, it appears that the delegate hearing the opposition raised the objection that the innovation claimed was not a manner of manufacture (see [62]).
The delegate noted (at [45]) that the Full Federal Court's decision in Grant had qualified NRDC so that a manner of manufacture required
'an artificial state of affairs, in the sense of a concrete, tangible, physical, or observable effect'
A physical effect in the sense of a concrete effect or phenomenon or manifestation or transformation is required.
This had been further explained by the Deputy Commissioner in Invention Pathways:
I do not take [Grant] to suggest that patentability is merely determined on the presence of a physical effect. Rather it clearly must be an effect of such substance or quality that the method considered as a whole is "proper subject of letters patent according to the principles which have been developed for the application of s. 6 of the Statute of Monopolies".
Accordingly, the Deputy Commissioner considered:
… the "concrete effect or phenomenon or manifestation or transformation" referred to must be one that is significant both in that it is concrete but also that it is central to the purpose or operation of the claimed process or otherwise arises from the combination of steps of the method in a substantial way. Consequently while the step of building a house involves a concrete physical effect it is peripheral to the method of acquiring a house and indeed could hardly be said to characterise the subject matter of the method such that it is considered an artificially created state of affairs. I consider the same to apply to a business scheme implemented in some part by computer and do not believe the patentability of such a method can arise solely from the fact that, in a general sense, it is implemented in or with the assistance of a computer or utilises some part a computer or other physical device in a incidental way.
In this case, the delegate considered at [54]:
54. It is clear that the claim relates to gathering information where the internet is used for transmitting and receiving data, and a computer is used for data retrieval, processing, presentation and storage of information in a well known manner. The process of automatically generating questions based on assessable criteria, as stated in the description, includes applying a question template into which the text of the relevant assessable criteria may be merged. In a simple form, a question may be generated from the criteria by prepending the text such as 'How can you show evidence that'.
and so found at [55]:
55. The claimed invention defines a method for gathering information where the data retrieval, processing and storage of information appear to have no physical effect other than that would arise in the computer with standard software in conventional use. Furthermore, there is no substantial effect or transformation in generating the questions by concatenating text matters. While the internet and the computer facilitate the operation of the claimed method by retrieving, generating and conveying information, they are not central to the purpose of the claimed invention. [So unlike the loyalty card in Welcome-Catuity], the claimed invention simply monopolises a scheme where the internet and the computer are used for mere convenience for operating the scheme.
Accordingly, the delegate rejected the application in that form. However, at [59], the delegate considered there could nonetheless be patentable subject matter:
the description of the opposed patent contemplates generating the questions automatically based upon the identification of particular keywords within the assessable criteria, and upon additional contextual information obtained from Evidence Guides, Range Statements and Employability Skills associated with Qualifications and Units of Competency without requiring human intervention.
and so allowed 60 days for amendments to be brought in (the delegate also flagged a potential fair basing problem for the proposed amendment).
Myall Australia Pty Ltd v RPL Central Pty Ltd [2011] APO 48 (link to be provided when available)
Lid dip: Patentology
Leave a Comment on A business method patent (not yet)/ business method,manner of manufacture,novelty,opposition,patentable subject matter,Patents
Streamlining patent applications
war / 5 July 2011 5 July 2011 / Patents
Back in February, the Prime Ministers of Australia and New Zealand issued a statement about harmonising IP laws (and procedures) between their countries.
Today, the Minister for Innovation, Industry and Science and the New Zealand Commerce Minister issued a further statement "revealing" that:
there will be a single application process for seeking patents in both countries from 2013; and
a single examination of the application for both countries will be introduced from 2014.
The two countries' patent laws will continue to be different but, when these steps are introduced, there will be an integrated application and examination process to get a patent in both countries.
According to the Australian Minister:
The savings in professional costs alone could be as high as $2,000 to $5,000 per invention — money that should and will stay in the pocket of the innovator.
Not entirely clear from the information released officially how things have progressed since February. Press Release here; Fact Sheet there (pdf).
Leave a Comment on Streamlining patent applications/ application,Australia,examination,harmonisation,New Zealand,Patents,SEM,Single Economic Market,Trans Tasman
Scope of disclosure in an innovation patent
war / 4 July 2011 / Patents
Patentology has a nice summary of the innovation patentee's successful appeal in Seafood Innovations Pty Ltd v Richard Bass Pty Ltd [2011] FCAFC 83.
One point: it seems like the disclosure in the body of the specification supporting the broadest claim was at a level of generality similar to that upheld by the High Court in the first round of Lockwood. Wonder how that will hold up for future application under the (proposed) Raising the Bar legislation?
Secondly, the (now) infringer's best line of defence (apart from invalidity) seems to have turned on trying to argue (now unsuccessfully) that incorporating additional integers in the accused gutting machine to those specified in the claim. Bennett J gave it short shrift.
Leave a Comment on Scope of disclosure in an innovation patent/ Australia,fair basis,infringement,innovation patent,Patents,sufficiency,validity

redpajama_set_name: RedPajamaCommonCrawl


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Home-About Terri-Personal Blog
Personal BlogTerri Paddock2021-09-01T11:15:11+01:00
Occasionally, I have time to write about productions where I haven't chaired post-show talks and perhaps even thoughts on topics that have nothing to do with theatre. You can find these ramblings here, including my still very popular What I Learned Today series.
11:14 am12 April 2017
Press Pass: Everything you need to know about Alfie Boe & Katherine Jenkins in Carousel
Terri Paddock2020-03-27T12:48:34+00:0012 April 2017|Features, Musicals, Personal, Reviews, Terri's Blog|
Carousel at the London Coliseum After successes with Sunset Boulevard and Sweeney Todd, English National Opera continues its
12:06 pm5 April 2017
Press Pass: Everything you need to know about 42nd Street (and Kate Middleton!)
Terri Paddock2020-03-27T12:48:41+00:005 April 2017|Features, Interviews, Musicals, Opinion, Personal, Reviews, Terri's Blog|
The cast of 42nd Street. © Brinkhoff & Mogenburg The second massive New York import in less than a
7:14 pm2 April 2017
Bra-waving, toe-tapping and female solidarity: The Girls and Stepping Out
Terri Paddock2020-01-08T09:50:54+00:002 April 2017|Features, Musicals, Opinion, Personal, Plays, Reviews, Terri's Blog|
Joanna Riding in The Girls; Dominic Rowan and Tracy-Ann Oberman in Stepping Out If you're seeking life affirmation,
2:38 pm28 March 2017
Take two: Tom Stoppard's Rosencrantz & Travesties, plus future fantasy casting
Terri Paddock2020-03-27T12:48:51+00:0028 March 2017|Features, Opinion, Personal, Plays, Reviews, Terri's Blog|
Doubling up with Stoppard: Joshua McGuire & Daniel Radcliffe in R&G, Freddie Fox & Tom Hollander in Travesties
12:52 pm22 March 2017
Press Pass: Everything you need to know about An American in Paris
Terri Paddock2020-03-27T12:49:02+00:0022 March 2017|Features, Musicals, Opinion, Personal, Reviews, Terri's Blog|
An American in Paris has burst onto the West End stage Here's what London is crying out for:
11:09 am21 March 2017
Off-West End musicals diary: The Sorrows of Satan and The Wild Party
The Wild Party at The Other Palace and The Sorrows of Satan at Tristan Bates Theatre Not long
Limehouse asks both 'what if?' and 'why not now?' Urgent questions
Terri Paddock2020-01-08T09:47:34+00:0019 March 2017|Features, Opinion, Personal, Plays, Politics, Terri's Blog|
Paul Chahidi, Roger Allam, Shirley Williams and Tom Goodman-Hill play the "Gang of Four" in Limehouse. © Jack Sain
The 'will of the people'? London has no place in My Country
Terri Paddock2020-01-08T09:47:05+00:0012 March 2017|Features, Opinion, Personal, Plays, Politics, Reviews, Terri's Blog|
My Country; a work in progress I've never voted Conserative. Indeed, the morning after the 2015 General Election,
1:29 am8 March 2017
My theatre diary: Off-West End plays Diary of a Teenage Girl, In Other Words and La Ronde
Terri Paddock2020-03-27T12:49:27+00:008 March 2017|Features, Opinion, Personal, Plays, Terri's Blog|
Clockwise: In Other Words, Diary of a Teenaged Girl and La Ronde Two Off-West End plays have recently
11:59 am22 February 2017
Press Pass: Everything you need to know about Gary Barlow & The Girls
Terri Paddock2020-03-27T12:49:39+00:0022 February 2017|Features, Musicals, News, Opinion, Personal, Reviews, Terri's Blog|
The cast of The Girls includes Joanna Riding, James Gaddas, Claire Moore, Claire Machin, Sophie-Louise Dann and Debbie Chazen
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Carmen Monarcha Soprano, page 2
On November 14th 2014, Carmen Monarcha performed a show in TOM JAZZ (the venue) in Sao Paulo, Brazil. Angela Caracik recorded her songs and Celeste Rabello put them together to one video. Duration: one hour. For the fans, to enjoy! Thanks Angela and Celeste!
On November 26th 2014 Carmen posted on Facebook: How kind of you Angela Caracik! Thank you all, but especially to you, our dear Ineke Cornelissen, Rudolf Schneider and Celeste Rabello, for the tremendous dedication and brilliant portrayal in all your recordings photos and aid in resolving certain issues !! Your sharing of my work, both in and out of Brazil, inspires me and for that you have earned my eternal gratitude and friendship. Thank you; it's wonderful to know that my art is so endearing a part of your lives. Thousand kisses.
Febr. 2014: Rudolfo Schneider from Brazil started a petition to promote a campaign, with the idea that Carmen Monarcha should perform the National Anthem during the opening ceremony of the World Cup (Football/Soccer) 2014 in Brazil. If you share the same idea, you can sign up by clicking on THIS LINK. Fill in your e-mail address and optional why you think it has to be Carmen. Press "Assine" and you have signed the petition.
February 2014. Carmen's new DVD. To our joy we received another DVD, sent to us by Rudolfo Schneider, including a lovely hand written message by Carmen herself. We'll cherish this DVD and Carmen's big kiss. Thank you Carmen and Thank you Rudolfo.
January 2014: New DVD by Carmen Monarcha. The DVD is sold via Carmen's web site: www.carmenmonarcha.com . On the website, first select the language (Brazilian - BR-, English -EN- or Spanish -ES-). The DVD (which includes a CD with the same songs) can be ordered through the 'Store'. Below the trailer of this wonderful DVD. Thanks to Silvana for sending us this DVD.
Before each song on the DVD, Carmen introduces the song in an elegant way. The DVD is recorded in Brazilian language (Portuguese), but contains English and Spanish subtitles. These subtitles cannot be selected from the DVD menu, but you have to use the 'Subtitle' function on the remote control of your DVD player.
Carmen's first solo DVD/CD. Carmen on Facebook: "On December 18th 2013 my first solo CD will be available for download in the iTunes!! After so much work it's finally coming true and I can share this with you all. I hope you like it!!" Note: If you order the DVD above, you also will receive a copy of this DVD. If you just want a copy of this CD, you may also visit the iTunes store
Track list: 1. Over the Rainbow, 2. Fascinacao, 3. Smile, 4. Minha Voz Minha Vida, 5. La vie en Rose, 6. Voz de Mulher, 7. Miss Celie's Blues, 8. Quando m'en Vo, 9. Summertime, 10. Valsinha, 11. Gitana, 12. Coco Penerue, 13. Tamba-Taja, 14. Papa, can you hear me, 15. Ave Maria, 16. Estrada Branca, 17. Moon River, 18. Oh mio babbino caro, 19. Azulao.
Lunch with the Brazilian fan group. On the 11th of January Carmen and Octavio had a lunch with the Brazilian fan group. Thanks to Rudolfo for sending us the photos.
October 2013. New DVD by Carmen Monarcha will be released on November 12th. Thanks to Rudolfo Schneider for drawing our attention to this wonderful video of the release of her DVD.
Click HERE to return to page 1 or HERE to go to page 3


Looking to buy a Chicago area house or condo?
Our property search results are updated from the MLS every 2 hours. Get access to more listings with more data, schools nearby, integrated Walk Scores, Google Maps, Sold Properties, Open Houses and more. We've made it easy for you to search homes for sale by budget, type of property or by neighborhood. See something you like? Contact us with just one click to set up a private appointment.


text:
Game Of Thrones Trivia #21: Did You Know? Daenerys Targaryen aka Emilia Clarke Has Indian Roots
Emilia Clarke is born and brought up in London, The UK, but she is like the one-eighth Indian.
Game Of Thrones Trivia #21: Did You Know? Daenerys Targaryen Aka Emilia Clarke Has Indian Roots ( Photo Credit – Episode Still )
American fantasy drama Game of Thrones has become a global phenomenon. The show has united fans across the world who are eagerly waiting for the prequel series, House of Dragon. While Emilia Clarke, who played the role of Daenerys Targaryen, will be missing from the series but we have an interesting 'family secret' to be revealed.
The 'Mother of Dragon' actress received International acclamation for her role in the hit HBO series. She is also the recipient of several accolades, including four Emmy Award nominations. But not many know that she has Indian roots and we bet many GOT fans will be thrilled to know this.
Emilia Clarke is born and brought up in London, The UK, but she is like the one-eighth Indian. During a conversation with Harpers Bazaar in 2017, the British actress revealed the big family secret. she said, "It's this epic story: My grandmother was colonial Indian, and it was a big old family secret because her mum had an affair with someone in India. She would wear makeup to make her skin look white."
Game Of Thrones Trivia #20: 'Arya' Maisie Williams Once Revealed How The Queen Of England Wasn't A Fan Of The Show
Not just that the Me Before You actress, Emilia Clarke also flew down to India when she was 16-years old with her then-boyfriend to scatter the ashes. She said, "She loved India more than she loved England. Fuck, yeah. I love that part of me—I'm like one-eighth Indian."
Well, that's interesting! Khaleesi is a bit Indian!
Emilia Clarke had also visited India with her Game of Thrones co-star Rose Leslie back in 2019.
HBO's epic fantasy series Game of Thrones, which enjoys a huge fan following across the globe, is based on a book titled A Song Of Ice And Fire, written by George RR Martin. The show gave Maisie Williams, Kit Harington, Sophie Turner, amongst others, enormous popularity.
GoT has eight seasons and the most appealing part of these eight sequences has to be the unpredictability factor in them.
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Feel good Friday: When what you do makes a difference!
We've been supporting local charity Time to Talk Befriending over the winter months to make sure lonely elderly people of the community aren't alone.
Emily Kenward, the founder of TTTB shared what she loves about Sussex with The Argus last month. From this, one of the people she has helped got in touch with them to share how the charity has changed her life.
If this makes you feel as warm and fuzzy as us, find out more about the amazing work they do and how you can volunteer with their befriending service here.


Do you want to bridge the gap between the public and professional world? Public relations is unique and distinguishable from marketing and advertising due to its emphasis on building and maintaining mutually beneficial relationships with an organization's internal and external publics as well as specialization in media relations and journalistic writing. Concord University public relations students have opportunities to build a strong portfolio of work which gives them a substantial edge in increasingly competitive job markets upon graduation. Furthermore, participating in the public relations workshop gives students the chance to practice skills learned in the classroom by allowing them to work with clients within the university and community to promote events through writing press releases, creating posters, working events, and creating other promotional materials.
As one of the fastest growing areas in the field of Communication, public relations offers a wide variety of careers to choose from and areas to work in including nonprofit work, humans relations, corporate communication, government agencies, and public relations firms. Interested in joining? For more information contact Cory Williams.
Want more info? See information from the Communication Arts Emphasis in Public Relations 2018-19 Progression Sheet.


The reasons why we are the denominated people of God are to be repeated and repeated. Deuteronomy 4:1–13; 5:1–33." Manuscript Releases, volume 8, 427.
Listen to God's Denominated People and follow along with this handout.
DENOMINATED: By a most solemn covenant"Thus the children of Israel were denominated as a special people. By a most solemn covenant they were pledged to be true to God. Then the people were bidden to prepare themselves to hear the law. On the morning of the third day the voice of God was heard. Speaking out of the thick darkness that enshrouded Him, as He stood upon the mount, surrounded by a retinue of angels, the Lord made known His law." Fundamentals of Christian Education, 506.


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\section{Introduction}

The famous Banach–Mazur game was proposed in 1935 by Stanislaw Mazur and recorded in the Scottish Book (\cite{Mauldin1981}, Problem 43). Let $(X, \tau)$ be a topological space. The Banach-Mazur game $\mathsf{BM}(X)$ played on $(X, \tau)$ is played between two players, \textsc{Alice} and \textsc{Bob}, who, alternately, select non-empty open subsets of $X$. \textsc{Alice} goes first and chooses a non-empty open subset $A_{0}$ of $X$. \textsc{Bob} must respond by selecting a non-empty open subset $B_{0}\subseteq A_{0}$. Following this, \textsc{Alice} must select another non-empty open subset $A_{1}\subseteq B_{0} \subseteq A_{0}$ and in turn \textsc{Bob} must again respond by selecting a non-empty open subset $B_{1}\subseteq A_{1}\subseteq B_{0} \subseteq A_{0}$. In general, \textsc{Alice} selects any non-empty open subset $A_{n}$ of the last move $B_{n-1}$ of \textsc{Bob} and the latter player answers by choosing a non-empty open subset $B_{n}$ of the set $A_{n}$, just chosen by \textsc{Alice}. Acting in this away, the players \textsc{Alice} and \textsc{Bob} produce a sequence of non-empty open sets
$$  A_{0}\supseteq B_{0} \supseteq A_{1}\supseteq B_{1} \supseteq \cdots \supseteq A_{n} \supseteq B_{n} \supseteq \cdots $$
We shall declare that \textsc{Bob} wins a play of the Banach-Mazur game $\mathsf{BM}(X)$ if $ \bigcap_{n\in \omega} B_{n} = \bigcap_{n\in \omega} A_{n}\not=\emptyset$. Otherwise, \textsc{Alice} is said to be the winner of this play.

Remember that a topological space is a Baire space provided countable intersections of dense open subsets are dense. Baire spaces can be characterized via the Banach–Mazur game. In fact, Oxtoby showed that a topological space $X$ is Baire if and only if \textsc{Alice} does not have a winning strategy in $\mathsf{BM}(X)$ (\cite{Kechris1995}, Theorem 8.11). Products of Baire spaces are not always Baire (\cite{Fleissner1978}, Example 1 and Example 4). A Baire space $X$ is productively Baire if $X\times Y$ is Baire, for each Baire space $Y$. Another application of the Banach-Mazur game to Baire spaces, is the following: if \textsc{Bob} has a winning strategy in $\mathsf{BM}(X)$ then $X$ is productively Baire. The reciprocal of this result does not hold. For example a Bernstein subset of the real line is productively Baire but the Banach-Mazur game is undertermined, that is, neither \textsc{Alice} nor \textsc{Bob} have a winning strategy in the Banach-Mazur game on such a space.

That is why the following question is natural: Is there a game-theoretical characterization for the property of being productively Baire?

In this paper we show some other game-theoretical conditions on the space to be productively Baire. For this we introduce two variations of the Banach-Mazur game.

This paper is organized as follows. In Section 2 we present the game $\mathsf{BM}_\mathrm{fin}$, our first variation of the Banach-Mazur game, where the second player may choose finitely many open sets each inning. With the game $\mathsf{BM}_\mathrm{fin}$ we present a game-theoretical condition on the space to be productively Baire, which is more general than the original one using the Banach-Mazur game.

In Section 3 we present another variation of the Banach-Mazur game, the game $\mathsf{BM}_\mathrm{{\omega}}$. This game is similar to the previous one, with the difference that the second player chooses a countable family of non-empty open sets.
It is worth mentioning that both of these game, as the original Banach-Mazur game, provide a  characterization for the Baire property.

Finally, Section 4 is dedicated to present an example, assuming the Continuum Hypothesis, that shows the games $\mathsf{BM}_\mathrm{fin}$ and $\mathsf{BM}_\mathrm{{\omega}}$ are not equivalent. We also present some open questions.

\section{The game $\mathsf{BM}_\mathrm{fin}(X)$}

In this section we introduce our first variation of the Banach-Mazur game. The difference with the classic Banach-Mazur game is that in this version $\textsc{Bob}$ has the advantage of choosing a finite number of non-empty open sets instead of just one.

\begin{definition}
	Given a topological space $X$, we define the game $\mathsf{BM}_\mathrm{fin}(X)$ played as follows: \textsc{Alice} plays $A_0$ a non-empty open set. Then \textsc{Bob} plays $\mathcal{B}_0$ a finite collection of non-empty open subsets of $A_0$. Then, for each $B \in \mathcal{B}_0$, \textsc{Alice} plays $A_B \subset B$ a non-empty open set. Let $\mathcal{A}_1 = \{A_B: B \in \mathcal{B}_0\}$. Then \textsc{Bob} plays $\mathcal{B}_1$ a finite collection of non-empty open subsets of $\bigcup \mathcal{A}_1$ and so on. For each $n \in {\omega}$, let $B_n = \bigcup \mathcal{B}_n$. \textsc{Bob} is declared the winner if $\bigcap_{n \in {\omega}} B_n \neq \emptyset$. \textsc{Alice} is declared the winner otherwise.
\end{definition}

Sometimes it is easier to see this game with a minor change in the
rules for \textsc{Bob}. In each inning, instead of picking finitely many
non-empty open subsets of  $\bigcup \mathcal{A}_n$, Bob can pick finitely many
open subsets of each open set played by \textsc{Alice} - including picking none
for some of them. These two versions of the rules are easily seen to
be equivalent.

We present the following technical lemma which tells us that we can assume that each move of \textsc{Bob} can be formed by a finite family of non-empty open sets  pairwise disjoint.

\begin{lemma}\label{lema disjoint open}
	Let $X$ be a Hausdorff space without isolated points. We may suppose that each $\mathcal{B}_n$ as in the definition of $\mathsf{BM}_\mathrm{fin}(X)$ is made of pairwise disjoint sets.
\end{lemma}

\begin{proof}
	We will show that if $\sigma$ is a winning strategy for \textsc{Bob} then there exists another winning strategy $\hat{\sigma}$ for \textsc{Bob} such that every move from \textsc{Bob} is 	formed by non-empty pairwise disjoint open sets.

	Indeed, in the first inning, \textsc{Alice} plays $A_{0}$. Next \textsc{Bob} responds $\sigma(\langle A_{0} \rangle)=\mathcal{B}_{0}=\{B^{0}_{0}, \cdots,B^{0}_{n_{0}}\}$ for some $n_{0}\in\omega$. So let $x^{0}_{i}\in B^{0}_{i}$ for each $i\in \{0,\cdots, n_{0}\}$ such that $x^{0}_{i}\not=x^{0}_{j}$ if $i\not=j$. Note that this is possible because $X$ has no isolated points. Now, since $X$ is Hausdorff, there is a pairwise disjoint family of non-empty open sets $\hat{\mathcal{B}_{0}}=\{\hat{B^{0}_{0}}, \cdots, \hat{B^{0}_{n_{0}}} \}$ such that $\hat{B^{0}_{i}} \subseteq B^{0}_{i} $ for each $i\in \{0,\cdots, n_{0}\}$. So define $\hat{\sigma}(\langle A_{0} \rangle)=\hat{\mathcal{B}_{0}}$.

	Next, in the second inning, \textsc{Alice} plays $\mathcal{A}_{1}=\{A_{B}:B\in \hat{\mathcal{B}_{0}}\}$, then \textsc{Bob} responds $\sigma(\langle A_{0}, \mathcal{A}_{1}\rangle)=\mathcal{B}_{1}=\bigcup_{A\in \mathcal{A}_{1}} \mathcal{F}^{1}_{A}$, where each $\mathcal{F}^{1}_{A}$ is a finite family of non-empty open subsets of $A\in\mathcal{A}_{1}$. Now for each $\mathcal{F}^{1}_{A}=\{B^{1}_{0}, \cdots, B^{1}_{n_{1}}\}$, we apply the above argument, that is, there is a family $\hat{\mathcal{F}^{1}_{A}} = \{ \hat{B^{1}_{0}}, \cdots, \hat{B^{1}_{n_{1}}}\}$ of pairwise disjoint non-empty open sets such that $\hat{B^{1}_{i}}\subseteq B^{1}_{i}$ for each $i\in\{0, \cdots, n_{1}\}$, so define $\hat{\sigma}(\langle A_{0}, \mathcal{A}_{1} \rangle)= \bigcup_{A\in\mathcal{A}_{1}} \hat{\mathcal{F}^{1}_{A}}$ and so on.

	As $\sigma$ is a winning strategy then there exists $x\in \bigcap_{n \in {\omega}} \bigcup\sigma(\langle A_{0}, \mathcal{A}_{1}, \cdots, \mathcal{A}_{n} \rangle)=\bigcap_{n \in {\omega}} \bigcup\mathcal{A}_{n}=\bigcap_{n \in {\omega}} \bigcup\hat{\sigma}(\langle A_{0}, \mathcal{A}_{1}, \cdots, \mathcal{A}_{n} \rangle)$. Therefore $\hat{\sigma}$ is a winning strategy for \textsc{Bob} in $\mathsf{BM}_\mathrm{fin}(X)$.
\end{proof}

Now we can generalize Oxtoby's theorem.

\begin{proposition}
  Let $X$ be a Hausdorff space without isolated points. If \textsc{Bob} has a winning strategy on $\mathsf{BM}_\mathrm{fin}(X)$, then $X$ is productively Baire.
\end{proposition}

\begin{proof}
	Let $\sigma$ be a winning strategy for \textsc{Bob} in the $\mathsf{BM}_\mathrm{fin}(X)$ game. Suppose that $X$ is not productively Baire, that is, there exists a Baire space, call it $Y$, such that $X\times Y$ is not a Baire space.

	Suppose that $\rho$ is a winning strategy for \textsc{Alice} in the $\mathsf{BM}(X \times Y)$ game. We will define a winning strategy $f$ for \textsc{Alice} in the $\mathsf{BM}(Y)$ game. Note that this is enough. Also, notice that we can suppose that $\rho$ always answer with basic open sets in the product.

	In the first inning, in $\mathsf{BM}(X\times Y)$, \textsc{Alice} plays $A_0 \times B_0 = \rho(\seq{})$. Then, in $\mathsf{BM}(Y)$, \textsc{Alice} plays $f(\seq{}) = B_0$. Next \textsc{Bob} responds $W_0 \subset B_0$ a non-empty open set. We now will proceed to define $f(\seq{W_0})$. In $\mathsf{BM}_\mathrm{fin}(X)$, \textsc{Alice} plays $A_{0}$ and \textsc{Bob} responds $\{V_0, ..., V_n\} = \sigma(\seq{A_0})$. By Lemma \ref{lema disjoint open}, we may suppose that $V_i \cap V_j = \emptyset$ if $i \neq j$.

	In the second inning, in $\mathsf{BM}(X\times Y)$, \textsc{Alice} could play in the following $n + 1$ ways:

	\[A_1^0 \times C_0 = \rho(\seq{V_0 \times W_0})\]
	\[A_1^1 \times C_1 = \rho(\seq{V_1 \times C_0})\]
	\[\vdots\]
	\[A_1^n \times C_n = \rho(\seq{V_n \times C_{n - 1}})\]

	Note this is valid since $C_0 \subset W_0 \subset B_0$ and each $C_k \subset C_{k - 1} \subset B_0$ for $0<k\leq n$, also $A^{k}_{1}\subset V_{k}$ for each $0\leq k \leq n$. Let $B_1 = C_{n}$. Finally, in $\mathsf{BM}(Y)$, \textsc{Alice} plays $f(\seq{W_0}) = B_1$. Next \textsc{Bob} responds $W_1 \subset B_1$ a non-empty open set and let us define $f(\seq{W_0, W_1})$. Now, in $\mathsf{BM}_\mathrm{fin}(X)$, \textsc{Alice} plays $\{A^{0}_{1}, \cdots, A^{n}_{1}\}$ next \textsc{Bob} responds
	\[\{U_0, ..., U_k\} = \sigma(\seq{A_0, \{A_1^0, ..., A_1^n\}}).\]

	By Lemma \ref{lema disjoint open} we may suppose that $U_0, ..., U_k$ and $A_1^0, ..., A_1^n$ are pairwise disjoint sets. Then for each $i = 0, ..., k$, there is only one $g(i)\in \{0, \cdots, n  \}$ such that \[U_i \subset A_1^{g(i)}.\]

	In the third inning, in $\mathsf{BM}(X\times Y)$, \textsc{Alice} could play in the following $k + 1$ ways:

	\[A_2^0 \times D_0 = \rho(\seq{V_{g(0)} \times C_{g(0) - 1}, U_0 \times W_1})\]
	\[A_2^1 \times D_1 = \rho(\seq{V_{g(1)} \times C_{g(1) - 1}, U_1 \times D_0})\]
	\[\vdots\]
	\[A_2^k \times D_k = \rho(\seq{V_{g(k)} \times C_{g(k) - 1}, U_k \times D_{k - 1}}).\]

	On the previous equations, use $C_{-1} = W_0$ if neccessary. As before, $D_i \subset D_{i -1} \subset W_1$. Let $B_{2}=D_{k}$ and, finally, in $\mathsf{BM}(Y)$, \textsc{Alice} plays $f(\seq{W_0, W_1})=D_k$. Next in $\mathsf{BM} (Y)$, \textsc{Bob} responds $W_2 \subset B_2$ a non-empty open set. Now, in $\mathsf{BM}_\mathrm{fin}(X)$, \textsc{Alice} can play $\{A^{0}_{2}, \cdots, A^{k}_{2}\}$ next \textsc{Bob} responds $\sigma(\langle A_{0},\{A_1^0, ..., A_1^n\} , \{A^{0}_{2}, \cdots, A^{k}_{2}\}\rangle)$. The remaining construction of $f$ is in similar fashion.

	Let us prove now that $f$ is winning for \textsc{Alice} in $\mathsf{BM}(Y)$. So fix a play $$\seq{B_0, W_0, B_1, W_1, ..., B_n, W_n, ...}$$ following $f$ (with notation similar to what we presented in the construction of $f$). We need to show that $\bigcap_{n \in {\omega}} B_n = \emptyset$.  Suppose not and let $y \in \bigcap_{n \in {\omega}} B_n$.

	Let $\mathcal{A}_n = \{A_n^0, ..., A_n^{k_n}\}$. Since $\sigma$ is winning for \textsc{Bob} in $\mathsf{BM}_\mathrm{fin}(X)$, there is an
	\[x \in \bigcap_{n \in {\omega}} \bigcup \mathcal{A}_n.\]
	Since we are supposing that each $\mathcal{A}_n$ is made by mutually disjoint open sets, there is only one $A_n$ in each $\mathcal{A}_n$ such that $x \in A_n$.

	In particular, in $\mathsf{BM}(X\times Y)$, in the first inning, $\langle x, y\rangle\in A_{0}\times B_{0}=\rho(\seq{})$. In the second inning, there is only one $A^{n_{0}}_{1}\in \mathcal{A}_{1}$ such that $x\in A^{n_{0}}_{1}$. Then $\langle x, y\rangle\in A^{n_{0}}_{1}\times C_{n_{0}}=\rho(\langle V_{n_{0}}\times C_{n_{0}-1}  \rangle)$. In the third inning, as $x\in \bigcup\mathcal{A}_{2}$, there is only one $A^{n_{1}}_{2}\in\mathcal{A}_{2}$ such that $x\in A^{n_{1}}_{2}$. Then $\langle x, y\rangle\in A^{n_{1}}_{2}\times D_{n_{1}}=\rho(\langle
	V_{n_{0}}\times C_{n_{0}-1}, U_{n_{1}}\times D_{n_{1}-1}  \rangle)$. Procceding like this, we can find a play of $\mathsf{BM}(X \times Y)$ that has $\seq{x, y}$ in all its movements but it is compatible with $\rho$, which is a contradiction.

\end{proof}

We will show in Section 3 that the game $\mathsf{BM}_\mathrm{fin}$ is equivalent to the Banach-Mazur game in the point of view of \textsc{Alice} (i.e she has a winning strategy in one of the games if, and only if, she has one in the other). But this is not the case for \textsc{Bob}. We will end this section showing that a Bernstein set is a witness for this.

Remember that a subset $X\subseteq \mathbb{R}$ is a Bernstein set if both $X$ and $\mathbb{R}\setminus X$ meets every uncountable closed subset of the real line. A Bernstein set has no isolated points and is dense in $\mathbb{R}$. To see that \textsc{Bob} has no winning strategy for the $\mathsf{BM}$ game in a Bernstein set just recall the following:

\begin{proposition}
	If $X \subset \mathbb{R}$ has no isolated points and \textsc{Bob} has a winning strategy on $\mathsf{BM}(X)$, then $X$ contains a Cantor set.
\end{proposition}

The previous result shows that \textsc{Bob} does not have a winning strategy for the $\mathsf{BM}(X)$ game played on a Bernstein set.

On the other hand, \textsc{Bob} has a winning strategy for $\mathsf{BM}_\mathrm{fin}$ in a Bernstein set

\begin{proposition}\label{2bmfinberstein}
  Suppose that $X \subset \mathbb{R}$ is a Bernstein set. Then \textsc{Bob} has a winning strategy on $\mathsf{BM}_\mathrm{fin}(X)$.
\end{proposition}

\begin{proof}
  For every open set $V \subset X$, let $V^*$ be an open set in $\mathbb{R}$ such that $V = V^* \cap X$. On each inning, \textsc{Bob} will split every open set played by \textsc{Alice} in two pieces, therefore, we will enumerate all the open sets played with sequences in $2^{<{\omega}}$. On the inning of number $n$, let
  \[\{A_s: s \in 2^n\}\]
  be the open sets played by \textsc{Alice}. For each $s \in 2^n$, let $B_{s \smallfrown 0}$ and $B_{s \smallfrown 1}$ be two open sets such that:
  \begin{itemize}
  \item $\emptyset \neq B_{s \smallfrown i} \subset \overline{B_{s \smallfrown i}^*}^\mathbb{R} \subset A_s^*$;
  \item $\overline{B_{s \smallfrown 0}^*}^\mathbb{R} \cap \overline{B_{s \smallfrown 1}^*}^\mathbb{R} = \emptyset$;
  \item the diameter of $B_{s \smallfrown i}$ is less than $\frac{1}{n + 1}$.
  \end{itemize}

  Then, in the next inning, we define $A_s$ as the open set given by \textsc{Alice} such that $A_s \subset B_s$.

  For each $n > 0$, let $B_n = \bigcup\{B_s: l(s) = n\}$. Note that, by the way we defined it, $C = \bigcap_{n \in {\omega}} \overline{B_n^*}$ is a Cantor set. Therefore, $C \cap X \neq \emptyset$ and thus \textsc{Bob} wins the game.
\end{proof}

The two previous results show that $\mathsf{BM}$ and $\mathsf{BM}_\mathrm{fin}$ are not equivalent for \textsc{Bob}.

\section{The game $\mathsf{BM}_\mathrm{{\omega}}(X)$}

In this part we present the game $\mathsf{BM}_\mathrm{{\omega}}$. The difference from the game $\mathsf{BM}_\mathrm{fin}$ is basically that in each inning \textsc{Bob} plays a countable number of non-empty open subsets contained in the union of the last move of \textsc{Alice}. Instead of just finitely many as the previous game.

\begin{definition}
  Given a topological space $X$, we define the game $\mathsf{BM}_\mathrm{{\omega}}(X)$ played as follows: \textsc{Alice} plays $A_0$ a non-empty open set. Then \textsc{Bob} plays $\mathcal{B}_0$ a countable collection of non-empty open subsets of $A_0$. Then, for each $B \in \mathcal{B}_0$, \textsc{Alice} plays $A_B \subset B$ a non-empty open set. Let $\mathcal{A}_1 = \{A_B: B \in \mathcal{B}_0\}$. Then \textsc{Bob} plays $\mathcal{B}_1$ a countable collection of non-empty open subsets of $\bigcup \mathcal{A}_1$ and so on. For each $n \in {\omega}$, let $B_n = \bigcup \mathcal{B}_n$. \textsc{Bob} is declared the winner if $\bigcap_{n \in {\omega}} B_n \neq \emptyset$. \textsc{Alice} is declared the winner otherwise.
\end{definition}

The following results show that the games $\mathsf{BM}, \mathsf{BM}_\mathrm{fin}$  and $\mathsf{BM}_\mathrm{{\omega}}$ are all equivalent for \textsc{Alice}.

\begin{proposition}\label{n1B}
If \textsc{Alice} does not have a winning strategy on $\mathsf{BM}_\mathrm{{\omega}}(X)$, then $X$ is a Baire space.
\end{proposition}

\begin{proof}
We will show that if $X$ is not a Baire space then \textsc{Alice} has a winning strategy $\delta$ on $\mathsf{BM}_\mathrm{{\omega}}(X)$.

Since, $X$ is not Baire, there are a sequence $\langle D_{n} : n\in\omega\rangle$ of open dense subsets of $X$ and a non-empty open subset $A$ of $X$ such that $\bigcap_{n \in {\omega}}D_{n}\cap A=\emptyset$. Then, in the first inning, \textcolor{black}{\textsc{Alice} plays $\delta(\seq{})=A_{0}=A$} and \textsc{Bob} responds $\mathcal{B}_{0}=\{B^{0}_{n} : n\in\omega  \}$, a countable collection of non-empty open subsets of $A_{0}$. In the second inning, \textcolor{black}{\textsc{Alice} plays $\delta(\langle \mathcal{B}_{0} \rangle)= \mathcal{A}_{1}=\{D_{0}\cap B^{0}_{n} :n\in\omega\}$}. Note that this move is valid because $D_{0}$ is open dense. Next \textsc{Bob} responds $\mathcal{B}_{1}=\{B^{1}_{n}:n\in\omega\}$ with $B^{1}_{n}\subseteq \bigcup\mathcal{A}_{1}$ for each $n\in\omega$. In the third inning, as $D_{1}$ is dense, \textcolor{black}{\textsc{Alice} can play $\delta(\langle \mathcal{B}_{0}, \mathcal{B}_{1} \rangle)= \mathcal{A}_{2}=\{D_{1}\cap B^{1}_{n} :n\in\omega\}$}. Next \textsc{Bob} responds $\mathcal{B}_{2}=\{B^{2}_{n}:n\in\omega\}$ with $B^{2}_{n}\subseteq \bigcup\mathcal{A}_{2}$ for each $n\in\omega$, and so on. The remaining construction of $\delta$ is in similar fashion.

Let us prove now that $\delta$ is winning for \textsc{Alice} in $\mathsf{BM}_\mathrm{{\omega}}(X)$. So fix a play $$\seq{A_0, \mathcal{B}_0, \mathcal{A}_1, \mathcal{B}_1, ..., \mathcal{A}_n, \mathcal{B}_n, ...}$$ following $\delta$ (with notation similar to what we presented in the construction of $\delta$). We need to show that $\bigcap_{n \in {\omega}} B_n = \emptyset$, where $B_n=\bigcup{\mathcal{B}}_n$ for every $n\in\omega$. In fact, note that $B_{0}\subseteq A$ and for each $n>0$ we have $B_{n}\subseteq D_{n-1}$. Then $\bigcap_{n \in {\omega}}B_{n}\subseteq \bigcap_{n \in {\omega}}D_{n}\cap A=\emptyset$. Therefore $\delta$ is winning for \textsc{Alice}.

\end{proof}

\begin{corollary}
	Let $X$ be a non-empty topological space. Then the following are equivalent:
	\begin{enumerate}
		\item $X$ is Baire

		\item \textsc{Alice} does not have a winning strategy on $\mathsf{BM}(X)$

		\item \textsc{Alice} does not have a winning strategy on $\mathsf{BM}_\mathrm{fin}(X)$

		\item \textsc{Alice} does not have a winning strategy on $\mathsf{BM}_\mathrm{{\omega}}(X)$

    \end{enumerate}
\end{corollary}

\begin{proof}
The implication $(1)\Rightarrow (2)$ is the classic Oxtoby's result (\cite{Kechris1995}, Theorem 8.11). Note that if $\textsc{Alice}$ has a winning strategy on $\mathsf{BM}_\mathrm{fin}(X)$	then $\textsc{Alice}$ has a winning strategy on $\mathsf{BM}(X)$, because the game $\mathsf{BM}_\mathrm{fin}$ is more difficult for \textsc{Alice}. This proves $(2)\Rightarrow (3)$.
Also, if $\textsc{Alice}$ has a winning strategy on $\mathsf{BM}_\omega(X)$, then $\textsc{Alice}$ has a winning strategy on $\mathsf{BM}_\mathrm{fin}(X)$, because the game $\mathsf{BM}_\omega$ is more difficult for \textsc{Alice}. This proves $(3)\Rightarrow (4)$. Finally, by Proposition \ref{n1B}, we have $(4)\Rightarrow (1)$.

\end{proof}

We end this section showing that $\mathsf{BM}_\omega$ is determined for a broad class of spaces.

\begin{definition}
A family $\mathcal{B}$ of non-empty open sets in a topological space will be called a \textbf{$\pi$-base} if every non-empty open set contains at least one member of $\mathcal{B}$. A $\pi$-base $\mathcal{B}$ is called \textbf{locally countable} if each member of $\mathcal{B}$ contains countably many members of $\mathcal{B}$.
\end{definition}

\begin{proposition}[\cite{Oxtoby1961}]
	The cartesian product of a Baire space $X$ and a Baire space $Y$ having countable $\pi$-base is a Baire space.
\end{proposition}

\begin{corollary}\label{cpbpb}
	If $X$ is a second countable Baire space then $X$ is productively Baire.
\end{corollary}

\begin{remark}\label{jogopibase}
Let $\mathcal{B}$ be a $\pi$-base for the topology of the space $X$. Note that, in the $\mathsf{BM}_\mathrm{{\omega}}$ game on $X$, we can assume that, both \textsc{Alice} and \textsc{Bob} must necessarily choose elements from $\mathcal{B}$ in their moves. This is also valid if we change $\mathsf{BM}_\mathrm{{\omega}}$ by $\mathsf{BM}$. This fact will be used freely from now on without mentioning it.
\end{remark}

\begin{theorem}
  If $X$ is a Baire space with a \textcolor{black}{locally} countable $\pi$-base, then \textsc{Bob} has a winning strategy on $\mathsf{BM}_\mathrm{{\omega}}(X)$.
\end{theorem}

\begin{proof}
  First, we will define a strategy $\sigma$ for \textsc{Bob} in the $\mathsf{BM}_\mathrm{{\omega}}(X)$ game using the fact that $X$ has a \textcolor{black}{locally} countable $\pi$-base $\mathcal{B}$.

 Indeed, in the first inning, \textcolor{black}{\textsc{Alice} plays $A_0$} be any element from $\mathcal{B}$ and let $\mathcal{A}_0 = \{A_0\}$ next \textcolor{black}{\textsc{Bob} responds $\sigma(\langle A_{0}  \rangle) = \{B \in \mathcal{B}: B \subset A_0\}$}, note that this is a valid move since $\mathcal{B}$ is locally countable. In the second inning, \textcolor{black}{\textsc{Alice} plays $\mathcal{A}_{1}=\{A_{B}\in\mathcal{B} : B\in \sigma(\langle A_{0}  \rangle) \}$} and \textcolor{black}{\textsc{Bob} responds $\sigma(\langle A_{0}, \mathcal{A}_{1} \rangle) = \bigcup_{A\in \mathcal{A}_{1}}\{B \in \mathcal{B}: B \subset A   \}$}, note that this is a valid move since $\mathcal{A}_{1}$ is countable and $\mathcal{B}$ is locally countable.

 Then, if $\seq{\mathcal{A}_n: n \leq k}$ are the plays of \textsc{Alice}, define \textcolor{black}{$\sigma(\seq{\mathcal{A}_0, ..., \mathcal{A}_n}) = \bigcup_{A\in\mathcal{A}_{n}}\{B \in \mathcal{B}: B\subset A \}$}. This completes the definition of the strategy $\sigma$.

 \begin{center}
 	\begin{table}
 		\centering
 			\begin{tabular}{c|c}
 				\multicolumn{2}{c}{$\mathsf{BM}_\omega(X)$ }\\
 				\hline
 				\hline
 				\multicolumn{2}{c}{\textcolor{white}{$GS_{\lambda}(A,k)$}} \\
 				$\textsc{Alice}$ & $\textsc{Bob}$ \\
 				\hline
 				$\textcolor{black}{A_{0}}$ & \\
 				&  $\textcolor{black}{\sigma(\langle A_{0}  \rangle)=\{B \in \mathcal{B}: B \subset A_0\}} $\\
 				$\textcolor{black}{\mathcal{A}_{1}}$ & \\
 				& $\textcolor{black}{\sigma(\langle A_{0}, \mathcal{A}_{1} \rangle) = \bigcup_{A\in \mathcal{A}_{1}}\{B \in \mathcal{B}: B \subset A   \}   \}}$ \\
 				\textcolor{white}{a}\\
 				\vdots & \vdots
 			\end{tabular}
 	\end{table}
 \end{center}
 Assume by way of contradiction that $\sigma$ is not winning, then there is a sequence $\seq{\mathcal{A}_n: n \in {\omega}}$ of plays for \textsc{Alice} that played against $\sigma$ will turn in a win for \textsc{Alice} - \emph{i.e.} $\bigcap_{n \in {\omega}} \bigcup \mathcal{A}_n = \emptyset$.

 Now we will define a winner strategy $\rho$ for \textsc{Alice} in the $\mathsf{BM}(X)$ game.

 Indeed, in the first inning, \textcolor{black}{\textsc{Alice} plays $\rho(\seq{}) = A_0$}  next \textcolor{black}{\textsc{Bob} responds $B_0 \subset A_0$} be any element from $\mathcal{B}$, note that $B_0\in\sigma(\seq{\mathcal{A}_{0}})$. In the second inning, in $\mathsf{BM}_\mathrm{{\omega}}(X)$, by definition, there is an $A_1 \in \mathcal{A}_1$ such that $A_1 \subset B_{0}$. Then, in $\mathsf{BM}(X)$, \textcolor{black}{\textsc{Alice} plays $\rho(\seq{B_{0}}) = A_1$} next \textcolor{black}{\textsc{Bob} plays $B_{1}$} be any element from $\mathcal{B}$ contained in $A_1$, note that $B_{1}\in \sigma(\seq{A_{0},\mathcal{A}_{1}})$, and so on. This completes the definition of the strategy $\rho$.

\begin{center}
	\begin{table}[h!]
		\centering
		\begin{tabular}{c|c}
			\multicolumn{2}{c}{$\mathsf{BM}(X)$ }\\
			\hline
			\hline
			\multicolumn{2}{c}{\textcolor{white}{$GS_{\lambda}(A,k)$}} \\
			$\textsc{Alice}$ & $\textsc{Bob}$ \\
			\hline
			$\textcolor{black}{\rho(\seq{})  =A_{0}}$ & \\
			&  $\textcolor{black}{B_{0}} $\\
			$\textcolor{black}{\rho(\seq{B_{0}})=A_{1}}$ & \\
			& $\textcolor{black}{B_{1}}$ \\
			\textcolor{white}{a}\\
			\vdots & \vdots
		\end{tabular}
	\end{table}
\end{center}

Let us prove now that $\rho$ is winning for \textsc{Alice} in $\mathsf{BM}(X)$. So fix a play $$\seq{A_0, B_0, A_1, B_1, ..., A_n, B_n, ...}$$ following $\rho$ (with notation similar to what we presented in the construction of $\rho$). We need to show that $\bigcap_{n \in {\omega}} B_n = \emptyset$. In fact, note that for each $n\in\omega$, $B_{n}\subset \bigcup\mathcal{A}_{n}$, then $\bigcap_{n\in\omega}B_{n}\subset\bigcap_{n \in {\omega}}\bigcup\mathcal{A}_{n}=\emptyset$. Thus, \textsc{Alice} wins the game.
\end{proof}

\begin{corollary}
	For a space $X$ with locally countable $\pi$-base, the game $\mathsf{BM}_\mathrm{{\omega}}(X)$ is determined.
\end{corollary}

\section{A relation between $\mathsf{BM}_\mathrm{fin}$ and $\mathsf{BM}_\mathrm{{\omega}}$.}

In this part we will see that, assuming the Continuum Hypothesis, there is a space in which the games $\mathsf{BM}_\mathrm{fin}$ and $\mathsf{BM}_\mathrm{{\omega}}$ are different. We begin with a compilation of some known facts about subsets of the real line and we present some technical lemmas that will help us to conclude our objective.

\begin{lemma}
  Let $\mathcal{B}$ be an infinite countable base for the topology on $\mathbb{R}$. Then the set of strategies for \textsc{Bob} in the game $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$ has cardinality at most $\mathfrak c$.
\end{lemma}

\begin{proof}
Remember that a strategy for \textsc{Bob} on $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$, where both players only play open sets from $\mathcal{B}$, is a function $$\sigma:S\subseteq \mathcal{B}^{<\omega}\to \mathcal{B}^{<\omega}$$
As $|\mathcal{B}^{<\omega}|=|\mathcal{B}|=\omega$, we have that $|S|\leq \omega$, on the other hand, note that $ \{ \seq{B} : B\in \mathcal{B}  \} \subseteq S$, so $|S|=\omega$. Finally,  $|\{ \sigma \in \mathcal{B}^{S}: \sigma\hspace{0.1cm}\mbox{is strategy for}\hspace{0.1cm}\textsc{Bob}  \}|\leq |\mathcal{B}|^{|S|}=\omega^{\omega}=\mathfrak{c}$.

\end{proof}

\begin{definition}
Let $A\subseteq \mathbb{R}$, we say that $A$ is
\begin{itemize}
	\item \textbf{nowhere dense}, if $\text{int}\hspace{0.05cm}(\overline{A})=\emptyset$.
	\item \textbf{meager}, if $A$ is a countable union of nowhere dense sets.
	\item \textbf{residual}, if A is the complement of a meager set.

\end{itemize}

\end{definition}

\begin{theorem}[Oxtoby]
Let $A\subseteq \mathbb{R}$, then $A$ is residual if and only if $A$ contains a dense $G_{\delta}$ subset of $\mathbb{R}$.
\end{theorem}

Remember the following cardinalities of the following families of subsets of the real line:

\begin{itemize}

	\item  $|  \{M\subseteq \mathbb{R} : M \hspace{0.1cm}\text{is meager}\}|=2^{\mathfrak{c}}$.

	\item $\mathcal{G}_{\delta}=\{G\subseteq \mathbb{R} : G \hspace{0.1cm}\text{is a}\hspace{0.1cm} G_{\delta}\text{-set} \}$, then $|\mathcal{G}_{\delta}|=\mathfrak{c}$.
	\item $\mathcal{F}_{\sigma}=\{F\subseteq \mathbb{R} : F \hspace{0.1cm}\text{is a}\hspace{0.1cm} F_{\sigma}\text{-set} \}$, then $|\mathcal{F}_{\delta}|=\mathfrak{c}$.
	\item $\mathcal{D}\mathcal{G}_{\delta}=\{D\subseteq \mathbb{R} : D \hspace{0.1cm}\text{is a dense}\hspace{0.1cm} G_{\delta}\text{-set} \}$, then $|\mathcal{D}\mathcal{G}_{\delta}|=\mathfrak{c}$, because $\{\mathbb{R}\setminus\{x\}:x\in\mathbb{R}   \}\subseteq \mathcal{D}\mathcal{G}_{\delta}$.
    \item Let $\mathcal{B}=\{B_{n} :n\in\omega  \}$ be a basis for the topology on $\mathbb{R}$, then the set $\{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \}$ has cardinality $\mathfrak{c}$, because for each $n\in \omega$, $\{B_{n}\setminus\{b\} : b\in B_{n}\} \subseteq \{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \}   $.
\end{itemize}

\begin{lemma}\label{easytec}
	Let $\mathcal{B}=\{B_{n} :n\in\omega  \}$ be a basis for the topology on $\mathbb{R}$. For each $G\in \mathcal{D}\mathcal{G}_{\delta}$ and $m\in\omega$, we have $G\cap B_{m} \in \{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \}$.
\end{lemma}
\begin{proof}
	Note that $G\cap B_{m}  \in \mathcal{G}_{\delta}$ and $G\cap B_{m}\subseteq B_{m}$ . Also, $B_{m}\subseteq \overline{G\cap B_{m}}$. In fact, otherwise, there is a $z\in B_{m}$ such that $z\not\in \overline{G\cap B_{m}}$, so there is an $n\in \omega$ with $z\in B_{n}$ such that   $B_{n}\cap B_{m}\cap G=\emptyset$, contradiction, because $G$ is dense.

\end{proof}

\begin{proposition}\label{hrusak}
The following properties of a topological space $X$ are equivalent:

\begin{enumerate}
	\item [(1)] $X$ is a Baire space.
 	\item [(2)] For every countable closed cover $\{H_{n} : n\in\omega   \}$ of $X$, the set $\bigcup_{n\in\omega}\text{int}(H_{n}) $ is dense in $X$.
	\item [(3)] For every sequence $V_{0}, V_{1}, \cdots$ of open sets with the same closure $K$, we have $K=\overline{\bigcap_{n\in\omega}V_{n}}$.
	\item [(4)] Every meager $G_{\delta}$-set in $X$ is nowhere dense.
	\item [(5)] Every meager set has empty interior.

\end{enumerate}
\end{proposition}

Also, every topological space which has a dense Baire subspace is evidently a Baire
space. The converse is not true: for instance, the real line is a Baire space but the
subspace of rationals is not.

\begin{lemma}\label{densoBairechar}
	Let $X\subseteq\mathbb{R}$ be dense, then
	$X$ is Baire if and only if $G\cap X$ is dense in $X$, for each $G\in\mathcal{D}\mathcal{G}_{\delta}$.
\end{lemma}

\begin{proof}
Suppose that $X$ is Baire and let $G=\bigcap_{n \in \omega}G_{n}$ be a dense $G_{\delta}$-set in $\mathbb{R}$. Then for each $n\in\omega$, $G_{n}$ is an open dense set in $\mathbb{R}$, therefore $G_{n}\cap X$ is an open dense set in $X$. As $X$ is Baire. Then $\bigcap_{n \in \omega}G_{n}\cap X=G\cap X$ is dense in $X$.

Now, let $\langle A_{n}:n\in\omega\rangle$ be a sequence of open dense sets in $X$. Then $A_{n}=A^{\prime}_{n}\cap X	$ with $A^{\prime}_{n}$ a non-empty open set in $\mathbb{R}$ and, as $A_{n}$ is dense in $X$, $A^{\prime}_{n}$ is open and dense in $\mathbb{R}$. Therefore $\bigcap_{n \in \omega} A^{\prime}_{n}$ is a dense $G_{\delta}$-set in $\mathbb{R}$, then by hypothesis, $\bigcap_{n \in \omega}A_{n}=\bigcap_{n \in \omega}A^{\prime}_{n}\cap X$ is dense in $X$.

\end{proof}

\begin{definition}
	Given $\sigma$ a strategy for \textsc{Bob} in the $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$ and $s = \seq{\mathcal{A}_n: n \in {\omega}}$ a sequence of finite collections of open sets, we define $\sigma * s$ as the set
	\[\bigcap_{n \in {\omega}} \bigcup \sigma(\seq{\mathcal{A}_0, ..., \mathcal{A}_n}).\]

\end{definition}

We are assuming that if we use this notation, then $\sigma$ and the sequence are compatible, in the sense that the previous intersection is well defined.

\begin{lemma}\label{lematec}
		Let $\sigma$ be a strategy for $\textsc{Bob}$ in the $\mathsf{BM}_\mathrm{fin}({\mathbb R})$, $X$ be a countable set of points in $\mathbb{R}$ and $\mathcal{Y}$ be a countable family of nowhere dense sets in $\mathbb{R}$. Then there exists a sequence $s$ of finite collections of basic open sets (which will be $\textsc{Alice}$'s choices) such that $\sigma*s$ is a closed nowhere dense set disjoint from $X\cup \bigcup \mathcal{Y}$.
\end{lemma}

\begin{proof}
Let $X=\{x_{n}:n\in\omega \}$, $\mathcal{Y}=\{Y_{n}:n\in\omega   \}$ and fix $\mathcal{B}=\{B_{n}: n\in\omega   \}$ be a basis for $\mathbb{R}$.

Let $x_{0}\in X$, and note that $(\mathbb{R}\setminus\{ x_{0} \}) \cap (\mathbb{R}\setminus \overline{Y_{0}})\cap B_{0}$ is a non-empty open set. So consider $z_{0}\in (\mathbb{R}\setminus\{ x_{0} \}) \cap (\mathbb{R}\setminus \overline{Y_{0}})\cap B_{0}$ and note that $(\mathbb{R}\setminus\{ z_{0} \})\cap (\mathbb{R}\setminus\{ x_{0} \}) \cap (\mathbb{R}\setminus \overline{Y_{0}})\cap B_{0}$ is a non-empty open set. Then choose $\mathcal{A}_0=\{A_0\}$ such that $\overline{A_{0}}\subseteq  (\mathbb{R}\setminus\{ z_{0} \})\cap (\mathbb{R}\setminus\{ x_{0} \}) \cap (\mathbb{R}\setminus \overline{Y_{0}})\cap B_{0}$ and note that $x_{0}, z_{0}\not\in A_{0}$.

Now, in the game $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$, we have that in the first inning $\textsc{Alice}$ can play $A_{0}$. Note that $B_{0}\not\subseteq A_{0}$ and $(\{x_{0}\}\cup Y_{0}) \cap A_{0}=\emptyset$. Next $\textsc{Bob}$ responds $\sigma(\langle A_{0}  \rangle)=\{B^{0}_{0}, \cdots, B^{0}_{n_{0}}\}$.

Let $x_{1}\in X$ and $j\in\{0,\cdots, n_{0} \}$, consider $B^{0}_{j}$ and note that $(\mathbb{R}\setminus \{x_{0}  \})\cap (\mathbb{R}\setminus\overline{Y_{0}} ) \cap (\mathbb{R}\setminus \{x_{1}  \}) \cap  (\mathbb{R}\setminus\overline{Y_{1}} ) \cap B_{1}$ is a non-empty open set. So let $z_{1} \in (\mathbb{R}\setminus \{x_{0}  \})\cap (\mathbb{R}\setminus\overline{Y_{0}} ) \cap (\mathbb{R}\setminus \{x_{1}  \}) \cap  (\mathbb{R}\setminus\overline{Y_{1}} ) \cap B_{1}$ and note that $(\mathbb{R}\setminus \{z_{1} \}) \cap (\mathbb{R}\setminus \{x_{0}  \})\cap (\mathbb{R}\setminus\overline{Y_{0}} ) \cap (\mathbb{R}\setminus \{x_{1}  \}) \cap  (\mathbb{R}\setminus\overline{Y_{1}} ) \cap B^{0}_{j}$ is a non-empty open set. Then choose $A^{1}_{j}\in\mathcal{B}$
such that $\overline{A^{1}_{j}}  \subseteq (\mathbb{R}\setminus \{z_{1} \}) \cap (\mathbb{R}\setminus \{x_{0}  \})\cap (\mathbb{R}\setminus\overline{Y_{0}} ) \cap (\mathbb{R}\setminus \{x_{1}  \}) \cap  (\mathbb{R}\setminus\overline{Y_{1}} ) \cap B^{0}_{j}$ and note that for each $j\in\{0, \cdots, n_{0} \} $, $z_{1}\not\in A^{1}_{j}$.

Then in the second inning, $\textsc{Alice}$ plays $\mathcal{A}_{1}=\{A^{1}_{0}, \cdots, A^{1}_{n_{0}}\}$ and note that $B_{1}\not\subseteq \bigcup \mathcal{A}_{1}$ and $(\{x_{0}, x_{1}\} \cup Y_{0}\cup Y_{1}) \cap \bigcup \mathcal{A}_{1}=\emptyset$. Next $\textsc{Bob}$ responds $\sigma(\langle A_{0}, \mathcal{A}_{1}\rangle) = \{ B^{1}_{0}, \cdots, B^{1}_{n_{1}}\} $.

In general, for each inning $m\in\omega$, suppose $\textsc{Bob}$ played $\{B^{m-2}_{0}, \cdots, B^{m-2}_{n_{m}}\}$ in the last inning. Let $x_{m}\in X$ and $j\in\{0, \cdots, n_{m}  \}$, consider $B^{m-2}_{j}$ and note that,
$$ \bigcap_{i=0}^{j} (\mathbb{R}\setminus \{x_{i}   \}    )  \cap  \bigcap_{i=0}^{j} (\mathbb{R}\setminus \overline{Y_{i}}) \cap B_{j}$$ is a non-empty open set. So put $z_{m}\in \bigcap_{i=0}^{j} (\mathbb{R}\setminus \{x_{i}   \}    )  \cap  \bigcap_{i=0}^{j} (\mathbb{R}\setminus \overline{Y_{i}}) \cap B_{j}      $ and note that $(\mathbb{R}\setminus \{z_{m}\})\cap \bigcap_{i=0}^{j} (\mathbb{R}\setminus \{x_{i}   \}    )  \cap  \bigcap_{i=0}^{j} (\mathbb{R}\setminus \overline{Y_{i}}) \cap B^{m-2}_{j}$ is a non-empty open set. Then fix $A^{m-1}_{j}\in\mathcal{B}$ such that
$$\overline{A^{m-1}_{j}}\subseteq(\mathbb{R}\setminus \{z_{m}\})\cap \bigcap_{i=0}^{j} (\mathbb{R}\setminus \{x_{i}   \}    )  \cap  \bigcap_{i=0}^{j} (\mathbb{R}\setminus \overline{Y_{i}}) \cap B^{m-1}_{j}      $$
Then in the inning $m$, $\textsc{Alice}$ plays $\mathcal{A}_{m-1}=\{A^{m-1}_{0}, \cdots, A^{m-1}_{n_{m}}\}$ and note that $$B_{m-1}\not\subseteq \bigcup \mathcal{A}_{m-1} \hspace{0.1cm}\text{and}  \hspace{0.1cm}(\{x_{0}, \cdots, x_{m-1}\}\cup  \bigcup_{i=0}^{m-1} Y_{i})\cap \bigcup \mathcal{A}_{m-1} =\emptyset  $$

For each $n\in\omega$, denote $\tilde{\mathcal{A}}_{n}=\{\overline{A} : A\in \mathcal{A}_{n}\}$, note that $\bigcup \tilde{\mathcal{A}}_{n}$ is a closed set, because $\mathcal{A}_{n}$ is finite.

\begin{claim}
Let $s=\langle \mathcal{A}_{n} : n\in \omega\rangle$. Then $\sigma*s$ is a closed set with empty interior.
\end{claim}
	\begin{proof}
	\textcolor{white}{text}
	\begin{enumerate}
		\item $\sigma*s$ is closed.

		In fact, we will show that $\sigma*s=\bigcap_{n\in\omega}\bigcup \tilde{\mathcal{A}}_{n}$.

		$(\subseteq)$ Let $z\in \sigma*s = \bigcup_{n\in\omega} \sigma(\langle A_{0}, \cdots, \mathcal{A}_{n}\rangle)$. Then there is a $B^{n}_{j}\in \sigma(\langle A_{0}, \cdots, \mathcal{A}_{n}\rangle)$ such that $z\in B^{n}_{j}\subseteq \bigcup \mathcal{A}_{n} \subseteq \bigcup \tilde{\mathcal{A}}_{n}$.

		$(\supseteq)$ Let $z\in \bigcap_{n\in\omega}\bigcup \tilde{\mathcal{A}}_{n}$. Then for every $n\in\omega$, we have $z\in \overline{A^{n+1}_{j}}\subseteq B^{n}_{j} \subseteq \bigcup \sigma(\langle A_{0}, \cdots, \mathcal{A}_{n} \rangle)$.

		\item $\sigma*s$ has empty interior.

		Suppose otherwise, then there exists $n_{0}\in \omega$ such that  $B_{n_0}\subseteq \sigma*s = \bigcap_{n\in\omega}\bigcup \tilde{\mathcal{A}}_{n}\subseteq \bigcap_{n\in\omega}\bigcup \mathcal{A}_{n}\subseteq \bigcup \mathcal{A}_{n_{0}}$, contradiction by our construction.
	\end{enumerate}

\end{proof}

\begin{claim}
	$(\sigma*s) \cap (X\cup \bigcup\mathcal{Y} ) =\emptyset$
\end{claim}
\begin{proof}
Suppose otherwise, there exists $$z\in (\sigma*s) \cap (X\cup \bigcup\mathcal{Y}) = \bigcap_{n \in {\omega}} \bigcup \sigma(\seq{\mathcal{A}_0, ..., \mathcal{A}_n}) \cap (X\cup \bigcup\mathcal{Y})$$ Then there exists $n\in\omega$ such that $$z\in (\{x_{0}, \cdots, x_{n}\}  \cup \bigcup_{j=0}^{n}Y_{j}  ) \cap \bigcup \mathcal{A}_{n},$$
but by our construction $ (\{x_{0}, \cdots, x_{n}\}  \cup \bigcup_{j=0}^{n}Y_{j}  ) \cap \bigcup \mathcal{A}_{n}=\emptyset$, contradiction.

\end{proof}

Therefore $(\sigma*s)$ is a closed nowhere dense set disjoint from $X\cup \bigcup \mathcal{Y}$.

\end{proof}

\newpage

The following is immediate

\begin{lemma}\label{2inducesws}
	Let $Y$ be a dense subspace of $X$ in which $\textsc{Bob}$ has a winning strategy in $\mathsf{BM}_\mathrm{fin}(Y)$, then $\textsc{Bob}$ has a winning strategy in $\mathsf{BM}_\mathrm{fin}(X)$.
\end{lemma}

\begin{proposition}\label{propfinal}
  (CH) There is a Baire subspace $X$ of $\mathbb{R}$ such that \textsc{Bob} has no winning strategy for $\mathsf{BM}_\mathrm{fin}(X)$.
\end{proposition}

\begin{proof}
Fix $\mathcal{B}=\{B_{n}:n\in\omega\}$ an infinite basis for the topology of $\mathbb{R}$,
let $\{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \} = \{G_{\xi}:\xi\in\omega_{1}\}$, and fix an enumeration $\{\sigma_{\xi} : \xi<\omega_{1} \}$ of $\textsc{Bob}$'s strategies in the game $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$.

By transfinite induction on $\xi < \omega_{1}$ we will construct a sequence $\{x_{\xi} : \xi < \omega_{1}  \}$ of points of $\mathbb{R}$ such that for every $\xi < \omega_{1}$

\begin{itemize}
\item [(a)] $x_{\xi}\in G_{\xi}$
\item [(b)] there exists a sequence of finite collections of open sets $s_{\xi}$ such that
$$(\sigma_{\xi}*s_{\xi})\cap \left(\{x_{\beta}:\beta\leq\xi   \} \cup \bigcup_{\beta<\xi}(\sigma_{\beta}*s_{\beta})    \right) =\emptyset$$

\end{itemize}

Consider $x_{0}\in G_{0}$ and $\sigma_{0}$. By Lemma \ref{lematec}, there exists $s_{0}$ such that $\sigma_{0}*s_{0}$ is a closed nowhere dense set and $(\sigma_{0}*s_{0})\cap \{x_{0}\}=\emptyset$. Set $x_{1}\in G_{1} \setminus (\{x_{0} \} \cup  (\sigma_{0}*s_{0})   )  $, by Lemma \ref{lematec}, there exists $s_{1}$ such that $\sigma_{1}*s_{1}$ is a closed nowhere dense set and $(\sigma_{1}*s_{1})\cap (\{x_{0}, x_{1}\} \cup (\sigma_{0}*s_{0})  )  =\emptyset$.

Now suppose that we have built $\{x_{\beta} : \beta<\alpha\}$ with $\alpha<\omega_{1}$, note that $$  \bigcap_{\beta<\alpha}\mathbb{R}\setminus (\sigma_{\beta}*s_{\beta})$$
is a dense $G_{\delta}$-set.
We claim that $G_{\alpha} \setminus \left( \{x_{\beta} : \beta<\alpha     \} \cup \bigcup_{\beta<\alpha} (\sigma_{\beta}*s_{\beta})       \right) \not=\emptyset$. In fact, suppose otherwise, $G_{\alpha}\subseteq
\{x_{\beta} : \beta<\alpha     \} \cup \bigcup_{\beta<\alpha} (\sigma_{\beta}*s_{\beta})$, so $G_{\alpha}$ is a meager $G_{\delta}$-set in $\mathbb{R}$, then, by Proposition \ref{hrusak}, $G_{\alpha}$ is nowhere dense, contradiction, because $G_{\alpha}\in \{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \}$. Therefore, choose $x_{\alpha} \in
G_{\alpha} \setminus \left( \{x_{\beta} : \beta<\alpha     \} \cup \bigcup_{\beta<\alpha} (\sigma_{\beta}*s_{\beta})       \right)$.
By Lemma \ref{lematec}, there exists $s_{\alpha}$ such that $\sigma_{\alpha}*s_{\alpha}$ is a closed nowhere dense set and $$(\sigma_{\alpha}*s_{\alpha})\cap \left(\{x_{\beta}:\beta\leq\alpha   \} \cup \bigcup_{\beta<\alpha}(\sigma_{\beta}*s_{\beta})    \right) =\emptyset$$

Finally denote $X=\{x_{\xi} : \xi\in\omega_{1}\}$, note that $X$ is a dense set. Indeed, let $m\in\omega$ and consider $b\in B_{m}$, so $(B_{m}\setminus \{ b \}) \in \{G\in \mathcal{G}_{\delta} \mid \exists n\in\omega : G\subseteq B_{n} \subseteq \overline{G}  \} = \{G_{\xi}:\xi\in\omega_{1}\}$. Then there is a $\xi_{m}\in\omega_{1}$ such that $B_{m}\setminus\{b\} = G_{\xi_{m}}$, by construction $x_{\xi_{m}} \in G_{\xi_{m}}\cap X \subseteq B_{m}\cap X$.

\begin{claim}
		$X$ is a Baire space. In particular $\textsc{Alice}$ has no winnnig strategy in $\mathsf{BM}_\mathrm{fin}(X)$.
\end{claim}

\begin{proof}
Suppose otherwise, then by Lemma \ref{densoBairechar}, there exists $G\in\mathcal{D}\mathcal{G}_{\delta}$ such that $G\cap X$ is not dense in $X$. So there is an $m\in\omega$ such that $(B_{m}\cap X) \cap (G\cap X) = B_{m} \cap (G\cap X) =\emptyset$. Then, by Lemma \ref{easytec}, there is a $\xi<\omega_{1}$ such that $G_{\xi}= B_{m}\cap G$, but by construction $x_{\xi}\in G_{\xi}\cap X = B_{m}\cap G \cap X =\emptyset$, contradiction.

\end{proof}

\begin{claim}
		$\textsc{Bob}$ has no winning strategy in $\mathsf{BM}_\mathrm{fin}(X)$.
\end{claim}

\begin{proof}
Suppose otherwise, that is, $\textsc{Bob}$ has a winning strategy $\tilde{\sigma}_{_{X}}$ in $\mathsf{BM}_\mathrm{fin}(X)$. As $X$ is dense, by Lemma \ref{2inducesws}, $\tilde{\sigma}_{_{X}}$ induces a winning strategy $\tilde{\sigma}$ for $\textsc{Bob}$ in $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$. Also, $\tilde{\sigma}$ satisfies that for each sequence  $s=\langle\mathcal{A}_{n} : n\in\omega\rangle$ of $\textsc{Alice}$'s moves in $\mathsf{BM}_\mathrm{fin}(\mathbb{R})$,
$$\tilde{\sigma}_{_{X}}*s^{\prime}  =  \bigcap_{n \in \omega}\bigcup \tilde{\sigma}_{_{X}}(\langle \mathcal{A}^{X}_{0}, \cdots, \mathcal{A}^{X}_{n} \rangle) \subseteq  \bigcap_{n \in \omega} \bigcup \tilde{\sigma}(\langle \mathcal{A}_{0}, \cdots, \mathcal{A}_{n} \rangle) \cap X  = (\tilde{\sigma}*s )\cap X   $$
where $s^{\prime}=\langle  \mathcal{A}_{n}\cap X  : n\in\omega  \rangle $ with $\mathcal{A}_{n}\cap X = \{A\cap X : A\in \mathcal{A}_{n} \}$. In particular, there is a $\xi<\omega_{1}$ such that $\tilde{\sigma}=\sigma_{\xi}$.

Let $s_{\xi}=\langle \mathcal{A}^{\xi}_{n} : n\in \omega \rangle$ of our construction.

\begin{claim}
	$(\sigma_{\xi} * s_{\xi})\cap X=\emptyset$.
\end{claim}

\begin{proof}
In fact, suppose otherwise, there is an $\alpha<\omega_{1}$ such that $x_{\alpha} \in (\sigma_{\xi}*s_{\xi})$. We have the following cases:
\begin{itemize}
	\item $\alpha\leq \xi$: by part (b),
	$(\sigma_{\xi}*s_{\xi})\cap \{x_{\beta}:\beta\leq\xi \}=\emptyset$, contradiction.
	\item $\alpha>\xi$: by construction, $x_{\alpha} \in
	G_{\alpha} \setminus \left( \{x_{\beta} : \beta<\alpha     \} \cup \bigcup_{\beta<\alpha} (\sigma_{\beta}*s_{\beta})       \right)$, contradiction.
\end{itemize}

\end{proof}

So $\emptyset\not= \tilde{\sigma}_{_{X}}*s^{\prime}_{\xi}  \subseteq    (\sigma_{\xi} * s_{\xi})\cap X =\emptyset$, contradiction. Then, $\textsc{Bob}$ has no winning strategy in $\mathsf{BM}_\mathrm{fin}(X)$.

\end{proof}

Therefore $X$ is the desired space.
\end{proof}

\begin{corollary}
		(CH) There exists a subspace $X$ of the real line in which the game $\mathsf{BM}_\mathrm{fin}(X)$ is not determined.
\end{corollary}

\begin{corollary}\label{chcor}
		(CH) There is a productively Baire space such that \textsc{Bob} has no winning strategy for the $\mathsf{BM}_\mathrm{fin}$ game.
\end{corollary}

\begin{proof}
	Let $X$ as in Proposition \ref{propfinal}, then by Corollary \ref{cpbpb}, $X$ is productively Baire.
\end{proof}

\begin{question}
	Corollary \ref{chcor} holds if we change $\mathsf{BM}_\mathrm{fin}$ to $\mathsf{BM}_\mathrm{{\omega}}$?
\end{question}

\begin{question}
Is there any relationship between the game $\mathsf{BM}_\mathrm{{\omega}}$ and the productively Baire spaces?
\end{question}

\begin{corollary}
(CH) There exists a space $X$ such that Bob has a winning strategy in $\mathsf{BM}_\omega$ but $\textsc{Bob}$ has no winning strategy in $\mathsf{BM}_\mathrm{fin}$.
\end{corollary}

redpajama_set_name: RedPajamaArXiv


text:
Q: Overriding order relation while looking for lower bounds Let $(\mathbb Z \times \mathbb Z^*, \pi)$ be a poset defined as follows:
$$\begin{aligned} (a,b) \pi(c,d)\Leftrightarrow (a,b) = (c,d) \text{ or } r(a,b) <r(c,d)\end{aligned}$$
whereas $r(x,y)$ is the reminder of the division of $x$ by $y$.
Let $X=\{(3,2),(12,5),(6,2),(11,9)\} \subset \mathbb Z \times \mathbb Z^*$ and find maximum, minimum, maximal elements, minimal elements, upper bounds, lower bounds, supremum and infimum.
I noticed that in gereral $\nexists(a,b)\in \mathbb Z \times \mathbb Z^* : r(a,b) <0$ and the only elements that have $0$ as reminder are in $S=\{(na,a)\in \mathbb Z \times \mathbb Z^* : n\in \mathbb Z\}$.
Generally speaking, let $P$ be a (partially) ordered set, and let $A$ be a subset of $P$ then we say $y\in P$ is a lower bound for $A$ if and only if $y\leq a$ for all $a\in A$.
As I come to find all the lower bounds for my specific poset, do I need to look for $(x,y)\in \mathbb Z \times \mathbb Z^*$ so that $(x,y)\leq (a,b)$ for all $(a,b)\in X$, meaning looking for $(x,y)\in\mathbb Z \times \mathbb Z^* : r(x,y) \leq r(a,b)$ or shall I substitute $\pi$ for $\leq$ used in the general definition and therefore look for $(x,y)\in \mathbb Z \times \mathbb Z^*$ so that $(x,y) \pi (a,b)$?
If the former is true then all lower bounds are in $S$, otherwise if the latter is true then there are no lower bounds because by very definition of $\pi$, if $(a,b)\neq(x,y)$ and $r(a,b)=r(x,y)$ then $(a,b)$ and $(x,y)$ are not comparable. Is that correct?

A: The relation $\,\pi\,$ determines a partial order on your set, in such a way that to say that $\,(a,b)\,$ "is less", or is dominated", by $\,(c,d)\,$ means precisely $\,(a,b)\pi(c,d)\,$ .
So yes: you can try to take lower/upper bounds for $\,X\,$ either from within or out $\,X\,$ . Sometimes you'll be able to find some of these bounds within $\,X\,$ (for example, when $\,X\,$ is finite we have only a finite number of checks to do) , and sometimes you won't. It usually isn't important unless some other specifications are given.
As you said, since we always have $\,r(a,b)\geq 0\,$ , if there's an element $\,(a,b)\,$ s.t. $\,r(a,b)=0\,$ we have then hit on a minimal element.
For example, since $\,6=2\cdot 3\Longrightarrow r(6,2)=0\Longrightarrow (6,2)\,$ is a minimal element in $\,\Bbb Z\times\Bbb Z\,$ and thus in any subset. Thus, this element is  the minimum of $\,X\,$ as there's no other element less than it in $\,X\,$ . If, say, we also had $\,(6,3)\in X\,\,,\,\,or\,\,(30,5)\in X\,$ , then we'd have two minimal elements and the set wouldn't have minimum...Now you try to continue with this.

redpajama_set_name: RedPajamaStackExchange


This queen sized room is situated on the first floor of the Manor House, offering a clean and bright space and boasting phenomenal views of the Stellenbosch Winelands and its majestic mountainous backdrop. It features a 40in. flat screen TV, minibar, air conditioning, Nespresso machine and has an en-suite bathroom with a shower.
This room can be made into a private wing, by joining the Nozellara room as both these rooms share a common reading lounge.


text:
Der Wenatchee Confluence State Park ist seit 1990 ein Erholungs- und Vogelschutzgebiet an der Mündung des Wenatchee River in den Columbia River. Er liegt nördlich der Stadt Wenatchee im Chelan County des US-Bundesstaates Washington. Der 80 ha große State Park ist durch den Wenatchee River in zwei Hälften geteilt.

Die beiderseitigen Uferzonen und der südliche Teil sind als Naturschutzzone mit Marschland, Kanälen, Inseln und Wanderwegen gestaltet. Bisamratte, Biber und über 200 Vogelarten kommen in dem als Horan Natural Area eingetragenen Gebiet vor. Darunter befinden sich Schwalbenmöwen, Sturmmöwen und Aztekenmöwen, sowie Küstenseeschwalben, Krabbenreiher, Grünreiher und Magnolien-Waldsänger. Weißkopfseeadler, Eistaucher, Nordamerikanische Pfeifente, Krickente und Brautente können von der Fußgängerbrücke beobachtet werden, die beide Gebiete verbindet. Königstyrann, Arkansaskönigstyrann, Katzendrossel, Dunenspecht, Zedernseidenschwanz, und Schwalbenarten wie Veilchenschwalbe, Rauchschwalbe, Sumpfschwalbe kommen dort auch vor.

Im nördlichen Bereich sind verschiedene Bade-, Sport- und Erholungsanlagen sowie Übernachtungsmöglichkeiten für Camper vorhanden. In den Jahren von 2002 bis 2006 lagen die jährlichen Übernachtungszahlen zwischen 36.053 und 40.267 Gästen. Die Anzahl der Tagesbesucher lag in dem Zeitraum zwischen 209.080 und 480.886 Menschen.

Weblinks

 Internetpräsenz des State Parks

Einzelnachweise

State Park in Washington
Vogelschutzgebiet in den Vereinigten Staaten
Columbia River
Chelan County

redpajama_set_name: RedPajamaWikipedia


As I shared in my mail on May 18 (see attached), we have better options than a full combination with Yahoo! at the price it suggested, and we have moved forward on our strategy to grow our online business.
Let me share a little background with you. When we made our original proposal on February 1st to combine with Yahoo!, we offered a 62 percent premium that was based on a desire to reach an agreement in short order. The faster we could reach an agreement, the sooner we could begin the regulatory process and create value through this combination.
In a March 10th meeting in Palo Alto, we explained to Yahoo! management the importance of reaching an agreement by the end of April in order to have an opportunity to complete the regulatory process by the end of this calendar year. Because we could not come to an agreement on price by the end of April and given our concerns about Yahoo!'s business performance, we elected to withdraw our bid and pursue better options for Microsoft.
Microsoft and Yahoo! would have entered into a long-term search partnership, where Microsoft would have provided favorable economics to Yahoo! search, including a three-year guarantee of higher monetization than Yahoo!'s Panama paid search system currently provides.
New Transfer of Cash to Yahoo! Shareholders. This proposal would have transferred $9 billion from Microsoft to Yahoo!, which could have been used by Yahoo! to reward their shareholders.
A More Profitable Ongoing Business. This proposal would have resulted in higher operating income on an annual basis for Yahoo!, with our projections more than doubling Yahoo!'s operating income in the first year of operation, and increasing it by more than $1 billion above its current operating income level.
A More Compelling Search Offering. The combination of the search platforms would have unlocked new R&D innovation, eliminated redundant engineering efforts and allowed for greater scale in serving our customers.
Unfortunately Yahoo! has chosen a different course, and yesterday announced an agreement that would start to consolidate over 90 percent of the paid search advertising market in Google's hands. This will make the market far less competitive. There are many experts who suggest that a host of legal and regulatory problems lie ahead for Google and Yahoo!.
Since my mail on May 18, we have been making great progress. At our advance08 conference, we announced Live Search cashback and Live Search Farecast, and the initial response to these user experience and business model innovations in search has been very positive. On June 2nd, we also announced a distribution deal with HP, the world's largest PC manufacturer, to install a Live Search-enabled toolbar on all HP consumer PCs planned to ship in the United States and Canada, beginning in January 2009.


I offered affordable mini-sessions to help raise money for my dear friend Donna Gerdau, who was just diagnosed with stage 3 breast cancer. Donna is an amazing lady who I have known since I was about 15 years old. She's done so much for me throughout the years, so I want to do something for her. 100% of the proceeds from the sessions went to Donna to help her during this difficult time.


Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting Americans With Disabilities Act – Titles, Litigation and How To Comply at the 1993 American Management Association Conference held in Chicago, Illinois.
Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting Tooling Up for Assistive Technology at the 1992 Illinois Occupational Therapy Association Annual Conference.
Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting Case Management Strategies at the 1992 American Management Association Annual Convention.
Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting ADA How to Keep Claims From Deteriorating in Bloomington, IL on July 29, 1992.
Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting ADA Programs and How They Effect Your Business in Tinley Park on December 9, 1991.
Mr. Ed Steffan, CPC, LPC and founder of EPS Rehabilitation, Inc. will be presenting Rehabilitation Practices: Small Company Perspectives at the 1990 Illinois Rehabilitation Counselor Association Annual Conference.
Mr. Edward Steffan, CRC, LPC and founder of EPS Rehabilitation, Inc. presents "Structured Transitional Employment Program: How To Develop, Implement and Achieve Success" at the 1990 Core*Mark International Conference in Anaheim California.
Mr. Edward Steffan, CRC, LPC and founder of EPS Rehabilitation, Inc. presents "Counseling Individuals to Maximize Their Vocational Potential" to doctoral candidates and "Developing Student Careers Through Counseling Programs" to university staff in Mexico.
Mr. Edward Steffan, CRC, LPC and founder of EPS Rehabilitation, Inc. presents "How to Purchase and Oversee Rehabilitation Services" at a conference held by the Illinois Department of Mental Health division.


I was at TCAF (Toronto Comics Arts Festival) this past Sunday, wandering around looking but not really interested in the things I was passing when I came to an abrupt halt! It was the artwork I noticed first, (which is why it's a comic arts festival) I had seen it before on the web and remembered it immediately. It is one of those art styles I am drawn too (no pun intended). Of course the book was Shi Long Pang, the Wandering Shaolin Monk. I came across this comic on the web before and remembered just loving the art. I can only guess the reason for me not diving in and reading the archives when I first discovered it is because I have two kids in diapers, and time to go through a comic archive is limited to say the least. Also I was hesitant only because it seemed like everything I wanted in a comic, art I loved, and a subject I loved, but I was afraid to be disappointed by the story. Now here I was at TCAF and there was the book in beautiful hardcover right in front of me, I stopped, picked it up and bought it with a hope the story would be good.
It was everything I hoped it would be and more. The author Ben Costa intertwines fiction and history wonderfully, with great footnotes to educate the reader. I love the tidbits of history brought into the story, a history that is fairly unknown to the western reader, that Costa explains making context easier to understand, which I appreciated. Costa does make the reader work and you do have to pay attention, but that is just another thing that is great about this book.
I'm a shy guy, but I did find myself babbling a bit to Ben as he signed my book, something I didn't expect cause I have never bought a book from an artist before. He was nice and gave me a great picture of Pang in my book, great stuff!
I couldn't recommend this book more highly. It's rare to find a comic that exceeds expectations, and Pang the Wandering Shaolin Monk does this with ease. Also it's nice to read a graphic novel that doesn't take 10 min to finish, I mean it actually takes a long time to read and when I started to get near the end I was really sad that it would be over. Thank goodness there is more coming!


In this short tutorial, we'll take a quick look at installing Ubuntu Operating System on a virtual machine.
Motivation or the lack thereof… Every time I get an invite to conduct a workshop on Network Simulation, the first thing people ask me is How to install ns3? I am almost always tempted to say RTFM politely, but I understand that it can be overwhelming for people using either ns3 or Linux for the first time.


No more than dark mages.
They seem to appear much sooner though. and the fact that they target your head repeatedly makes them more dangerous (IMO).
I agree. Since they appear quite early and your endurance might not be too trained (or not at all) they become very dangerous because one hit can knock you unconscious and then kill you. Also, they attack from a distance, which in early game is more dangerous than ever, since you're less likely to have esp and even if you do your int is probably too low to be of any help unless the worm is close. On top of that, their sprite is so hard to spot you can probably mistake them for blood in the floor, specially if it's standing in a corridor where some monster escaped bleeding.
I'm not a fan of their addition, honestly (as if from the previous parragraph it wasn't obvious). Also, it doesn't make much sense to me that a worm could have psychic powers.
A regular worm, sure. But these are mind worms. Clearly they have mental abilities unknown to mere humans.
Yeah, with their brains the size of a nail.
Im not a huge fan of their addition either (at least not in their current incarnation). I feel they need to be nerfed a bit. or have their generation be made more rare.
I feel I must point out that any opinions I make about the current state of the game are uninformed. I haven't played this version, and I didn't play the last version much either. That said, I don't think that magic, or psychic powers, really correlates well with size. If it did, polar bears would be far more intimidating.
How early are they appearing?
Well, it's not that size correlates with psychic powers or intelligence, but a worm seems as the most basic creature. I'd prefer a will-o-wisp for that matter.
I've run into this worms in the early levels of the gloomy cave.
How long do you take to finish UT? If you're playing too slowly and have too much HP, you will happen upon much more difficult monsters in early GC.
BTW, mind worms were already in 0.50, just not generated. Original devs "borrowed" them from from some sci-fi, I think.
I don't take that long. I'm finding them more often on the final levels before Elpuri now and they're much more easier to kill when well equipped and with decent esp range. Still, they demand immediate attention. Quick question, when they're hatchlings, does it mean they've hatched from someone's head or can they spawn randonly as hatchlings too? Also, are hatchlings weaker/do less damage than regular ones?
* They are extremely fragile and slow, so you will kill them immediately when you find them.
* They are cowardly and will try to run away from you, so if you're attacked and don't want to deal with them, just head back and they will not pursue.
If anything, we could do something with the HP threshold for their generation.
Quick question, when they're hatchlings, does it mean they've hatched from someone's head or can they spawn randonly as hatchlings too? Also, are hatchlings weaker/do less damage than regular ones?
I think they can both be hatched from someone's head, or (very rarely) generated. They are weaker and cannot infect you with mind worm eggs.
Yes, I think the hp needed for them to spawn should be revised, because they're only really dangerous at low endurance since the head has only a few hit points and they attack exclusively there.
There is something that needs to be addressed with these mind worms. They need to make a special sound when attacking your brain and another one when entering your skull (possibly with an attack animation too), because when you're at the end game and have high int and esp your message bar gets spammed with actions done by all the creatures in the level you sense, so spotting either of these messages among all the things going on is pretty damn hard. And you may not even notice the damage in your head until it's too late. I believe it's currently the only damage type that doesn't make a sound or animation of some sort. Since these little bastards are so deadly (they can knock you unconscious and that's basically game over if there's enemies nearby), the player needs to be warned somehow.
I just got infected without noticing it before killing one of these. My head was moderately damaged so I retreated to a previous level to heal and bam, a mind worm got out of my head. While doing so, in the same turn, it knocked me unconscious and the rest is history. Needless to say I was pretty pissed because I died not for making a poor decision but for missing a single line of text, whereas obviously my character would've noticed a freaking worm entering and leaving his skull.
This is specially true now that there are sounds and animations for combat, both changes I love. Now you rely a lot more on visuals rather than text to understand what's going on, which is a good thing as it helps to navigate through the game more fluently. But it's also more likely you won't notice a message in your dialog box.
I agree 100%. for a random spawn, they have the difficulty of a boss monster. probably more difficult because of the head only/unconsious attack. Its really annoying and unbalancing, and really, not enjoyable.