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August 19

Best Mathematicians of 2010

Does anyone know who won the fields medal? I can't seem to find this information anywhere! ... If it hasn't been released, when will it be? I heard that it would be released on 19 August,2010 but I don't want to be waiting till midnight for the declaration of the world's best mathematicians! Please guys, put me out of my misery and tell me how many more hourse I need to wait? I've waited for 4 years already ... Thanks! Signed:THE DUDE. (BTW, I fixed my typsetting bug!) —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

I think I know now ... Damn, why do applied mathematicians always get the awards? Where are the algebraists, topologists and geometers (and number theorists for that matter)? And why on Earth hasn't anyone received it for logic and set theory yet? (FOrget the the first question in above paragraph, I think I know who won it.) Thanks guys ... —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

There is already a Wikipedia User:THE DUDE, which apparently is not you. Both impersonating other users and pretending to be a user which doesn't really exist is forbidden. Your continued failure to sign your posts, followed by vandalizing your automatically-supplied signatures, is disruptive. Please stop. You can create an account if you wish. -- Meni Rosenfeld (talk) 08:16, 19 August 2010 (UTC)[reply]

Meni Rosenfeld, I just don't want to be known to the public. I want to be anonymous. You have a name, Meni Rosenfeld. If THE DUDE is taken, can I have another name. I never vandalized signatures. I just didn't want to be made public. Isn't that a simple rquest? Or are the people here at Wikipedia so serious. What's a change in signature between friends? —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

Dmcq has advised you to create an account if you want to remain anonymous. You ignored his advice, lied about your motivation, and now you are playing dumb. We have no reason to believe your intentions are in any way friendly. -- Meni Rosenfeld (talk) 09:44, 19 August 2010 (UTC)[reply]

OK, I fixed up all signatures? Now can you please answer my question Meni Rosenfeld? —Preceding unsigned comment added by 114.72.209.200 (talk) 11:34, 19 August 2010 (UTC)[reply]

I don't know the answer, but I'm sure that now those who do will be much happier to assist you. (Your next step is to sign your posts by typing 4 tildes, ~~~~ ). -- Meni Rosenfeld (talk) 11:45, 19 August 2010 (UTC)[reply]

114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) Is this right? 114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) How did you type 4 tildas without a signature coming up?[reply]

Creating an account is the only way to hide you IP address. No matter what edit you do to this page its and easy matter just to look at the history to find your IP. You can create an account at Special:UserLogin/signup --Salix (talk): 08:46, 19 August 2010 (UTC)[reply]

Using the nowiki tag. -- Meni Rosenfeld (talk) 13:24, 20 August 2010 (UTC)[reply]

Why isn't anyone answering my questions? Please answer guys ... Thanks guys ... 114.72.244.249 (talk) 08:37, 20 August 2010 (UTC)[reply]

Can't speak for anyone else, but I haven't attaempted to answer your original questions because (a) they call for opinion, which is not the best use of Wikipedia's reference desks and (b) you come across as bad-tempered and argumentative, so even if I did venture an opinion, I think there is a good chance you would just use it as a springboard for an argument, so it would be a waste of my time. Gandalf61 (talk) 08:53, 20 August 2010 (UTC)[reply]

Please answer my question? I won't attack you. What makes you think I'm bad tempered? I just like to express my opinion 'is all. I'm as polite as I can be and yet I'm misunderstood.I won't argue. Come on? —Preceding unsigned comment added by 110.20.55.15 (talk) 11:38, 20 August 2010 (UTC)[reply]

Fields Medal winners are listed at Fields Medal. Dragons flight (talk) 08:54, 20 August 2010 (UTC)[reply]

Is the argument of a given unary function a singleton?

The argument of a given binary function is an (ordered) pair, so the argument of a given unary function is a singleton, isn't it? Eliko (talk) 08:45, 19 August 2010 (UTC)[reply]

Yes--Shahab (talk) 10:38, 19 August 2010 (UTC)[reply]

Unfortunately, it's mentioned nowhere in Wikipedia, as far as I know. Eliko (talk) 10:47, 19 August 2010 (UTC)[reply]

Well, technically, it's an individual rather than a singleton. But you could just as easily take it to be a singleton, and nothing important would change. --Trovatore (talk) 10:47, 19 August 2010 (UTC)[reply]

I'm looking for the formal definition. "technically" (as you've put it), a binary function doesn't have two arguments each of which is an individual, but rather has an argument being an (ordered) pair of individuals (because the order of both individuals must be preserved), so I would expect that, "technically", a unary function won't have one argument which is an individual, but rather will have an argument being a singleton of an individual. Eliko (talk) 12:33, 19 August 2010 (UTC)[reply]

There isn't necessarily a "the" formal definition. You're talking about details at the "implementation" level, which may vary from formalization to formalization, but don't change the conceptual content for most purposes.

I would say that a unary function takes an individual, and a binary function, at the most fundamental level, takes two individuals (distinguishing between them, as you note). For many purposes it's convenient to package those two individuals into a single individual, that we then call an ordered pair. But at the very fine level you seem to be examining this, that technically makes it a unary function that takes as its sole argument an ordered pair, rather than a binary function with two arguments. --Trovatore (talk) 22:11, 19 August 2010 (UTC)[reply]

You seem to distinguish between "the most fundamental level" and the "very fine level". So would you agree with the following? "At the most fundamental level", a binary real function takes two arguments each of which is a real number, while a unary real function takes one argument which is a real number. On the other hand, i.e. "at the very fine level", every real function is unary, the following way: a function that would be called "binary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument an ordered pair" of real numbers, while a function that would be called "unary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument a singleton" of a real number. Eliko (talk) 07:15, 20 August 2010 (UTC)[reply]

Dude, fair warning: If you persist in trying to parse my rhetorical flourishes for some precise technical meaning, I'm going to get upset with you and leave the conversation. I doubt I'm the only one who'll react that way. Phrases like "at a very fine level" have a function, which you as a presumably competent social actor should be able to discern, but the indication of some exact mathematical or philosophical category is not it.

That said, here's how I see it: A binary function, conceptually, is a way of (but not necessarily an expressible rule for) taking two things and getting a third thing, in such a way that the third thing depends only on the two inputs (with the slots for the two inputs being distinguishable).

How you code that concept into some specific framework, such as set theory, is another matter. You have lots of choices, and most of the time it doesn't matter even slightly which choice you pick. Occasionally it will matter (say, when you don't have a pairing function available), and for those instances it's good to practice mental flexibility and not get too hung up on the formalism. --Trovatore (talk) 08:54, 20 August 2010 (UTC)[reply]

Monoliths in 2001:A Space Odyssey

In 2001: A Space Odyssey, the Monoliths are 1:4:9. That would be 1 by 4 by 9. But 1 what by 4 what by 9 what? Centimetres? Metres? Yards? Feet? Inches? Can somebody help? --Editor510 ^{drop us a line, mate} 08:58, 19 August 2010 (UTC)[reply]

PS: I know this is about an entertainment film but it is a mathematical question, so please don't ask me to move the question.

The author carefully avoided that question by using the smaller side of the monolith as unit of measurement, and then noticing that the larger side was 9 units and the intermediate side was 4 units. Bo Jacoby (talk) 10:56, 19 August 2010 (UTC).[reply]

(EC) Monolith (Space Odyssey)#Appearance and capabilities discusses this, the monoliths are not all the same size.81.98.38.48 (talk) 10:58, 19 August 2010 (UTC)[reply]

Ratios don't require units. HiLo48 (talk) 11:09, 19 August 2010 (UTC)[reply]

Good estimation of number of divisors d(n)

Hello! Let d(n) be a number of positive divisors of n, for example, d(6)=4. In 1907 S. Wigert proved that for any $\epsilon >0$ holds two statements:
1) for infinitely many n $d(n)>2^{(1-\epsilon ){\frac {\ln n}{\ln \ln n}}}$ ;
2) for all sufficiently large n $d(n)<2^{(1+\epsilon ){\frac {\ln n}{\ln \ln n}}}$ ;

Does anybody know more precise estimations than these?
Actually I am able to prove better estimation than 1), but I don't have any information about better results.
Thank you!
RaitisMath (talk) 12:52, 19 August 2010 (UTC)[reply]

If you don't get an answer here, you might try MathSciNet.—msh210℠ 17:32, 20 August 2010 (UTC)[reply]

Our divisor function function article discusses estimates a little bit and has some references. I'm not sure if there's enough to help you. 67.122.209.167 (talk) 09:28, 21 August 2010 (UTC)[reply]

Nowhere dense subsets

Hi, I have recently encountered this proposition that seems somewhat vague to me: Proposition: Let X be a topological space without isolated points having countable $\pi$ -weight and such that every nowhere dense subset in it is closed. Then it is a Pytkeev space.

The thing which is not clear to me is this, if every nowhere dense subset is closed, doesn't that means that it has to be discrete? and doesn't discrete means that every point is isolated? So, is the condition given in this proposition is that, there arn't any nowhere dense subsets in X? Which means that every subset of X is dense somewhere? which doesn't make sense.. I mean, what about subsets of X that contain one point for instance? Thanks! Topologia clalit (talk) 21:55, 19 August 2010 (UTC)[reply]

Consider an infinite set with the cofinite topology. This space is not discrete, but every nowhere dense set is closed: if a set is infinite, its closure is the entire space, so the only nowhere dense sets are finite. (This space also has no isolated points, and, if the underlying set is countable, has countable

\pi

-weight.) —Preceding unsigned comment added by 130.195.5.7 (talk) 00:09, 20 August 2010 (UTC)[reply]

Alternatively, you can start with any topology and make a finer one by defining all the nowhere dense sets to be closed (and then taking appropriate unions). This new space will have isolated points precisely if the old space did, will have no greater

\pi

-weight (maybe the same?), and all nowhere dense sets will be closed. —Preceding unsigned comment added by 130.195.5.7 (talk) 00:35, 20 August 2010 (UTC)[reply]

I see.. Thanks for your example. But I am still confused. I mean, here is the proof of this proposition. If we take your example of the cofinite topology, under consideration, how can I explain the emphesized remark in brackets? I mean, the nowhere dense subsets in your example are closed and not discrete.. Proof: Let $x\in Cl(A)\setminus A$ . Then $x\in Cl(Int(Cl(A)))$ , because every nowhere dense set is closed (and hence discrete). Let $P_{n}:n\in \omega$ be a list of elements of a countable $\pi$ -base in the space, which are contained in $Int(Cl(A))$ . Let $M_{n}\cap P_{n}$ . Then $\{M_{n}:n\in \omega \}$ is a countable $\pi$ -net, at x, and each $M_{n}$ is infinite.

The comment is saying that the nowhere dense subsets are discrete. I gave you an example where the total space isn't discrete. It's not hard to see that every finite set in my example is discrete, however.

To see that nowhere dense sets are discrete under the hypothesis, suppose one wasn't. Then it would have an accumulation point. Remove this accumulation point, and the resulting set would still be nowhere dense, but would not be closed, contrary to assumption.

Also, something in your second remark bothers me. Suppose for example that I take $\mathbb {R} ^{2}$ with the usual topology and try to define all the nowhere dense sets to be closed. Then, for example, the nowhere dense set $\{{1 \over n}\}_{n\in \mathbb {N} }$ will turn closed? But it cant be since it doesn't contain it's accumulation point 0.. what am I missing here? Thanks! Topologia clalit (talk) 08:10, 21 August 2010 (UTC)[reply]

Could you please supply an explicit reference to the text? — Fly by Night (talk) 16:11, 21 August 2010 (UTC)[reply]

Ya sure, it's "Weakly Frechest-Urysohn and Pytkeev spaces" by V.I. Malykhin and G. Tironi. From Topology and its Applications 104 (2000) 181-190 Let me know if you can't find it. Thanks! Topologia clalit (talk) 16:20, 21 August 2010 (UTC)[reply]

No, I can't find it. If it's from a journal then it's highly unlikely that I'll be able to find it. I was thinking more along the lines of a book. — Fly by Night (talk) 16:35, 21 August 2010 (UTC)[reply]

When you define the nowhere dense sets to be closed, you're changing the topology. Not everything which was an accumulation point still is.

I see.. OK thanks. I'll think about it. Prove to myself that this is a topology... Here is a direct link to the article: http://www.f2h.co.il/307676563971 Proposition 2.1 there.. What do you think? Topologia clalit (talk) 17:07, 21 August 2010 (UTC)[reply]

I don't speak Hebrew, and I don't want to download the application because I don't understand the site. Sorry. — Fly by Night (talk) 17:14, 21 August 2010 (UTC)[reply]

Nothing to worry about, just your average infested-with-extremely-intrusive-ads-and-possibly-malware file-sharing site...

Transferbigfiles below is more friendly. Anyway what the OP linked to is a pdf, not an application. -- Meni Rosenfeld (talk) 18:34, 21 August 2010 (UTC)[reply]

Sorry, I haven't even noticed that this link is in Hebrew.. Here is another link in: https://www.transferbigfiles.com/53c3b8c5-f0f5-4bbc-93d2-a0030eaae50a?rid=gcIEkCL2J4VQPfWN7RipdA%3d%3d It's in English.. —Preceding unsigned comment added by Topologia clalit (talk • contribs) 17:51, 21 August 2010 (UTC) Topologia clalit (talk) 18:07, 21 August 2010 (UTC)[reply]