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August 14

Cardinality of Topology

Given a set X, what is the cardinality of the set of subsets of X which form a topology for X? In other words: what is the cardinality of the topologies of X? — Fly by Night (talk) 01:07, 14 August 2010 (UTC)[reply]

P.S. It's not just the cardinality of the power set P(X). The topology whose open sets are all subsets is the discrete topology. So this topology counts as a single member of the set of topologies. — Fly by Night (talk) 01:22, 14 August 2010 (UTC)[reply]

The cardinality of P(P(X)) (which is

2^{2^{|X|}}

, with an appropriate interpretation of infinite exponents where necessary) is clearly an upper bound. A topology is an element of that power set of the power set of X. Some of those elements will not be valid topologies, of course. The question is what proportion actually are valid topologies. The cardinality of P(X) (ie.

2^{|X|}

) is an obvious lower bound (for any subset A of X, there is a distinct topology where a set is open iff it is empty or contains A). Where inbetween those bounds it actually is, I don't know. Some small finite X, we can just work it out by enumeration, of course. --Tango (talk) 02:25, 14 August 2010 (UTC)[reply]

Every filter is a topology. There are

2^{2^{|X|}}

many filters (see above for a proof by Trovatore). —Preceding unsigned comment added by 203.97.79.114 (talk) 11:31, 14 August 2010 (UTC)[reply]

Thanks for the credit, but the proof is actually due to Pospíšil. --Trovatore (talk) 07:18, 16 August 2010 (UTC) [reply]

How can this hold for a finite set without implying that every collection of subsets is a topology? -- 1.47.99.181 (talk) 23:44, 14 August 2010 (UTC)[reply]

It doesn't hold for finite sets. The reason is in infinite sets "very sparse subsets" (like cantor set in the real numbers) can still have the same cardinality as the whole set Money is tight (talk) 04:29, 15 August 2010 (UTC)[reply]

Oh, to be picky, every filter is almost a topology. Every topology has to have the empty set as an element (that is, the empty set is open), whereas no nontrivial filter has the empty set as an element. But that's OK; just adjoin the empty set as an element, and you have a topology. That's fine for cardinality purposes (it gives you an injective map from nontrivial filters to topologies). --Trovatore (talk) 07:41, 16 August 2010 (UTC)[reply]

On finite X, topologies are in 1-1 correspondence with preorders, and counting them appears to be a difficult enumeration problem. See [1].—Emil J. 13:34, 16 August 2010 (UTC)[reply]

Rings, subfields and integral domains

Hi all - I'm working on the following problem and was looking for some guidance:

"Let R be a ring, and K a subring of R which is a field. Show that if R is an integral domain and $dim_{K}(R)<\infty$ then R is a field."

The problems I'm working on are from a course lectured at my university about 10 years ago, so I don't have an explicit definition of the notation, but I presume $dim_{K}(R)$ is the dimension of R as a vector space over K - however, my first query is, is it necessarily the case that if K and R are related as described above, then R can -definitely- be treated as a vector space over K? I don't need a proof or anything, I just want to check that's definitely the case.

Secondly and more pertinently, where should I start? I don't want a complete answer and indeed I'd rather not have one, I want to work through it myself - but where to begin? I considered writing an n-dimensional basis for R; $e_{1},\,e_{2},...$ and then expanding a general $r\in R$ as $r=\sum _{1}^{n}r_{i}e_{i},\,r_{i}\in K$ , then finding a formula for the inverse, but I'm not sure if that's feasible or even a sensible way to go about the problem. Could anyone suggest a place to start?

Many thanks, 86.30.204.236 (talk) 19:45, 14 August 2010 (UTC)[reply]

You're over thinking the proof, Let r be in R and consider the set {1, r, r², ...}. R is finite dimensional so this set can't be linearly independent and there is a relation of the form a + br + cr² + ... + krⁿ. Assume n is minimal in this relation. Then divide by r once to get an expression for 1/r in R.--RDBury (talk) 21:46, 14 August 2010 (UTC)[reply]

Forgive me, but I'm not convinced that proof is entirely rigorous; I don't think you can just "divide by r", as you put it! In "dividing by r" you are multiplying by the inverse of r, which is the very thing you are trying to prove exists. This is pedantic, but instead you should state that as K is a field, you can take WLOG the constant term to be the multiplicative identity, then factorise to get something of the form 1 = r(b + cr + ... + kr²) to show that r has such an inverse before using it. Also, worth adding that the proof fails when R is not an ID as {1, r, r², ...} need not be an infinite set.--86.165.252.185 (talk) 21:30, 18 September 2010 (UTC)[reply]

Ah, I have a habit of going in over the top, whoops! Thankyou very much, that's great 86.30.204.236 (talk) 23:53, 14 August 2010 (UTC)[reply]

That R is a vector space over K is very simple too. A vector space is just an additive group with multiplication by elements of the underlying field. Since R is a ring, it is an additive group, and obviously you can multiple elements of R by elements of a subring of R. We don't immediately have a basis for R over K, but RDBury's proof doesn't require a basis. I'm not sure if there is any general form for a basis of a vector space of that type - does anyone know of one? --Tango (talk) 20:13, 15 August 2010 (UTC)[reply]

Well for the finite dimensional case, pick any element e₁ in R for your first basis element, and then pick any element e₂ not in Ke₁ for your second one, then pick any element not in the span of e₁ and e₂ for the third one, etc. Rckrone (talk) 06:07, 16 August 2010 (UTC)[reply]

Well, yes, that's the standard algorithm for finding a basis of any finite dimensional vector space. I was wondering if there was any more definite result for the special case of a ring over a subring that is a field. --Tango (talk) 14:13, 16 August 2010 (UTC)[reply]

There is no general way of describing the basis explicitly if the dimension is infinite. Think cases like R = R, K = Q. Any such basis is nonmeasurable, and cannot be proven to exist without the axiom of choice.—Emil J. 13:30, 16 August 2010 (UTC)[reply]