Wikipedia:Reference desk/Archives/Mathematics/2008 November 15

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November 15

Countable compactness => Lindelof (under what conditions)

This may be a silly question but please give me some sleep (and an answer too if possible):

Is there a countably compact, separable, first countable space that is not Lindelof?

Seperable, first countable spaces that are not Lindelof are easy to find (for instance the Sorgenfrey plane). But does adding countable compactness ensure the Lindelof condition? It seems (if the example I have in mind, is correct) that one can find a counterexample assuming the continuum hypotheses. Otherwise, the problem is a 'little' less trivial. I find it amazing that an 'unrelated axiom' (countable compactness) can make finding a counterexample a lot harder.

If this problem is so obvious and I have not been thinking properly a reference would be much appreciated (or a proof if it is a 'textbook theorem').

Thanks in advance.

(Some immediate 'narrowing down' can be done by eliminating topological groups and more importantly metric spaces but this 'narrowing down' is not that useful if my guess that the answer to the question is 'no', is correct).

Topology Expert (talk) 12:40, 15 November 2008 (UTC)

Have you checked Counterexamples in Topology? I'm in a different city from my copy at the moment. Algebraist 12:51, 15 November 2008 (UTC)

Yep, I checked there (first) but it is not there. Topology Expert (talk) 08:09, 16 November 2008 (UTC)

Novak space might be a possibility? Dmcq (talk) 10:20, 22 November 2008 (UTC)

Sounds good. According to CiT, the Novak space X is countably compact, but X x X isn't. It follows that X isn't quasicompact (Tychonoff). And unless I am being unusually dense, it follows directly from the definitions that a space is quasicompact iff it is both countably compact and Lindelöf. --Hans Adler (talk) 14:22, 22 November 2008 (UTC)

Thankyou for the responses. Novak space is trivially separable, countably compact and not Lindelof. The first countable bit is what worries me (obviously not second countable because it is not Lindelof). By the way, quasicompact and Tychonoff are different (non-equivalent properties). I think in recent years, the word compact (instead of 'quasicompact') is more common anyhow.

Topology Expert (talk) 00:54, 23 November 2008 (UTC)

Aargh. I didn't even see "first countable" in your question. Sounds like a question for Saharon Shelah, but I will think about it anyway. By "Tychonoff" I meant the Tychonoff theorem --- I wasn't familiar with the term Tychonoff space. I think compact vs. quasicompact is a cultural, rather than temporal, question. In Germany and particularly France and probably some other countries it's still very common to include Hausdorff in the definition of compactness. To avoid misunderstandings, in topology I only ever use "quasicompact" and "compact Hausdorff", except where both are equivalent. (Similar to using $\subsetneq$ and $\subseteq$, and reserving $\subset$ for cases where we know it's a proper subset, but we don't use the fact.) --Hans Adler (talk) 01:30, 23 November 2008 (UTC)

Number of terms

Could you please explain in detail how to deduce the formula to find the number of terms in the expansion of ( 1+x+x^2+x^3+... )^n (in simplified form)Kasiraoj (talk) 15:12, 15 November 2008 (UTC)

To what power of x is the part in brackets meant to go up to? If it goes on forever (which is what having "..." at the end usually means), then there will be infinitely many terms in the expansion. If you mean it to stop at a certain point, say x^m, then the expansion will have mn+1 terms. The lowest order term is going to be 1 (=1ⁿ) and the highest order term is going to be x^mn (=(x^m)^n), and there aren't going to be any terms missing inbetween since any integer between 0 and mn can be made by adding together n integers from 0 to m (just add together as many m's as you can without going over, then add on the amount necessary to get to the desired number, and then make the rest of them 0). --Tango (talk) 16:36, 15 November 2008 (UTC)

If $|x|<1$ then the sum of the geometric series inside the parenthesis is ${\frac {1}{1-x}}$ so the value of the expression in question can be found in closed form. Topology Expert (talk) 08:14, 16 November 2008 (UTC)

I guess that depends on whether you consider a Laurent polynomial to be a simpler form than a power series, which probably depends on context. --Tango (talk) 13:48, 16 November 2008 (UTC)