Talk:First-countable space

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Please can we have a nice simple example of a space which fails to be first-countable? Lupin 15:21, 6 Sep 2004 (UTC)

[edit] Malformed wiki bracket or math notation?

Hi, I know zip about this subject, and was directed here by a project to help clean up malfomred Wiki brackets. the example here was this - [0,ω1). I am not sure if this is normal notation for something like this or not, but in any case I changed the [ to a (. If it's supposed to be a [, go ahead and change it back. Thanks! --Sparky the Seventh Chaos 20:21, Oct 30, 2004 (UTC)

[edit] Use of 'counterexample'?

I am a bit concerned about the use of the term counterexample in the second section. I think of a counterexample as being an example which shows that a proposed result is false. In this case, there is no 'theorem' or propsed result - so are we not just talking about "examples" (of non first-countable spaces)? If I am being over-pedantic, please ignore this comment! Madmath789 21:54, 18 May 2006 (UTC)

I suppose you could argue that it's a counterexample to the statement that all topological spaces are first-countable. -- Fropuff 02:56, 19 May 2006 (UTC)

[edit] Uncountable products

Have we got an example of an uncountable product of first-countable spaces that is not itself first-countable? -GTBacchus(talk) 03:44, 3 November 2007 (UTC)

Sure: ZR, RR, and II are all not first countable, although each factor is. -- Fropuff 06:42, 3 November 2007 (UTC)
How can you see that the product spaces aren't first countable? (Sorry if this is a dumb question.) -GTBacchus(talk) 19:27, 3 November 2007 (UTC)
Not a dumb question. Actually, one can show that a product of first-countable spaces is first-countable if and only if all but countably many of the factors have the trivial topology. Let Y be a product of uncountably many spaces X_a. For any open set U in Y we have p_a(U) = X_a for all but finitely many a (here p_a is the projection onto the ath factor). Given any countable family {U_i} of open sets in Y, one can always find an a such that p_a(U_i) = X_a for all i and X_a is not trivial (there are only countably many a for which this won't be true). Take any proper neighborhood V of x in X_a. It's preimage in Y won't contain any U_i, so {U_i} cannot be a basis. -- Fropuff 02:24, 5 November 2007 (UTC)