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)

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 - ${\displaystyle [0,\omega _{1})}$. 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)

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)Reply

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)Reply

Could we also add more examples of first countable spaces? We currently have only 1 class of examples, namely metric spaces, but 3 non-examples. Oh there you go Fropuff, you can use the word non-example. —Preceding unsigned comment added by Setitup (talk • contribs) 22:48, 26 March 2010 (UTC)Reply

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)Reply

Sure: Z^R, R^R, and I^I are all not first countable, although each factor is. -- Fropuff 06:42, 3 November 2007 (UTC)Reply

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)Reply

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)Reply

This could be added to the counter-examples section. It's generic enough to not be obscure. 67.198.37.16 (talk) 21:52, 1 December 2023 (UTC)Reply

In Properties is said (correctly) that in firs-coutable spaces sequential and countable compactness are equivalent. Then is said However, there exist examples of sequentially compact, first-countable spaces which aren't compact (these are necessarily non-metric spaces). what means here compact (sure nor sequentially or countable)? This is not compatible with the definition in Compactness Page, where Compact is defined as Countable Compact. --MeMedesimoMeco (talk) 13:11, 8 June 2011 (UTC)Reply

Id like to ask if there should be added that in sequence from "That is, for each point x in X there exists a sequence N1, N2, … of neighbourhoods of x such.." all Ni's should be non empty. Theonepercentrule (talk) 12:28, 5 June 2019 (UTC)Reply

The article switches between spellings, even in the first paragraph. This is distracting. 14:08, 1 November 2023‎ User:Cerberus0