Talk:Second-countable space

Counterexample

I think it would be nice to add to the Introduction an example of a topological space that isn't second countable.

• Just take $\mathbb {R} \,$ with the discrete topology. A less trivial example would be an uncountable disjoint union of any space with at least one non-trivial open set. Albmont (talk) 22:02, 9 April 2008 (UTC)

There are so many spaces that are not second countable. Both your examples are quite trivial. Perhaps a slightly less trivial example would be the Sorgenfrey line or even an uncountable set with the finite complement topology, or even(!) the box product of R with itself countably (infinitely) many times.

Topology Expert (talk) 09:23, 4 November 2008 (UTC)

Necessary and sufficient condition for a manifold to embbed in some finite dimentional Euclidean space?

Is is true that second-countability is a necessary and sufficient condition for a manifold to embbed in some finite dimentional Euclidean space? 131.111.28.33 (talk) 15:06, 3 November 2008 (UTC)

YES (if you assume that 'manifolds' are locally Euclidean Hausdorff)!!! Try proving it yourself assuming that the manifold is compact (Hint: For each point in the manifold, choose a locally Euclidean neighbourhood. The collection of all neighbourhoods forms an open cover of the compact manifold and therefore has a finite subcover. Assuming that in a normal space, any finite open cover dominates a partition of unity, construct an imbedding). The proof for the non-compact case is a little less trivial.

Topology Expert (talk) 09:20, 4 November 2008 (UTC)

Choice

"Specifically, every second-countable space is separable (has a countable dense subset) and Lindelöf (every open cover has a countable subcover)."

Every second-countable space being separable is equivalent to countable choice, but does ZF prove that every second-countable space is Lindelof?

JumpDiscont (talk) 02:25, 13 October 2009 (UTC)

Could you please ask this question at Wikipedia:Reference desk/Mathematics? There, I am sure that quite a few people will answer your question in great detail. You may add information to the article if you feel that some ideas are not explained in depth. --PST 03:57, 13 October 2009 (UTC)

"Open quotient"

"Quotients of second-countable spaces need not be second-countable; however, open quotients always are."

What is meant by open quotient? Does this mean the equivalence classes used to make the quotient are all open? That would do it. I don't think this terminology is standard. 2001:468:D01:60:C62C:3FF:FE22:BEDC (talk) 00:29, 24 September 2013 (UTC)

Examples need clarification.

Under Examples is the following: "Define an equivalence relation and a quotient topology by identifying the left ends of the intervals - that is, identify 0 ~ 2 ~ 4 ~ … ~ 2k and so on. X is second countable, as a countable union of second countable spaces. However, X/~ is not first countable at the coset of the identified points and hence also not second countable." I am barely able to understand part of this and unable to understand the rest. I'm not a mathematician. I have never seen the word "identify" used the way it is here, and think it needs clarification. I am not sure - is 'identity' the exact same as 'equivalent'? One change I think would help would be "Define an equivalence relation ( ~ ) and..." What would be the effect if instead of "...by identifying the left ends..." it said "by defining the left end of each interval as equivalent - that is: 0 ~ 2 ~ 4 ~ ... ~ 2k. ??? It sure makes it clearer to me. Is there any value in using the term "identify" or is it unnecessary (and uncommon?) jargon. ----------- I also do not understand at all the notation X/~ the forward slash throws me - does this have anything to do with it being a quotient topology (which I have zero understanding of, another issue here). Does this mean the set X with the equivalence relation? Or something else? ( X\~ would be without the equivalence relation or the quotient of the equivalence relation (however absurd that sounds to me) ?) I understand two sets X\Y but not X/Y nor X\<operator>, nor X/<operator>. Anyone besides me think this needs clarification?Abitslow (talk) 05:14, 9 November 2013 (UTC)