Does a second countable space have a converging sequence in its closure?

[R136a1]

Hello everybody!

I hope I'm posting in the correct forum. Apparently, I can post questions from grad books in this forum, so I decided to post here!

The topology book I'm using asks me to prove that if $X$ is a second countable topological space, if $x\in \overline{A}$, then there exists a sequence $(x_n)$ in $A$ converging to $x$.

I'm pretty lost at how to prove something like this. Any hint would be appreciated.

 

[WannabeNewton]

Welcome to the forum friend! Actually that is true for any first countable space $X$; $X$ doesn't have to be second countable so it's weird that your book would restrict the proof to that; let's prove it assuming only first countable. Let $p\in \bar{A}$ then every neighborhood of $p$ contains a point of $A$. Also note that since $X$ is first countable by hypothesis, there exists a nested neighborhood basis for $p$ i.e. there exists a sequence $(U_{i})_{i = 1}^{\infty}$ of neighborhoods of $p$ such that $U_{i+1}\subseteq U_i,\forall i$, on top of the usual neighborhood basis property that every neighborhood of $p$ must contain $U_{i}$ for some $i$. Use these two facts to construct a sequence in $A$ converging to $p$.

 

[R136a1]

Oh right, the book says first countable. I was trying to write down the theorem from memory, so apparently that failed, haha

Thanks for the hint. It was helpful. I know the proof of the relevant theorem in the real numbers. In that case, you can build a sequence $(x_n)$ because for any $n$, we can find a $y\in A$ such that $|x-y|<1/n$. So we just let $x_n$ be that element in $A$.

So it appears that we can do the same thing in a first countable space now. We just take $x_n\in U_n$. Then it's rather easy to see that $x_n$ converges to $x$ (because it's a neighborhood basis).

The crucial part was that you said there is a nested neighborhood basis. I didn't know that, although it seems obvious now.

So, this also shows the theorem for metric spaces since every metric space is first countable right? Not that it matters here, but just out of curiosity, is any metric space also second countable? I guess not since the proof that $\mathbb{R}$ is second countable requires the rational numbers, but I can't seem to found a decent counterexample.

Thanks again!

 

[WannabeNewton]

Yep that's all you have to do! Nice! The standard definition of first countable does not include the nested neighborhood basis property itself but it can be proven from said standard definition, which you can prove yourself because as you said it is easy to see. As for what you said, yes every metric space is first countable but no every metric space is not second countable however for metric spaces second countable, separable, and Lindelof are all equivalent. A trivial example of a metric space that is not second countable is an uncountable set endowed with the discrete metric.

 

[R136a1]

Alright, thanks! That counterexample is a bit easier than I expected. It's a shame I didn't find it myself

By the way, is the statement in my first post equivalent to first countable? Do you happen to know anything about that?

 

[WannabeNewton]

Not necessarily. See here: http://en.wikipedia.org/wiki/Sequential_space

 

[R136a1]

Thanks! That's very interesting. I don't know enough topology yet to understand everything though, but I hope that will change soon!