0
$\begingroup$

I just have started to learn basic topology and came across a question in my class note: Suppose that $U$ is a compact Hausdorff topological space and let $M$ be a countable infinite subset of $U$. Let $f:U\to V$ be a continuous function, where $V$ is a metric space. We know that $M$ must have a cluster point say $m$ in $U$.

What I am confused about is that :

There must exists a sequence $(u_n)$ in $M$ such that $f(u)$ is a sub-sequential limit of $(f(u_n))$.

What happens if we replace the compact by locally compact? What is the function of $U$ being Hausdorff?

Thanks in advance.

$\endgroup$
0
$\begingroup$

We use that $U$ is compact to show that $M$ has a cluster point. $\mathbb{R}$ with the standard topology is locally compact and Hausdorff, but if $M=\mathbb{Z}\subseteq\mathbb{R}$, then $M$ has no cluster point.

Of course, this doesn't directly apply to the part that you're confused about. But making $U$ compact is already a very strong condition! It implies, for example, that the range of $f$ is compact … so you should be able to take $U$ to be a compact metric space without loss of generality. Indeed, to make the above example more relevant, just let $U=V$, and take $f$ to be the identity.

(This sort of argument also shows that Hausdorffness is not in fact necessary — although it's tricky to concoct an interesting continuous function from a non-Hausdorff set to a metric space.)

$\endgroup$
2
  • $\begingroup$ Thankyou @JacobManaker for your answer. Here $U$ is a general compact topological space! why we should take it metrizable? $\endgroup$
    – user884919
    Apr 11 at 7:44
  • $\begingroup$ @user884919: My point is that you can replace $(f,U)$ with $(\text{id},f(U))$ in the problem, and then $f(U)$ is a subset of a metric space (and thus metric). $\endgroup$ Apr 11 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy