1
$\begingroup$

The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.

While studying Topological Spaces, I came across metrizable spaces. If I understand this correctly, A metrizable space is a topological space that is generated by a metric space. Thus metrizable spaces inherit topological properties from metric spaces by definition (if I am not mistaken).

Theorem 4.7 in the book states "Every metric space is first countable" (Croom 110); however, the author does not proof why this is true.

I want to prove the following:

Prove that a metrizable space is also first countable.

Since metrizable space inherit properties from metric spaces, I would assume my proof would be as follows:

Let $(X, \mathscr{T})$ be a metrizable topological space. By definition, the space is generated by a metric. Since every metric space is first countable by Theorem 4.7, then the metrizable space is first countable.

Is this sufficient?

Whether it is sufficient (or not), I am still extremely curious as to why every metric space is first countable. I would like to proof this and add it to my proof above as well. The only thing that comes to mind to show a metric space is first countable is by showing a collection of open balls $\{B(a,1/n)\}_{n=1}^{\infty}$ is a countable local basis at $a$ for each point $a$ in a metric space.

Am I on the right track and how would I proceed?


Thank you for taking the time to read this question. I greatly appreciate any assistance you may provide.

$\endgroup$
  • 4
    $\begingroup$ Yes, a space is metrizable if its topology can be generated by some metric. And yes, the first countability of metric spaces follows from the observation that $\{B(x,1/n):n\in\Bbb Z^+\}$ is a local base at $x$ for each point $x$ of a metric space $\langle X,d\rangle$. $\endgroup$ – Brian M. Scott Mar 16 '15 at 6:47

Your Answer

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

Browse other questions tagged or ask your own question.