Is the image of a continuous open map still satisfies countability axioms?

[Chris L T521]

Here's this week's problem!

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Problem: Let $f:X\rightarrow Y$ be a continuous open map. Show that if $X$ satisfies the first or the second countability axiom, then $f(X)$ satisfies the same axiom.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

No one answered this week's problem. You can find the solution below.

[sp]Proof: WLOG, assume that $Y=f(X)$ so that $f$ is surjective.

First countable: Let $y\in Y$. Then there is an $x\in X$ such that $f(x)=y$ since $f$ is surjective. Let $\{U_n\}_{n\in\Bbb{N}}$ be a local countable base for $x$. Then it follows that $\{f(U_n)\}_{n\in\Bbb{N}}$ is a countable local base for $y$ and hence $Y$ is first-countable.Second countable: Let $\{U_n\}_{n\in\Bbb{N}}$ be a countable base for $X$. Then $\{f(U_n)\}_{n\in\Bbb{N}}$ are all open (since $f$ is open) and are a base for $Y$. Thus, for any open set $O\subseteq Y$, we can express it's preimage $f^{-1}(O)$ as a union of some $U_n$, the same-indexed $f(U_n)$ union up to $O$. Therefore, $Y$ is second-countable.This completes the proof.$\hspace{.25in}\blacksquare$[/sp]