# The image of first countable space

I'm trying to show that the continuous first countable image of a space does not have to be first countable, but the continuous open image of first countable space is first countable.

So I let $X$ be a discrete topology so that any function defined on $X$ is continuous. Also let X be first countable space, and Y a space and let $f:X \to Y$ be a continuous function.

Let $\mathscr B_x$ be a local base (countable nbh base of $x$), and pick $y\in Y$ such that $x\in f^{-1}(y)$.

If I let $W$ be a neighborhood of $y$, then what relationship to the space $X$ I can get from here, since it's only a nbh?

If I let $V$ be an open neighborhood of $y$, then since $f$ is continuous, I can get an open neighborhood $B$ of $x$ such that $f(B)\subset V$, right?

Can I get a set $\mathscr {V}=\{f(B): B\in \mathscr{B}_x\}$ to be a neighborhood base of $y$ from here?

• Which part of the question are you trying to answer there at the end? Apr 6 '13 at 22:41
• the second part, but how do I show the first part, it's kinda obvious? Apr 6 '13 at 22:57
• I’ll write up an answer with some hints. Apr 6 '13 at 22:58

You have a reasonable idea for the first part, but you need actually to carry it out. Specifically, you need a space that isn’t first countable to use as $Y$, the codomain of your function. A very simple example is any uncountable set with the cofinite topology. And for $X$ you can then take the discrete topology on the same set. What should you use then for your continuous function $f$?
You’re on the right track for the second part. Start with an arbitrary $y\in Y$ and any $x\in X$ such that $f(x)=y$, and let $\mathscr{B}_x$ be a countable base at $x$, exactly as you’ve done. Let $$\mathscr{V}_y=\{f[B]:B\in\mathscr{B}_x\}\;,$$ since $f$ is open, you know that $\mathscr{V}_y$ is a family of open nbhds of $y$, whether it’s a base or not. Now try to prove that it is a base at $y$. Start with an arbitrary open nbhd $U$ of $y$.
• What does continuity of $f$ tell you about $f^{-1}[U]$?
• Now use the fact that $\mathscr{B}_x$ is a base at $x$.
• @IvordesGreenleaf: Exactly: use one uncountable set, give it the discrete topology for $X$, and give it the cofinite topology for $Y$. Apr 6 '13 at 23:51
• I ended up having that there exists $f(B)\in \mathscr V_y$ such that $y\in f(B)\subset U$, and that should do it right. Apr 7 '13 at 1:08
• @IvordesGreenleaf: It does indeed. The crucial facts here are that $f[B]$ is open, and that if $B\subseteq f^{-1}[U]$, then $f[B]\subseteq U$. Apr 7 '13 at 1:12