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?
Best Answer
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 $y$.