# Continuous bijection is a homeomorphism

Let $X$ be a Hausdorff countably compact space and $Y$ first countably. If $f:X\to Y$ is a continuous bijection then it is a homeomorphism.

Like in the case of compact spaces, I'm trying to show $f$ is closed.

If $A\subseteq X$ is closed, $A$ es countably compact and so is $f(A)$. I want to define a countable open cover for $f(A)$ by using $Y$ is first countably, but I don't know how to do this.

Would you give me a hint?

• This isn't true (consider when $X$ and $Y$ have two points); are you sure you have the hypotheses right? Maybe you want to assume $Y$ is Hausdorff as well? – Eric Wofsey Oct 17 '15 at 1:29
• @EricWofsey this is an exercise from Dugundji's Topology but I already see the mistake. Thank you. – Tanius Oct 17 '15 at 2:32

For this statement to be true, you want to require $Y$ to be Hausdorff, not $X$. Given that correction, the following should be helpful.
Here's one formulation of the proof in the compact case. Suppose you have a point $x$ which is not in $f(A)$. For each $y\in f(A)$, choose disjoint open neighborhoods $U_y$ and $V_y$ containing $x$ and $y$. By compactness, there are finitely many $y_1,\dots,y_n$ such that $V_{y_1},\dots,V_{y_n}$ which cover $f(A)$. The intersection $\bigcap_{i=1}^n U_{y_i}$ is then a neighborhood of $x$ disjoint from $f(A)$. Since $x\not\in f(A)$ was arbitrary, this means $f(A)$ is closed.
To modify this for your situation, try requiring all the sets $U_y$ to be members of some fixed countable base at $x$, and then appropriately combining the sets $V_y$ into only countably many open sets which still cover $f(A)$.
• how well is this? Let $\mathcal{B}(x)$ a countable base of $x$. Let $U_y$ and $V_y$ as above. For every $y\in Y$, let $B_y\in\mathcal{B(x)}$ such that $B_y\subseteq U_y$. Define $y_1\sim y_2$ iff $B_{y_1}=B_{y_2}$. Pick one element of each class and call $Y'$ the set of all these elements. $Y'$ es countable and I'm still trying to prove $\{V_y:y\in Y'\}$ covers $f(A)$. Is this true? – Tanius Oct 17 '15 at 2:30
• It won't work to just pick one element of each class--there's no reason to expect those to cover all of $f(A)$. – Eric Wofsey Oct 17 '15 at 2:36
• You mean we may take a countable subcover of $\{V_y:y\in Y\}$ or need to define new open sets? – Tanius Oct 17 '15 at 3:31
• There's no reason to expect that there exists a countable subcover, so try defining some new open sets in terms of the $V_y$'s. (The equivalence relation you defined before was a step in the right direction; you just used it in the wrong way.) – Eric Wofsey Oct 17 '15 at 4:22