# Continuous bijection between compact and Hausdorff spaces is a homeomorphism

Exercise: Let $$f:(X,\tau)\to (Y,\tau_1)$$ be a continuous bijection. If $$(X,\tau)$$ is compact and $$(Y,\tau_1)$$ is Hausdorff, prove that $$f$$ is a homeomorphism.

I tried to prove this on the following way:

First I proved the following Lemma:

Lemma: If $$(X,\tau)$$ and $$(Y,\tau_1)$$are compact Hausdorff spaces and $$f:(X,\tau)\to(Y,\tau_1)$$ is a continuous mapping then $$f$$ is a closed mapping.

Proof: If $$A\in X$$ is compact than it is closed in $$(X,\tau)$$. Then if $$\{a_n:n\in\mathbb{N}\}$$ is an arbitrary sequence in A then by the compactness there is a subsequence that converges in A such that $$\lim_{n\to\infty}a_{in}=a$$ where $$a \in A$$. By continuity of $$f$$, $$\lim_{n\to\infty} f(a_{in})=f(a)$$ so that $$f(a)\in f(A)$$. So $$f(A)$$ is compact since the space $$(Y,\tau_1)$$ is compact then $$f(A)$$ is closed. So $$f$$ is a closed mapping.

In the Exercise the function is continuous so if $$B\in\tau_1$$ then $$f^{-1}(B)\in\tau$$, now it is left to show that $$f$$ send open sets to open sets. This is where my problem begins:

Compactness is going to be preserved by continuity of $$f$$, then $$(Y,\tau_1)$$ must be compact as every image of a subset of $$(X,\tau)$$ that would imply that $$f$$ is a closed mapping by the Lemma. If $$C$$ is a closed set in $$X,\tau$$ then $$f(X\setminus C)=X\setminus f(C)$$ which must be open. However I am not certain about this last step.

Question:

How should I solve the question? Is my proof right?

• Showing $f$ closed, starts by taking a closed $A \subseteq X$ (which is then compact as $X$ is compact). $A \in X$ makes no sense here. – Henno Brandsma Dec 16 '18 at 15:09
• Continuity is also that the inverse image of closed sets is closed, just as the inverse image of open sets must be open. – Henno Brandsma Dec 16 '18 at 15:10

The continuous image of a compact space is compact. We don't need sequences to see this; in fact sequences don't even suffice to see it, in general. The definition of compactness is by open covers, so use that:

If $$f:X \to Y$$ is continuous, $$A \subseteq X$$ is compact, then consider an open cover $$O_i, i \in I$$ of $$f[A]$$. Then $$f^{-1}[O_i], i \in I$$ is a cover of $$A$$ (by basic set theory) and an open cover as $$f$$ is continuous. So finitely many $$f^{-1}[O_i], i \in F$$ (so $$F \subseteq I$$ finite) exist that also cover $$A$$ and again simple set theory tells us that the $$O_i, i \in F$$ is a finite subcover of the original cover for $$f[A]$$. Hence $$f[A]$$ is compact.

The lemma then follows from the basic fact that if $$Y$$ is Hausdorff, and $$B \subseteq Y$$ is compact, then $$B$$ is closed in $$Y$$. This too is shown using open covers and the definition of Hausdorffness. Plenty of proofs can be found online.

Now if a bijection $$f: X \to Y$$ is closed, this is the same as saying its inverse map $$g: Y \to X$$ is continuous: $$g$$ is continuous iff $$g^{-1}[C]$$ is closed for all closed $$C \subseteq X$$. And $$g^{-1}[C] = f[C]$$ because $$g$$ is the inverse of the bijection $$f$$. As $$f$$ is a closed map by the lemma, you're done.

• What if X = Y is a finite set with same topology on both X and Y which is not Housdorff? – Balasubramannyan S Jun 26 at 12:14
• @BalasubramannyanS we need Hausdorff. – Henno Brandsma Jun 26 at 15:21
• Any counter example? – Balasubramannyan S Jun 29 at 2:39

Your proof that $$f$$ is closed is (a little) bad. Because you suppose to begin with a closed $$A$$ not a compact $$A$$. However this is not a big deal because a closed subset of a compact set is compact. Moreover the fact that $$f(A)$$ is compact in a compact space $$Y$$ doesn't necessarily means that is closed$$^1$$. For this you need the fact that $$Y$$ is Hausdorff (compact set in an Hausdorff space is closed).

About the second, you're right. A bijection that is also closed is necessarily open because $$f(X\backslash C) = Y\backslash f(C)$$.

To prove this you can show that each of the sets is included in the other. Let $$y\in f(X\backslash C)$$ then clearly $$y\in Y$$ but $$y\not \in f(C)$$ because $$f$$ is injective.

On the other hand if $$y\in Y$$ then since $$f$$ is onto there exists $$x\in X$$ such that $$f(x)=y$$. Moreover if $$y\not\in f(C)$$ then $$x\not\in C$$ again because $$f$$ is injective.

1. For example if $$Y$$ is equipped with the trivial topology it is always compact (and every subset of it is compact) but no non-trivial subset is closed.