# A property about open subsets of compact metric spaces.

I've been looking at various problems from past Topology exams, and I came across a problem dealing with compact metric spaces that I have never seen before. The statment to the problem is as follows:

Let $X$ be a compact metric space. Show every open subset of $X$ is homeomorphic to a compact metric space.

I'm having difficulty showing why this is true. Can anyone help? Thank you in advance!

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Isn't there some kind of typo? This would mean that every open subset of $X$ is compact. $(0,1)$ is an open subset of compact metric space $[0,1]$, but it is not compact. – Martin Sleziak Oct 23 '12 at 15:17
I don’t blame you for having difficulty: it’s false. Compactness is a topological property, preserved by homeomorphisms, so if $U$ is a non-compact open subset of $X$, $U$ cannot be homeomorphic to a compact metric space. Example: $(0,1)$ in $[0,1]$. – Brian M. Scott Oct 23 '12 at 15:18
I wrote the problem down from one of my old exams, so there might be a typo. I will dig through my old notes and try to find it again. – josh Oct 23 '12 at 15:31

Let $X = \{ (x,y) \in {\mathbb{R}}^{2} \colon -1 \leq x,y \leq 1 \}$ and $B = \{ (x,y) \in \mathbb{R}^{2} \colon x^{2} + y^{2} < 1 \}$ with the usual topology. The set $X$ is compact. The set $B$ is not compact. Suppose that $B$ is homeomorphic to some compact metric space $Y$. Let's say $f \colon B \rightarrow Y$ is the homeomorphism. Let $\mathcal{O}$ be an open cover of $B$ that does not have a finite subcover. Then $\{ f(O) \colon O \in \mathcal{O} \}$ is an open cover of $Y$. Since $Y$ is compact there exists a finite $\mathcal{F} \subseteq \{ f(O) \colon O \in \mathcal{O} \}$ that covers $Y$. But then $\{ f^{-1}(U) \colon U \in \mathcal{F} \}$ is a finite cover of $B$. This is a contradiction of there not being a finite subcover in $\mathcal{O}$.