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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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    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.2012-10-23

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Can't be shown.

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 $\mathcal{O}$.

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    Thank you Jay for the formal argument. I copied the problem incorrectly. I meant complete instead of compact. I posted that question here: http://math.stackexchange.com/questions/219706/open-subsets-of-a-complete-metric-space2012-10-23