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I've been going over previous exams, and I came across a question that I missed. It is as follows:

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

I am having difficulty showing this. Any help would be greatly appreciated.

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Here’s an outline to get you started. Let $U$ be a proper open subset of $X$, and let $d$ be the given metric on $X$. Define

$$f:U\to\Bbb R:x\mapsto\frac1{d(x,X\setminus U)}\;,$$

and show that $f$ is continuous. Now define

$$\rho:U\times U\to\Bbb R:\langle x,y\rangle\mapsto d(x,y)+|f(x)-f(y)|\;,$$

and show that $\rho$ is a metric on $U$ that generates the same topology as $d$. Then show that $\langle U,\rho\rangle$ is complete.

The intuitive idea is that we want to kill off Cauchy sequences in $U$ whose limits are not in $U$; $f(x)$ is large when $x$ is close to the boundary of $U$, so adding the extra $|f(x)-f(y)|$ term stretches the metric and keeps sequences converging to points outside of $U$ from being $\rho$-Cauchy.

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  • $\begingroup$ @josh: I’m not sure why your edit was rejected: it wasn’t quite right, since the correct arrow there is plain, but it was certainly better than the typo that I had! $\endgroup$ Oct 23 '12 at 23:01
  • $\begingroup$ Haha, thanks. I realized that I put in the wrong type of arrow after I submitted it. $\endgroup$
    – josh
    Oct 23 '12 at 23:47

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