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The question I am trying to solve is the following from Metric Spaces: A Companion to Analysis:

  1. Let $X$ be a complete metric space. Let $A$ be a nonempty set in $X$. Show that $A$ is homeomorphic to a complete metric space. Hint: One way is to define $g : A \to X \times \mathbb{R}$, $g(x) = \left(x, \frac{1}{\operatorname{dist}(x, A^c)}\right)$

I try using the hint that the author has provided and tried to show that $g$ is a homeomorphism.

  • One-to-one: If we take $x = y \in A$, then $\left(x, \frac{1}{\operatorname{dist}(x, A^c)}\right) = \left(y, \frac{1}{\operatorname{dist}(y, A^c)}\right)$
  • Onto: Since $g(A) = X \times \mathbb{R}$, the image of $A$ is the same as $X \times \mathbb{R}$ so $g$ is onto. (I am a little bit unsure if this would be the correct justification...)
  • $g$-continuous and $g^{-1}$-continuous: I am having a little bit of trouble trying to show that both $g$ and $g^{-1}$ are continuous.

I appreciate the help!

Joshua
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1 Answers1

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As the comment from Lorago suggests, this exercise is asking you to prove something that is false, as the counterexample of $\mathbb Q\subseteq \mathbb R$ shows.

Based on the hint, however, it is highly likely that $A$ is meant to be open. In fact, the second coordinate of the map described in the hint will be finite everywhere if and only if $A$ is open, so this is the only context in which the hint makes sense.

Supposing that $A$ is assumed to be open, the claim is indeed true. In this case, (assuming with no loss of generality that $A\neq X$, as otherwise the proof is trivial) the map $g$ from the hint is well-defined, injective, and continuous (since each coordinate function is continuous), if $X\times \mathbb R$ is given the product topology.

Also as discussed in the comments, $g$ does not map surjectively onto $X\times \mathbb R$, but this is no matter, as it still maps surjectively onto its image $g(A)$ (as every map does).

Moreover, $g^{-1}\colon g(A)\to A$ is given by projection onto the first coordinate, which is continuous. Therefore $g$ is a homeomorphism onto its image.

It remains to show that $g(A)$ has a complete metric. To see this, give $X\times \mathbb R$ the metric $d((x_1,r_1),(x_2,r_2))=\sup(d(x_1,x_2),|r_1-r_2|)$. It is straightforward to verify that $X\times \mathbb R$ is complete with this metric (just use completeness in each coordinate), and that this metric induces the product topology.

Finally, observe that if $(x_n,r_n)\in g(A)$, and $(x_n,r_n)\to (x,r)$, then since $x_n\in A$ and $r_n=\frac{1}{d(x_n,A^c)}\to r\neq \infty$, we have $d(x_n,A^c)\not\to 0$, so $d(x,A^c)\neq 0$, whereby $x\notin A^c$ (since $A^c$ is closed), so that $x\in A$.

We also have $r=\frac{1}{d(x,A^c)}$ by continuity of the distance function, so that $(x,r)=g(x)\in g(A)$. Therefore $g(A)$ is a closed subset of a complete metric space, and is thus itself complete under this metric.

M W
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