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Is this space completely metrizable? The metric that is inherited from $\mathbb R$ is not complete on $[0,1) \cup [2,3]$ since there are Cauchy sequences that do not converge, e.g. $x_n=1-\frac{1}{n}$.

This question is similar to the question of why $(0,1)$ is completely metrizable which was answered positively in the following thread...

Show that $(0,1)$ is completely metrizable

user41728
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2 Answers2

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$[2,3]\cup[4,\infty)$ with the subspace metric is complete. And $$ f(x) = \begin{cases} x & \text{when }x\in[2,3] \\ 3+\frac{1}{1-x} & \text{when }x\in[0,1) \end{cases} $$ is a homeomorphism from your space to mine.

hmakholm left over Monica
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  • I'm not the downvoter, but I guess what you want to say is that you can choose a metric on $[0,1)\cup [2,3]$ such that $f$ is a homeomorphism, right? – user159517 Jun 30 '17 at 19:11
  • @user159517: Being a homeomorphism doesn't depend on a metric, only on the topology. The topology is, of course, the one induced on $[0,1)\cup[2,3]$ by the subspace metric -- otherwise it would be a different topological space than the one the OP is talking about. – hmakholm left over Monica Jun 30 '17 at 19:16
  • Ah yes, I read the question too sloppily, I thought the goal was just to find a metric on this set such that one has a complete metric space. – user159517 Jun 30 '17 at 20:46
  • @user159517: The discrete metric would do that much quicker. :-) – hmakholm left over Monica Jul 01 '17 at 09:11
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Hint: A subspace of a completely metrizable space $X$ is completely metrizable if and only if it is $G_{\delta}$ in $X$.

Sahiba Arora
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