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Here, under the Section 'Examples ', $(0,1)$ is not complete with its usual metric inherited from $\mathbb{R}$, but it is completely metrizable since it is homeomorphic to $\mathbb{R}$.

To show $(0,1)$ is not complete with its usual metric, define a sequence such that $a_n=\frac{1}{n}$ for all $n \in \mathbb{N}$. Clearly the sequence is Cauchy but it does not have a limit in $(0,1)$, hence $(0,1)$ is not complete.

I'm having trouble to prove the open set is completely metrizable by defintion. I don't know how to construct such a metric. Can someone give some hints?

Idonknow
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1 Answers1

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HINT: Find a homeomorphism $f:(0,1)\to\Bbb R$, and define a metric $d$ on $(0,1)$ by

$$d(x,y)=|f(x)-f(y)|\;.$$

(Of course you’ll have to prove that this is a metric on $(0,1)$ and that it generates the right topology.)

Brian M. Scott
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  • May I know what is the intuition of constructing such metric? I have verified that the open set equips with the metric is indeed complete. – Idonknow Dec 03 '14 at 05:13
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    @Idonknow: It’s just using a homeomorphism to transfer a metric from one space to another. The general principle is that if $f:X\to Y$ is a homeomorphism, where $\langle Y,d\rangle$ is a metric space, then $$\rho(x_0,x_1)=d\big(f(x_0),f(x_1)\big)$$ is a metric on $X$ compatible with the topology. Use the homeomorphism $f$ to push the points over to $Y$, find the distance between them over in $Y$, and call that the distance between them in $X$. – Brian M. Scott Dec 03 '14 at 05:16
  • Can I say that $(X,\rho)$ obtained above is always complete? – Idonknow Dec 03 '14 at 06:18
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    @Idonknow: Only if $\langle Y,d\rangle$ is. – Brian M. Scott Dec 03 '14 at 06:19