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Does there exist a complete metric on $(0,1)$ inducing the usual topology?

My problem is that I cant understand what will I have to do to answer the question.It's a problem of a competitive exam.

Learnmore
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  • Do you know a complete metric space that is homeomorphic to $(0,1)$? – Daniel Fischer Oct 22 '14 at 15:12
  • A subset of $\mathbb R$ is complete iff it is closed.Is it possible to find so?Please correct me if I sound wrong.I have just started completeness. – Learnmore Oct 22 '14 at 15:14
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    Consider $d(x,y)= \left|\tan(\pi(x-\frac12))-\tan(\pi(y-\frac12))\right|$. And how did I come up with this? – Hagen von Eitzen Oct 22 '14 at 15:16
  • Note that completeness is not a topological property, it is a metric [more generally, uniform] property. While a subset of $\mathbb{R}$ is complete in the induced metric if and only if it is closed, subsets of $\mathbb{R}$ can be complete in other metrics inducing the topology without being closed. In fact, every open subset of $\mathbb{R}$ is completely metrisable. – Daniel Fischer Oct 22 '14 at 15:17
  • Exactly I was going to ask this please explain @HagenvonEitzen sir – Learnmore Oct 22 '14 at 15:18
  • @learnmore: A subset of $\Bbb R$ is complete under the standard metric iff it is closed. But here the problem is asking to change the metric. – Andrea Mori Oct 22 '14 at 15:18
  • Can anyone please help me to interpret the problem in simpler terms – Learnmore Oct 22 '14 at 15:20

1 Answers1

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The usual topology on $(0,1)$ just happens to give a space which is homeomorphic to $\mathbb{R}$, the reals. So, let's use this fact. Let $f\colon (0,1)\to \mathbb{R}$ be a homeomorphism and let $d\colon\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be the usual metric on the reals. Let's then define a metric $\delta\colon (0,1)\times(0,1)\to\mathbb{R}$ by $$\delta(x,y)=d(f(x),f(y)).$$ Prove that $\delta$ induces the usual topology on $(0,1)$ by using the fact that $f$ is a homeomorphism and $d$ induces the usual topology on $\mathbb{R}$.

Dan Rust
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  • What to show to prove $\delta $ induces usual topology any hints – Learnmore Oct 22 '14 at 15:43
  • Show that an open ball in $((0,1),\delta)$ is an interval $(a,b)$ with $0\leq a<b\leq 1$. Hint: the preimage of an open ball in $(0,1)$ under $f^{-1}$ will also be an open interval (possibly infinite if $a=0$ or $b=1$), and $f$ is necessarily monotonic. – Dan Rust Oct 22 '14 at 15:58