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Can anyone construct a complete metric on $(0,1)$ which induces the usual subspace topology on $(0,1)$ ?

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    $\begingroup$ Hint: What complete metric space is $(0,1)$ homeomorphic to? $\endgroup$ – Ted Shifrin Dec 16 '13 at 3:59
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    $\begingroup$ What is the obvious space $(0,1)$ is homeomorphic to? What have you tried and where are you stuck? $\endgroup$ – Ted Shifrin Dec 16 '13 at 4:19
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    $\begingroup$ @TedShifrin : I see that $\mathbb{R}$ should be the choice.. I mean $\mathbb{R}$ is complete and homeomorphic to $(0,1)$.. Now... what should be the next step :O $\endgroup$ – user87543 Dec 16 '13 at 6:13
  • $\begingroup$ @TedShifrin : Sir, look at this.... May day... May day... :) $\endgroup$ – user87543 Dec 18 '13 at 9:33
  • $\begingroup$ Sorry. I missed your earlier response. Yes, it's right. Write down a homemorphism $f$ from $(0,1)$ to $\Bbb R$ and use $f$ and the metric on $\Bbb R$. $\endgroup$ – Ted Shifrin Dec 18 '13 at 12:11
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The map $$ f:(0,1)\to\mathbb{R}:x\mapsto\tan\pi\left(x-\frac{1}{2}\right) $$ is a bijection which allows you to define the metric $$ d(x,y)=|f(x)-f(y)| $$ which makes $((0,1),d)$ complete. Since $f$ maps intervals to intervals then both topologies are equivalent.

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Define the function $f$ on $X=(0,1)$ by $$f(x)=\dfrac{2x-1}{x(1-x)}=\dfrac1{1-x}-\dfrac1x, $$ for every $x$ in $X$, and use $f$ to define a metric $d$ on $X$ by $$d(x,y)=|f(x)-f(y)|,$$ for every $x$ and $y$ in $X$. Then the metric space $(X,d)$ is complete, and its topology is the usual one.

Every function $f:X\to\mathbb R$ would do as soon as $f$ is continuous, increasing, and has limits $-\infty$ at $0$ and $+\infty$ at $1$.

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More generally, if $O$ is an open subspace of a Polish space $X$, then $O$ is a Polish space. An argument is to consider the inclusion $$\left\{ \begin{array}{ccc} O & \to & X \times \mathbb{R} \\ x & \mapsto & \left( x, \frac{1}{d(x,X \backslash O)} \right) \end{array} \right..$$

Then $O$ is closed in $X \times \mathbb{R}$ (roughly speaking, $\partial O$ is "pushed to the infinity" in $X \times \mathbb{R}$) and $X \times \mathbb{R}$ is completely metrizable using the distance $$((x_1,y_1),(x_2,y_2)) \mapsto \max (d(x_1,x_2), |y_1-y_2|).$$

For the specific example $X= \mathbb{R}$ and $O= (0,1)$, it gives the distance $$(x,y) \mapsto \max \left( |x-y| , \left| \frac{1}{\min(x,1-x)}- \frac{1}{\min(y,1-y)} \right| \right).$$

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