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I want to show that $(0,1)$ is homeomorphic to $\mathbb{R}$ by finding a homeomorphism between the two. I think the function will be related to $tan(x)$ but I'm stuck on how to modify it to fit the domain $(0,1)$.

Any help would be appreciated!

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  • $\begingroup$ all you need to do is compress the real numbers into a finite length. Then it is trivial. $\endgroup$ – Jacob Wakem Jul 10 '16 at 22:07
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$\tan:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}$ is a homeomorphism between $(-\pi/2,\pi/2)$ and $\mathbb{R}$. Define $f:(0,1)\to(-\pi/2,\pi/2)$ by $f(t)=-(1-t)\frac{\pi}{2}+t\frac{\pi}{2}=-\frac{\pi}{2}+t\pi$. Then, $f$ is a homemorphism between $(0,1) $ and $(-\pi/2,\pi/2)$. Therefore, $h:(0,1)\to\mathbb{R}$ given by $h(x)=\tan(f(x))=\tan(-\frac{\pi}{2}+t\pi)$ works.

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Would an answer not involving the tangent function fit your needs? If so, try this:

$f:\mathbb{R}\to (-1,1),\quad x\longmapsto \dfrac{x}{1+\mid x\,\,\mid}\,\,$. It's a homeomorphism (check!).

$g:(-1,1)\to (0,1),\quad x\longmapsto\dfrac{x+1}{2}\,\,.$ It's a homeomorphism (check this too!).

Now, set $h:\mathbb{R}\to (0,1)$ by the equation $h(x)=g(f(x))$ for all $x\in\mathbb{R}$. It's a homeomorphism as a compose of two such functions.

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The function $$\frac x{\sqrt{1+x^2}}$$ should do nicely.

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  • $\begingroup$ This has same problem as $\tan$. Does not map $(0,1)$ to $\mathbb R$, so needs a change of variables. $\endgroup$ – GEdgar May 17 '16 at 14:32
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    $\begingroup$ It goes the opposite way. The inverse function is $\frac x{\sqrt{1-x^2}}$. $\endgroup$ – Lubin May 17 '16 at 17:07
  • $\begingroup$ this is my favorite map from the reals into open unit. I am not a fan of the trig ones. thanks for this. $\endgroup$ – Hossien Sahebjame Jul 21 at 1:16
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Wrap the interval into a semicircle in R^2 and map each point of the semicircle to the intersection of the diameter through that point with R^1.

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$h:\mathbb{R}\to (0, 1)$ given by $h(x)=\dfrac{(1+x+|x|)}{2(1+|x|)}$ is a homeomorphism.

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