4
$\begingroup$

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!

Improve this question
$\endgroup$
  • $\begingroup$ 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
13
$\begingroup$

$\tan:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}$ 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.

Improve this answer
$\endgroup$
12
$\begingroup$

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.

Improve this answer
$\endgroup$
9
$\begingroup$

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

Improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 5
    $\begingroup$ 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 '20 at 1:16
1
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.