Homeomorphism from $(0,1)$ to $\mathbb{R}$ 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!
 A: $\tan:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}$ 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 $h(t)=\tan(f(t))=\tan(-\frac{\pi}{2}+\pi t)$ works.
A: 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.
A: The function $$\frac x{\sqrt{1+x^2}}$$ should do nicely.
A: 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.    
A: $h:\mathbb{R}\to (0, 1)$  given by $h(x)=\dfrac{(1+x+|x|)}{2(1+|x|)}$ is a homeomorphism. 
A: This should work:
$$f:\mathbb{R} \to (0,1) \text{ s.t. } \forall x \in \mathbb{R}, f(x) = \frac{e^x}{1+e^x} $$
A: $(0,1)$ is an open subset of $\mathbb{R}$. Therefore it is a 1 dimensional manifold. Since it is not compact, it must be homeomorphic to $\mathbb{R}$.
