Show that (0, ∞) is homeomorphic to (0, 1)

[Mikaelochi]

TL;DR

(1) I need to find a function that maps (0, ∞) to (0, 1) or vice a versa. (2) Show f is a bijection (3) Show that f is continuous (4) that the inverse of f is continuous

So, I already have a function in mind: tan(pi*x - pi/2) that maps (0, 1) to (0, oo). I just forget how to rigorously show that a function is continuous. I was hoping to get some help on showing that this tangent function I just wrote is continuous (not the topological definition, just like the real analysis definition). Rigorously that is. Thanks!

 

[PeroK]

Summary:: (1) I need to find a function that maps (0, oo) to (0, 1) or vice a versa. (2) Show f is a bijection (3) Show that f is continuous (4) that the inverse of f is continuous

So, I already have a function in mind: tan(pi*x - pi/2) that maps (0, 1) to (0, oo). I just forget how to rigorously show that a function is continuous. I was hoping to get some help on showing that this tangent function I just wrote is continuous (not the topological definition, just like the real analysis definition). Rigorously that is. Thanks!

Are you sure that function works?

You mean an epsilon-delta proof that $\tan$ and $\arctan$ are continuous?

 

[Mikaelochi]

Yeah I mean an epsilon-delta proof that tan and arctan are continuous.

 

[PeroK]

Yeah I mean an epsilon-delta proof that tan and arctan are continuous.

I'm tempted to say that with a problem at this level you may assume the continuity of trig functions. Otherwise, you could use some trig identities to crank out a formal proof.

 

[nuuskur]

You have a composition of continuous maps, which is continuous. Can prove this fact in general using epsilon-delta trickery. If you want bijections, there is a good way of getting those via composition.

<br /> (0,1) \xrightarrow[]{f} (0,\pi/2) \xrightarrow[]{g} (0,\infty)<br />
Put
<br /> f(x) = \frac{\pi}{2}x \quad\mbox{and}\quad f^{-1}(x) = \frac{2}{\pi}x<br />
also
<br /> g(x) = \tan x \quad\mbox{and}\quad g^{-1}(x) = \arctan x<br />
Verify you do have inverses i.e $g\circ g^{-1}$ and $g^{-1}\circ g$ are the identities. A homeomorphism would then be $g\circ f$. As for continuity, note that in $(0,\infty)$
|\arctan a - \arctan b|\leqslant |\arctan (a-b)| \leqslant |a-b|
Lipschitz maps are continuous.

 

[pasmith]

Simpler examples: <br /> \begin{align*}<br /> x &amp;\mapsto -\ln x \\<br /> x &amp;\mapsto x^{-1} - 1 \\<br /> x &amp;\mapsto x/(1 - x) \\<br /> x &amp;\mapsto \operatorname{arctanh}(x)<br /> \end{align*}<br />

 

[JFerreira]

Summary:: (1) I need to find a function that maps (0, oo) to (0, 1) or vice a versa. (2) Show f is a bijection (3) Show that f is continuous (4) that the inverse of f is continuous

So, I already have a function in mind: tan(pi*x - pi/2) that maps (0, 1) to (0, oo). I just forget how to rigorously show that a function is continuous. I was hoping to get some help on showing that this tangent function I just wrote is continuous (not the topological definition, just like the real analysis definition). Rigorously that is. Thanks!

Useful functions to handle with similar question is $$f(x)=\frac{x}{1-|x|}$$ and $$g(x)=\frac{x}{1+|x|},$$ with some change of variable. Note that $x$ can be a vector, like one in $\mathbb{R}^2$, for instace. Try to search to "\(\frac{x}{1-|x|}\)" on SearchOnMath.