# Show that (0, oo) is homeomorphic to (0, 1)

• I
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!

PeroK
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2020 Award
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?

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

PeroK
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Gold Member
2020 Award
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.

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.

$$(0,1) \xrightarrow[]{f} (0,\pi/2) \xrightarrow[]{g} (0,\infty)$$
Put
$$f(x) = \frac{\pi}{2}x \quad\mbox{and}\quad f^{-1}(x) = \frac{2}{\pi}x$$
also
$$g(x) = \tan x \quad\mbox{and}\quad g^{-1}(x) = \arctan x$$
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.

Last edited:
dextercioby
pasmith
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Simpler examples: \begin{align*} x &\mapsto -\ln x \\ x &\mapsto x^{-1} - 1 \\ x &\mapsto x/(1 - x) \\ x &\mapsto \operatorname{arctanh}(x) \end{align*}

dextercioby