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For example how would I find a homeomorphism from $(0, 1)$ to the reals and one from $(0, 1)$ to $(0, ∞)$?

I know the definition of a homeomorphism is if a function is both one-to-one, onto, continuous, and has a continuous inverse. However, I do not know how to craft my own to fit these standards from the maps above.

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  • $\begingroup$ Do you know any functions $f(x)$ that are continuous bijections on some open interval $(0,b)$ which has a pole at $b$? If so you can you change it to $f(bx)$ and it will work on the interval $(0,1)$. You probably know of one such function from trigonometry. Can you find it? $\endgroup$ Commented Mar 15, 2021 at 17:22
  • $\begingroup$ In general it is a lot harder to prove that two spaces are homeomorphic than proving that two spaces are not homeomorphic. $\endgroup$
    – freakish
    Commented Mar 15, 2021 at 18:39

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There are some basic real homeomorphisms which we can use as building blocks to construct more with various desired properties. The most useful are:

  • Linear functions $f(x) = ax+b$ map interval endpoints to different interval endpoints.
  • $\exp : \mathbb{R} \to \mathbb{R}^{+}$ and its inverse $\ln$ go between the full set and one-ended open intervals.
  • $\tan : (-\frac{\pi}{2}, \frac{\pi}{2}) \to \mathbb{R}$ and its inverse $\arctan$ go between bounded and unbounded open intervals. (For a one-ended open interval, note $\tan : (0, \frac{\pi}{2}) \to \mathbb{R}^{+}$.)

Composing any two homeomorphisms gives another homeomorphism. (If you haven't seen the proof, it's straightforward from definitions.)

For the first example, to map from $(0,1)$ to the reals, $\tan$ has similar properties, but the wrong interval endpoints. So let's compose it with a linear map $L(x) = ax+b$, and see if we can get $f = \tan \mathrel{\circ} L$ with the desired domain and range.

$$ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \tan L(x) = \lim_{x \to 0^{+}} \tan(ax+b) $$

so we would like $a(0) + b = -\frac{\pi}{2}$. Similarly, for the limit as $x \to 1^{-}$, we would like $a(1) + b = \frac{\pi}{2}$. This gives $b= -\frac{\pi}{2}$, $a = \pi$, $L(x) = \pi(x - \frac{1}{2})$, and finally

$$ f(x) = \tan \left[ \pi \left(x - \frac{1}{2}\right) \right] $$

(Optionally, then notice $f(x) = -\cot(\pi x)$. Possibly drop the negative sign from this if you don't mind it decreasing rather than increasing. These usually won't matter much for further use.)

Can you write a $(0,1) \to (0,\infty)$ homeomorphism now?

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  • $\begingroup$ Would it just be adjusting tan further, to $f(x)=tan (x* (pi / 2))$ or is this incorrect because there is a residual negative piece to the function when less than 0? $\endgroup$
    – User
    Commented Mar 15, 2021 at 18:51
  • $\begingroup$ Yes, $f(x) = \tan(x \pi/2)$ is a very nice simple $(0,1) \to (0, \infty)$. $\endgroup$
    – aschepler
    Commented Mar 15, 2021 at 23:19

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