Show that the following subests of $(\mathbb{R},\tau_e)$ are homeomorphic.

$(0,1)$ , $(0,+\infty)$, $(-\infty,+\infty)$.

(i) If I take $\varphi:(\mathbb{R},\tau_e) \rightarrow (0,+\infty)$, $x \mapsto e^x$ this map is continuous, bijective, and the inverse is continuous and so is a homeomorphism.

(ii) $\varphi:(0,1) \rightarrow (0,+\infty)$, $x \mapsto \frac{x}{1-|x|}$

I think this should work, in fact if $x=\frac{1}{N}$, $N>0$, I have $\varphi(x)=\frac{1}{N-1}$ and for $N$ large enough this is really close to $0$.

It's a bijective and continuous map and the inverse is also continuous.

(iii) $\varphi:(0,1) \rightarrow (-\infty,+\infty)$... I don't know how to move... any hint ?

thanks :)

  • 3
    $\begingroup$ From (iI) you have $(0,1)\to(0,\infty)$ and from (i) you have $(0,\infty)\to(-\infty,\infty)$ $\endgroup$ Jun 26 '17 at 20:04
  • 1
    $\begingroup$ It's clear now: I can do that because composition of homeomorphisms in an homeomorphisms, right? $\endgroup$
    – VoB
    Jun 26 '17 at 20:27
  • $\begingroup$ Right.............. BTW: A common example in textbooks for $f:(-1,1)\to (-\infty,\infty)$ is $f(x)=\tan \pi x/2.$ So consider $g(x)=f(2x-1)$ for $x\in (0,1).$ $\endgroup$ Jun 28 '17 at 6:08

A function that goes from an interval to $\mathbb{R}$ is the tangent $x \mapsto \tan(x)$ that goes from $(-\frac{\pi}{2}, \frac{\pi}{2})$ onto all of $\mathbb{R}$ continuously and bijectively, with inverse $f(x) = \arctan(x)$, all of which are even differentiable. Just scale: the map $t \mapsto \pi t-\frac{\pi}{2}$ maps $(0,1)$ homeomorphically onto the domain of $\tan(x)$, so we can combine them to $f(x) = \tan( \pi x-\frac{\pi}{2})$ which is thus a homeomorphism from $(0,1)$ with $\mathbb{R}$.

  • $\begingroup$ but this function doesn't go from (0,1) to $\mathbb{R}$... and if I restrict it to $(0,1)$ its image doesn't cover all $\mathbb{R}$... I can't understand :( $\endgroup$
    – VoB
    Jun 26 '17 at 23:10
  • $\begingroup$ I made the compositions as suggested by @Hegen von Eitzen and got the map $\phi:(0,1) \rightarrow (-\infty,+\infty)$, $x \mapsto \frac{ln(x)}{1-ln(x)}$ $\endgroup$
    – VoB
    Jun 26 '17 at 23:26
  • $\begingroup$ @feddy another composition is what I meant, see above. Any two finite open intervals $(a,b)$ and $c,d)$ are homeomorphic by a linear map.. $\endgroup$ Jun 27 '17 at 4:17
  • $\begingroup$ oh now it's all clear! ;) I didn't know ho to scale. A last question: the map I found is an homeomorphism from $(0,1)$ to $\mathbb{R}$ ? $\endgroup$
    – VoB
    Jun 27 '17 at 6:17
  • $\begingroup$ @feddy Yes, the inverse of (i) followed by (ii). I just wanted a different idea (two asymptotes ). Both are fine. $\endgroup$ Jun 27 '17 at 6:20

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