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Any tips on finding a homeomorphism between the open interval $$(0,1)$$ and $$\mathbb{R}^1$$ The exercise begins by instructing the student to find an explicit function but how do I even start this? I know a homeomorphism is a continuous bijective function that maps the open interval $(0,1)$ into $R^1$. But how do I show two spaces are homeomorphic, and more specifically, how do I go about finding an explicit function? Thanks in advance!

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Or, you can directly look at $\tan(x):(-\frac {\pi}2, \frac {\pi}2) \to \Bbb R$, now can you prove that $(-\frac {\pi}2, \frac {\pi}2)$ is homeomorphic to $(0,1)$?

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  • $\begingroup$ Thank you! I can see that this function works by mapping the finite interval in the real line. I also can show that any two open intervals are homeomorphic to one another. Thanks again! $\endgroup$ – coreyman317 Feb 21 '18 at 11:45
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The map $x\mapsto \frac1x$ would map $(0,1)\to(1,\infty)$, where the left end $0$ is taken to infinity.. Alternatively, $x\mapsto \frac1{x-1}$ would map $(0,1)\to (-\infty,-1)$ with the right end $1$ being taken to infinity. By combining these (while ensuring continuity at the joint), we can arrive at something like $$f(x)=\begin{cases}\frac1x&0<x\le \frac12\\ \frac1{x-1}+4&\frac12<x<1\end{cases} $$

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Hint: What does $\arctan(x)$ look like? Perhaps you could do some transformations to make it fit?

Another example of such a function:

$$f(x) = \frac{x}{\sqrt{1+x^2}}$$

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