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So, this is in reference to a question I asked earlier.

"I have to show that the following function f:(0,1)→ℝ. I will use this function: $f(x)=\frac{1}{x}+\frac{1}{x−1}$." I've figured that this is a bijection.

Now, I am a little confused on how to show this is a homeomorphism with respect to the usual subspace topology on (0,1) and the usual topology on the real numbers. I know that all that remains is to show that both $f$ and $f^{-1}$ are continuous.

I started out with proving $f$ is continuous. In order to show that I want to show provided $f^{-1}(U)$ is open in the subspace topology for every open $U$ in the usual topology.

To begin: Let $U$ be an open under the usual topology. Consider $f^{-1}(U)$ in the subspace topology. Since $f$ is bijective, $f^{-1}(U) \subseteq (0,1)$.

At this point, I want to show that $f^{-1}(U)$ is open. But, I'm stuck here. Once I can show this, I can easily find the method to find $f^{-1}$ is also continuous.

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If using calculus is not prohibited, you can simply observe, that $f$ is continuous on $(0,1)$ as an elementary function on an interval, and, since the derivative is negative, $f$ is strictly decreasing, hence the inverse is continuous.

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  • $\begingroup$ This is correct, but I need to show this somehow relating to the topological spaces. So, although this is true for the function, I need to ensure that for all open segments under the usual topology the preimage of these segments are also open under the subspace topology. I think I am getting confused with the fact that I'm dealing with open sets and the function as well. $\endgroup$ – H Cruz Nov 18 '14 at 6:08
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    $\begingroup$ @HCruz Proof of the theorem, that inverse to continuous decreasing is continuous decreasing, uses intervals (prebasis), hence maybe it would be good attempt? $\endgroup$ – Przemysław Scherwentke Nov 18 '14 at 6:16
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    $\begingroup$ Derivative is non-positive. $\endgroup$ – copper.hat Nov 18 '14 at 6:17
  • $\begingroup$ @copper.hat Thank you again. I hope that now it is good. (I am thinking about $\tan$ as a model one, hence translating onto OP's functions is far from perfectness). $\endgroup$ – Przemysław Scherwentke Nov 18 '14 at 6:21
  • $\begingroup$ Looks good! ${}{}$ $\endgroup$ – copper.hat Nov 18 '14 at 6:32
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Showing that the preimage of every open interval is an open interval is equivalent to showing that the function satisfies the familiar $\epsilon$-$\delta$ definition of continuity, and this equivalence is not hard to prove. Once you’ve done that, let $\epsilon>0$ and $x\in(0,1)$ be given, and find $\delta_0>0$ and $\delta_1>0$ such that $$\frac1y\in\left(\frac1x-\frac{\epsilon}2,\frac1x+\frac{\epsilon}2\right)$$ whenever $y\in(x-\delta_0,x+\delta_0)$, and $$\frac1{1-y}\in\left(\frac1{1-x}-\frac{\epsilon}2,\frac1{1-x}+\frac{\epsilon}2\right)$$ whenever $y\in(x-\delta_1,x+\delta_1)$. Then take $\delta=\min\{\delta_0,\delta_1\}$.

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