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In James.E.Munkres There is a question that says prove that subspace $(a,b)$ of $\mathbb R $ is homeomorphic with $(0,1) $.

In the definition of homeomorphism it is given that Let $X$ and $Y$ be two topological spaces.

Let $f:X \rightarrow Y $.Now if both the function $f$ and it's inverse function are continuous, then the function $f$ is called homomorphism.

Now my question is shouldn't there be a function two prove if two sets are homeomorphic??? Any help would be appriciated. Thanks.

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  • $\begingroup$ "then the function is called a homeomorphism" and this is indeed the desired function to show that two topological spaces (sets with a topological structure) are homeomorphic. $\endgroup$ – Andres Mejia May 22 '17 at 12:37
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    $\begingroup$ So I have to find the desired function?? $\endgroup$ – user426700 May 22 '17 at 12:38
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    $\begingroup$ For your exercise, you just need to show a continuous function $f:(a,b)\rightarrow(0,1)$ whose inverse function is also continuous. Hint: don't look for smg too complicated ! $\endgroup$ – Evargalo May 22 '17 at 12:39
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    $\begingroup$ Oh. OK. Thanks. $\endgroup$ – user426700 May 22 '17 at 12:41
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Note: This was a point of confusion for me too, but that extra "e" does make a difference. Homomorphisms (which you talk about in algebra) are completely different from homeomorphisms, which is a central focus in topology. So I would be careful not to mix the two!

Intuitively, every interval $(a, b)$ in $\mathbb{R}$ is indeed homeomorphic to the unit interval, $(0, 1)$ (notice only open intervals are homeomorphic to each other; $[a, b)$, for instance, is not homeomorphic to $(a, b)$). You can see this because you can either stretch or shrink (ie, "continuously deform") $(a, b)$ into the unit interval, and vice versa.

You rigorously prove that two objects are homeomorphic to each other by finding an explicit map, a homeomorphism, between the two objects, which is the bicontinuous bijection defined in your question. For instance, if I were to prove that $(2, 4)$ is homeomorphic to $(1, 0)$, one way to go about finding a homeomorphism is parametrizing $(2, 4)$. In other words, I need to find $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto tx$, where $x \in (2, 4)$ and $tx \in (1, 0)$, where $0 \leq t \leq 1$ (this works in general for any homeomorphisms to a unit interval, circle, sphere, etc).

$$ x = 2 + t(4 - 2) = 2 + 2t \implies t = \frac{x - 2}{2} \implies f(x) = \bigg(\frac{x - 2}{2}\bigg) x$$. You can easily verify that $f$ and its inverse are continuous (they are just polynomial/ rational functions!) so we have a homeomorphism. You can just use the same idea for $(a, b)$.

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