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In the context of learning topology, I'm triyng to see how to show that there is no bijective continuous function $f:(0,1) \to [0,1]$ using some basic topology facts such as the definition of continuity.

In this setting, $f$ is continuous if for all open $U$ in $[0,1]$, the set $f^{-1}(U)$ is open in $(0,1)$. Equivalently, we could swap "closed" for "open" in this definition.

As a first thought, I don't see any problem with the fact that, if such an $f$ exists, it would satisfy $f^{-1}([0,1]) = (0,1).$ I think this is OK because the set $(0,1)$ is closed in $(0,1)$. Is this correct? If so, what basic topological theorems do we use to prove this statement? Do we need to go to compactness?

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marked as duplicate by Asaf Karagila general-topology Nov 18 '17 at 12:18

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    $\begingroup$ But such functions do exist. Have you left off some condition, like 'injective'? $\endgroup$ – JonathanZ Nov 18 '17 at 1:15
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    $\begingroup$ As an example, $f(x) = \sin^2(100*x)$ $\endgroup$ – JonathanZ Nov 18 '17 at 1:18
  • $\begingroup$ Also, you're right that $(0.1)$ is a closed subset of $(0,1)$, so that's not an objection. You've identified a common problem when starting in topology, forgetting that the ambient space isn't always $\mathbb{R}$. Your mind has to be re-trained to accept that sometimes $(0,1)$ is closed, like in your case here. $\endgroup$ – JonathanZ Nov 18 '17 at 1:28
  • $\begingroup$ I suppose I mean bijective, now that you've pointed this out. $\endgroup$ – theQman Nov 18 '17 at 1:31
  • $\begingroup$ I'm not sure if my idea will help, I thought of using the fact that there is an inverse right $ g : [0,1] \longrightarrow (0,1) $ of $ f $. Then $f\circ g = {Id}_{[0,1]}$. but this does not guarantee that $g$ is continuous. $\endgroup$ – Luiz Collovini Nov 18 '17 at 1:33
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The function has to be monotone, assume it's increasing, now consider $f^{-1}(1)$ and observe that $1$ is not in $(0,1)$.

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Let $f : (0, 1) \to [0, 1]$ is a continuous bijection and let's write $f[X]$ for the image of $X \subseteq (0, 1)$ under $f$. As $f$ is a bijection, there is a unique $x \in (0, 1)$ such that $f(x) = 0$. And then $f[(0, x]]$ and $f[[x, 1)]$ are connected subsets of $[0, 1]$ each having at least two elements and each containing $0$. But connected subsets of intervals are intervals. So, for some $\epsilon_0 > 0$ and $\epsilon_1 > 0$, $f[(0, x)] \supseteq (0, \epsilon_0)$ and $f[(x, 1)] \supseteq (0, \epsilon_1)$. So $f[(0, x)] \cap f[(x, 1)] \neq \emptyset$ contradicting the assumption that $f$ is bijective.

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    $\begingroup$ Why is zero in those two image sets? $\endgroup$ – William Elliot Nov 18 '17 at 3:10
  • $\begingroup$ @WilliamElliot: thanks! I've fixed a couple of typos that made a nonsense of my answer. Please have another look. $\endgroup$ – Rob Arthan Nov 18 '17 at 19:08
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Suppose $f:(0,1)\to [0,1]$ is a continuous surjection . For $n\in \Bbb N$ the subspace $S(n)=[2^{-n},1-2^{-n}]$ is connected so its image $f(S(n))$ is connected so $f(S(n))$ is an interval.

For some $n_1$ there exist $x,y\in S(n_1)$ with $f(x)=0$ and $f(y)=1,$ implying that $0$ and $1$ belong to the interval $f(S(n_1))$ , so $f(S(n_1)=[0,1].$

Then $\phi \ne f( (0,1)$ \ $S(n_1)) \subset [0,1]=f(S(n_1))$ so $f$ cannot be a bijection.

By contrast the three totally-disconnected spaces $\Bbb Q\cap (0,1),\;$ $ \Bbb Q \cap [0,1),\;$ $\Bbb Q\cap [0,1]$ are homeomorphic to each other. (The homeomorphisms do not preserve the "$<$" order.)

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