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As the topic is there exist a homeomorphism between either pair of $(0, 1),(0,1],[0,1]$

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  • $\begingroup$ Partial answer: math.stackexchange.com/questions/42308/… $\endgroup$
    – Matt
    Nov 19, 2012 at 6:32
  • $\begingroup$ The short answer is no, since homeomorphisms preserve open set structure: i.e. open sets are mapped to open sets and closed sets are mapped to closed sets. $\endgroup$
    – icurays1
    Nov 19, 2012 at 6:33
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    $\begingroup$ @icurays1: That isn’t really an argument: how does it explain why this situation differs from that with $(0,1)\cap\Bbb Q$, $(0,1]\cap\Bbb Q$, and $[0,1]\cap\Bbb Q$, which are homeomorphic? $\endgroup$ Nov 19, 2012 at 6:39
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    $\begingroup$ @Mathematics: No, a continuous function need not take open sets to open sets. What is true is that $f:X\to Y$ is continuous iff $f^{-1}[U]$ is open in $X$ for every open set $U\subseteq Y$. $\endgroup$ Nov 19, 2012 at 7:00
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    $\begingroup$ Does this answer your question? Continuous bijection from $(0,1)$ to $[0,1]$ $\endgroup$ Dec 17, 2021 at 6:07

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No two of the three spaces are homeomorphic. One way to see this is to note that $(0,1)$ has no non-cut points, $(0,1]$ has one non-cut point, and $[0,1]$ has two. (A non-cut point is one whose removal does not disconnect the space.) Another way to see that $[0,1]$ is not homeomorphic to either of the others is to note that $[0,1]$ is compact, and they are not. $(0,1)$ and $(0,1]$ can also be distinguished by the fact that the one-point compactification of $(0,1)$ is homeomorphic to the circle $S^1$, while that of $(0,1]$ is homeomorphic to $[0,1]$.

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    $\begingroup$ +1 nice answer very easy to understand $\endgroup$
    – jasmine
    Apr 7, 2018 at 10:34
  • $\begingroup$ The Compact concept doesn't work if we consider them as independent spaces rather than subspace of R . $\endgroup$ Nov 25, 2021 at 15:34
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    $\begingroup$ @GogolePi: Of course it does: compactness of a space $X$ is an inherent property of $K$, i.e., one that is independent of any space in which $X$ may happen to be embedded. Similarly, if $X$ and $Y$ are homeomorphic spaces, then $X$ has a one-point compactification iff $Y$ has one, and in that case their one-point compactifications are homeomorphic. $\endgroup$ Nov 25, 2021 at 21:57
  • $\begingroup$ I am not talking about one point compactification . If we consider (0,1] and [0,1] as independent spaces then both of them are compact as they are bounded and clopen ( every space is clopen as it and it's complement phi , both are in the topology) . So unless cou consider (0,1) and [0,1] as subspace of R , how can you say one is compact and the other is not ? I was talking about this . $\endgroup$ Nov 26, 2021 at 3:45
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    $\begingroup$ @GogolePi: I thought so, and I already told you in the first sentence of my previous comment; the rest was in case you were also asking about the rest of the answer. Once again: whether a space $X$ is compact depends only on the space $X$ itself and has absolutely nothing to do with any space in which $X$ may happen to be embedded. It is an intrinsic property of the space. It is trivial to show that $(0,1)$ is not compact without any reference to $\Bbb R$: the open cover $$\left\{\left(\frac1n,1-\frac1n\right):n\ge 3\right\}$$ of $(0,1)$ obviously has no finite subcover. It’s a little ... $\endgroup$ Nov 26, 2021 at 8:44
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No. The idea of the proof is that when you remove a point from $(0,1)$, you end up with a disconnected space. So first we shall show that $(0,1)\not\cong [0,1)$. Assume that $f:[0,1)\to (0,1)$ is a homeomorphism. Then, remove the point $\{0\}$ from the domain. That is $A=[0,1)-\{0\}$ meaning that $$f\vert_{A}:(0,1)\to (0,1)-\{f(0)\}$$ is also a homeomorphism since $(0,1), [0,1]$ and $[0,1)$ are all in the subspace topology of the usual topology on $\mathbb{R}$. However this is a continuous bijection from a connected to a disconnected space, which is impossible! Now I will show that $(0,1)\not\cong [0,1]$. We use a similar idea; that is we remove $\{0\}$ from $[0,1]$ meaning that if we have a homeomorphism $\varphi:[0,1]\to (0,1)$, then the restriction of this homeomorphism to $(0,1]$ $$\varphi\vert_{(0,1]}:(0,1]\to (0,1)-\{f(0)\}$$ is a continuous bijection from a connected to a disconnected space, a contradiction again! Finally we move onto the most difficult part of the proof: showing that $[0,1]\not\cong [0,1)$. The idea is the same, but it's slightly harder to execute. The idea is that when we remove $0$ from both spaces, we get $(0,1)$ and $(0,1]$ and then when we remove $1$ from both spaces, we get a disconnected space $(0,1)-\{1\}$ and a connected space $(0,1)$ and we use the same idea. If we remove any other points from $(0,1]$, we get a disconnected space. So overall we have it that if there were a homeomorphism $f:[0,1]\to [0,1)$, then there would be a homeomorphism $$f\vert_{[0,1]-\{0,1\}}(0,1)\to [0,1)-\{f(0),f(1)\}$$ which is a continuous bijection from a connected to a disconnected space which is impossible

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