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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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Partial answer: math.stackexchange.com/questions/42308/… –  Matt Nov 19 '12 at 6:32
    
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. –  icurays1 Nov 19 '12 at 6:33
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@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? –  Brian M. Scott Nov 19 '12 at 6:39
    
@BrianM.Scott I have a question, isn't it that every continuous function on a open set has open images? in another words we can show that there is no homeomorphism between them as long as some of them are open or close? There is a theorem talking about that but i am not sure where go wrong –  Mathematics Nov 19 '12 at 6:51
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@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 $U\subseteq Y$. –  Brian M. Scott Nov 19 '12 at 7:00

1 Answer 1

up vote 10 down vote accepted

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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