# Is there exist a homeomorphism between either pair of $(0,1),(0,1],[0,1]$

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

– Matt
Nov 19, 2012 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. Nov 19, 2012 at 6:33
• @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? Nov 19, 2012 at 6:39
• @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 $U\subseteq Y$. Nov 19, 2012 at 7:00
• Does this answer your question? Continuous bijection from $(0,1)$ to $[0,1]$ Dec 17, 2021 at 6:07

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 $[0,1]$.
• @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. Nov 25, 2021 at 21:57
• @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 ... Nov 26, 2021 at 8:44
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