As the topic is there exist a homeomorphism between either pair of $(0, 1),(0,1],[0,1]$
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Partial answer: http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1?rq=1 – Matt Nov 19 '12 at 06:32
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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 06:33
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2@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 06:39
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@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 06:51
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@BrianM.Scott You make a good point, my statement doesn't really offer any intuition about homeomorphisms in general. I spoke too soon. – icurays1 Nov 19 '12 at 06:55
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2@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 07:00
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i see. i misunderstood the theorem. – Mathematics Nov 19 '12 at 07:03
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1Does this answer your question? Continuous bijection from $(0,1)$ to $[0,1]$ – Jacob Manaker Dec 17 '21 at 06:07
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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 $[0,1]$.
Brian M. Scott
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The Compact concept doesn't work if we consider them as independent spaces rather than subspace of R . – R29nb2xlIFBp Nov 25 '21 at 15:34
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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. – Brian M. Scott Nov 25 '21 at 21:57
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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 . – R29nb2xlIFBp Nov 26 '21 at 03:45
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1@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 ... – Brian M. Scott Nov 26 '21 at 08:44
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... harder to prove that $[0,1]$ is compact, but the usual proofs of that fact can all be carried out entirely in $[0,1]$, with no reference to any other space. – Brian M. Scott Nov 26 '21 at 08:46
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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
Maths fan123
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Here is a proof for two of them.
(0,1) and (0,1] are not homemorphic
Proof: suppose they are, then let
$$f:(0,1] \to (0,1)$$
Be a homoemorphism. Then so is removing a points, i.e.,
$$g:(0,1] \setminus \{1\} \to (0,1) \setminus \{f(1)\}.$$
But $(0,1] \setminus \{1\}=(0,1)$ which is connected, but $(0,1) \setminus \{f(1)\}=(0,f(1))\cup(f(1),1)$ which is disconnected thus there does not exists a homeomorphism. QED.
Theorem: (0,1) and [0,1] are not homeomorphic
Proof: Suppose they are, then let
$$f:[0,1] \to (0,1)$$
be the desired homeomorphism. Then $f([0,1])$ is compact and equal to $(0,1)$ which is not compact. QED.
homosapien
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