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Under what conditions $X$ is homeomorphic to $X \times \mathbb{N}$? where $\mathbb{N}$ is the discrete space.

Lo52
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    What have you thought of so far? Any examples? – rfauffar Aug 16 '13 at 02:06
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    One example: If $X$ is infinite and has the discrete topology, then there is a bijection between $X$ and $X\times\mathbb{N}$. This bijection is a homeomorphism in the discrete topology. Have you thought of any other examples? – rfauffar Aug 16 '13 at 02:07
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    Let $f:X\approx X\times\mathbb N$. Then $X$ is partitioned into countably many open sets $Y_i$, such that $Y_i=f^{-1}(X\times{i})\approx X$. – Karl Kroningfeld Aug 16 '13 at 02:10
  • Nice! This produces some sort of fractal, since if $X$ has a smaller copy of itself inside, this itself has a copy of itself, etc... – rfauffar Aug 16 '13 at 02:17
  • @RobertAuffarth Indeed. I am not very confident in my abilities manipulating such things, though. Do you think there is a nondiscrete topology on a set $X$ of cardinality $\aleph_1$ satisfying $X\approx X\times\mathbb N$? Also, we should look for some data on which bases are possible. – Karl Kroningfeld Aug 16 '13 at 02:19
  • @RobertAuffarth. Well, Any topological space which is the sum of countably many subspaces satisfies this property. – Lo52 Aug 16 '13 at 02:20
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    Kho: Sum, as in disjoint union? No. $X=\bigcup_{i=0}^\infty(i,i+1)\subset\mathbb R$ does not work. Edit: Maybe you have a point, based on a bijection $\mathbb N\cong\mathbb N\times\mathbb N$. In fact, I think the counter-example $X$ is actually an example $X$. – Karl Kroningfeld Aug 16 '13 at 02:22
  • Take for example the unit square with vertical and horizontal lines of the form ${k/2^n}\times[0,1]$ and $[0,1]\times{k/2^n}$ removed, for $k,n\in\mathbb{N}$. This is a sort of fractal, and satisfies the condition you gave. – rfauffar Aug 16 '13 at 02:29
  • @user1, I think I have got what I am looking for. According to your answer and what I already have I can say that $X$ is homeomorphic to $X \times \mathbb{N}$ iff X is the disjoint union of countably many subspaces. – Lo52 Aug 16 '13 at 02:49
  • Kho: Cool. Feel free to use what I said in an answer to your own question. – Karl Kroningfeld Aug 16 '13 at 02:51
  • @user1, Please your notation $\approx$ stands for what? – Lo52 Aug 16 '13 at 03:08
  • Kho: Homeomorphism. So I used $f:X\approx Y$ to mean that $f$ is an explicit homeomorphism with domain $X$ and codomain $Y$. – Karl Kroningfeld Aug 16 '13 at 03:09
  • @user1, In your answer why $Y_{i} \approx X$? – Lo52 Aug 16 '13 at 03:18
  • Kho: For any homeomorphism $f:X\to Y$, and any subspace $Y'\subseteq Y$, the subspace $f^{-1}(Y')\subseteq X$ is homeomorphic to $Y'$. Apply to this case where $Y'=X\times{i}\approx X$. (Sorry for the late reply btw) – Karl Kroningfeld Aug 16 '13 at 03:45

2 Answers2

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Let $X = \bigsqcup_{i \in \mathbb{N}} X_{i}$ be the disjoint union of countably many subspaces and each $X_{i} \approx X$, then $X \approx X \times \mathbb{N}$. On the other hand, if $X \approx X \times \mathbb{N}$ then $X$ can be partitioned into countably many clopen susbsets of $X$ each of it is homeomorphic to $X$ which is equivalent to that $X$ is the disjoint union of countably many subspaces each of it is homeomorphic to $X$.

Lo52
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    We can go one step further and say $X\cong X\times \mathbb N$ if and only if there exists a $Y$ such that $X\cong Y\times \mathbb N$ – Aaron Aug 16 '13 at 05:20
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It's true that if $X\approx X\times\mathbb N$, then $X$ is the disjoint union of countably many subspaces. However, the converse isn't true! Here's a counterexample: $E=\mathbb ({-2},{-1})\cup\mathbb N$. All but one of the connected components of $E$ are singletons. The same can't be said of $E\times\mathbb N$.

Here's a more likely claim, which you may or may not consider trivial: $X\approx X\times\mathbb N$ if and only if $X\approx Y\times\mathbb N$ for some space $Y$. You can think of this as a formula that generates all spaces with the desired property. Can you prove it?

Perhaps more interesting: $X\approx X\times\mathbb N$ if and only if $X\approx A\times B$ where $A$ is a topological space and $B$ is an infinite discrete topological space.

Chris Culter
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  • I agree but my converse required that $X$ is the disjoint union of countably many subspaces which are homeomorphic to $X$ – Lo52 Aug 16 '13 at 05:36
  • @Kho I see that you've edited your answer to be correct, but I'm afraid I don't understand your comment here. In my counterexample, do you agree that $E$ is the disjoint union of countably many subspaces? – Chris Culter Aug 16 '13 at 05:39
  • $X \approx X \times \mathbb{N}$ iff $X$ is the disjoint union of countably many subspaces WHICH ARE HOMEOMORPHIC TO $X$. – Lo52 Aug 16 '13 at 05:48
  • @Kho Agreed! That qualification is not implicit, and when you leave it out, the statement becomes incorrect. – Chris Culter Aug 16 '13 at 05:54
  • Chris Culter: If you look at the edit history of Kho's answer, you will see that $X_i\approx X$ was always there, barring the five minute window after answering. – Karl Kroningfeld Aug 16 '13 at 06:10
  • @user1 I apologize if it seemed like I was accusing anyone of... well, anything! :) – Chris Culter Aug 16 '13 at 06:21