When $X$ is homeomorphic to $X \times \mathbb{N}$?

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

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What have you thought of so far? Any examples? – rfauffar Aug 16 '13 at 2:06
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 2:07
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 Kronenfeld Aug 16 '13 at 2:10
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 2:17
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 Kronenfeld Aug 16 '13 at 2:22

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 $X$.

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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 5:20

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.

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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 5:36
@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 5:39
$X \approx X \times \mathbb{N}$ iff $X$ is the disjoint union of countably many subspaces WHICH ARE HOMEOMORPHIC TO $X$. – Lo52 Aug 16 '13 at 5:48
@Kho Agreed! That qualification is not implicit, and when you leave it out, the statement becomes incorrect. – Chris Culter Aug 16 '13 at 5:54
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 Kronenfeld Aug 16 '13 at 6:10