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Let $\mathbb R ^\omega$ be the set of all sequences of real numbers in the product topology.

Let $X$ be the set of all sequences in $\mathbb R ^\omega$ which have at least one 0.

Let $Y$ be the set of all sequences in $\mathbb R ^\omega$ which have at least two 0's.

Are $X$ and $Y$ homeomorphic, and if so, is there a simple proof of this?

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    $\begingroup$ @Gro-Tsen That will not be the case because they are countable unions of nowhere dense subsets. I think that probably they are each homeomorphic to the pseudo-boundary of the Hilbert cube, but proving that would require some very deep theorems from infinite dimensional topology. I was hoping there would be a simple argument that I just hadn't noticed. $\endgroup$
    – D.S. Lipham
    Commented Aug 13, 2021 at 0:40
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    $\begingroup$ @RonniePavlov correct but $X$ is the union of sets where $x_0=0$, where $x_1=0$, etc. each of which is nowhere dense in $X$. by Baire's theorem $X$ cannot be completely metrizable, so it is not the same as $\mathbb R ^\omega$. $\endgroup$
    – D.S. Lipham
    Commented Aug 13, 2021 at 1:53
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    $\begingroup$ @RonniePavlov No, basic neighbourhoods in the product only look at finitely many coordinates at a time. It is a product of intervals where all but finitely many are the whole of $\mathbb{R}$: every nonempty open set contains points with infinitely many zeros. $\endgroup$
    – KP Hart
    Commented Aug 13, 2021 at 5:35
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    $\begingroup$ It might be insightful to answer the question first for the product of the discrete space $\mathbb{N}$. $\endgroup$
    – Joel David Hamkins
    Commented Aug 13, 2021 at 11:56
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    $\begingroup$ @JoelDavidHamkins In the Baire space $\omega^\omega$ both sets are open, hence Polish and nowhere compact. Therefore they are homeomorphic to the Baire space again (Alexandroff: doi.org/10.1007/BF01451582). Here the spaces are of first category; a different kettle of fish. $\endgroup$
    – KP Hart
    Commented Aug 13, 2021 at 19:02

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For every $n\in\mathbb N$ consider the space $$X_n=\{x\in\mathbb R^\omega:\lvert x^{-1}(0)\rvert\ge n\}.$$

Theorem. For any positive integer numbers $n<m$, the spaces $X_n$ and $X_m$ are not homeomorphic.

Proof. The spaces $X_n$ and $X_m$ are not homeomorphic because the space $X_m$ is a $\sigma Z_{m-1}$-space whereas $X_n$ is not.

Let us recall that a closed subset $A$ of a topological space $X$ is a $Z_k$-set in $X$ if the set $C([0,1]^k,X\setminus A)$ is dense in the function space $C([0,1]^k,X)$ endowed with the compact-open topology. A topological space $X$ is called a $\sigma Z_k$-space if $X$ is the countable union of $Z_k$-sets.

$Z_k$-sets are higher-dimensional counterparts of nowhere dense sets and $\sigma Z_k$ are counterparts of meager spaces.

Since any singleton in $\mathbb R^m$ is a $Z_{m-1}$-set, the space $X_m$ is a $\sigma Z_{m-1}$-space. Assuming that $X_m$ is homeomorphic to $X_n$, we would conclude that $X_n$ is a $\sigma Z_{m-1}$-space. Consider the closed subset $A=\{0\}^n\times\mathbb R^{\omega\setminus n}\subseteq X_n$ of $\mathbb R^\omega$ and observe that $A$ is not a $Z_n$-set in $\mathbb R^\omega$. Since $X_n$ is a $\sigma Z_{m-1}$-space, $A=\bigcup_{i\in\omega}A_i$ is a countable union of $Z_{m-1}$-sets $A_i$. Since $n\le m-1$, each $Z_{m-1}$-set $A_i$ is a $Z_n$-set and hence the set $C([0,1]^n,X_n\setminus A_i)$ is dense in $C([0,1]^n,X_n)$. It can be shown that the set $C([0,1]^n,X_n)$ is dense in $C([0,1]^n,\mathbb R^\omega)$ and hence $C([0,1]^n,X_n\setminus A_i)$ is dense in $C([0,1]^n,\mathbb R^\omega)$, which means that $A_i$ is a $Z_n$-set in $\mathbb R^\omega$ (being a closed subset of the closed subset $A$ of $\mathbb R^\omega$). By the proof of Lemma 2.7 in the paper Concerning locally homotopy negligible sets and characterization of $l_2$-manifolds of Toruńczyk, the completeness of $\mathbb R^\omega$ implies that the closed $\sigma Z_n$-set $A=\bigcup_{i\in\omega}A_i$ is a $Z_n$-set in $\mathbb R^\omega$, which is not true. $\quad\square$

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    $\begingroup$ Very surprising! Thank you $\endgroup$
    – D.S. Lipham
    Commented Jul 23, 2022 at 17:10
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    $\begingroup$ @D.S.Lipham You are wellcome. Today I looked at unanswered questions in tag "gn" and found this your interesting question at the very top of the stack. It is strange that I have not noticed it earlier (it was posed almost a year ago, probably, because of the summer and vacances). $\endgroup$
    – Taras Banakh
    Commented Jul 23, 2022 at 17:17

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