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?



Sequences with 0's in $\mathbb R ^\omega$
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](https://bibliotekanauki.pl/articles/1364141) 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. $\quad\square$


