# Sequences with 0's in $\mathbb R ^\omega$

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?

• @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. Commented Aug 13, 2021 at 0:40
• @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 $\mathbb R ^\omega$. Commented Aug 13, 2021 at 1:53
• @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. Commented Aug 13, 2021 at 5:35
• It might be insightful to answer the question first for the product of the discrete space $\mathbb{N}$. Commented Aug 13, 2021 at 11:56
• @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. Commented Aug 13, 2021 at 19:02

## 1 Answer

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 $$n, the spaces $$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. $$\quad\square$$

• Very surprising! Thank you Commented Jul 23, 2022 at 17:10
• @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). Commented Jul 23, 2022 at 17:17