# Scattered separators in Erdős space

Let $$X$$ be the set of all points in $$\ell^2$$ with all rational coordinates. $$X$$ is known to be totally disconnected, but $$X$$ is not zero-dimensional. For instance, the empty set does not separate the point $$\langle 0,0,0,...\rangle\in X$$ from the closed set $$\{x\in X:\|x\|\geq 1\}$$ because $$\{\|x\|:x\in A\}$$ is unbounded for every clopen set $$A\subseteq X$$.

The set $$S:=\{x\in X:\|x\|=1/2\}$$ separates $$\langle 0,0,0,...\rangle$$ and $$\{x\in X:\|x\|\geq 1\}$$. That is, $$X\setminus S$$ is the union of two disjoint open sets, one containing $$\langle 0,0,0,...\rangle$$, and the other containing $$\{x\in X:\|x\|\geq 1\}$$. Note that $$S$$ has no isolated points; $$\overline {S\setminus \{s\}}=S$$ for every $$s\in S$$.

My question is:

Does there exist a closed scattered separator between $$\langle 0,0,0,...\rangle$$ and $$\{x\in X:\|x\|\geq 1\}$$? A set is scattered if every non-empty subset has an isolated point.

Observe that scattered subsets in metrizable separable (more generallity hereditarily Lindelof) spaces are at most countable.

On the other hand, any closed separator $$S$$ between zero and the set $$F=\{x\in X:\|x\|\ge 1\}$$ has cardinality continuum and hence cannot be scattered.

To see that $$|S|=\mathfrak c$$, write $$X\setminus S$$ as the union $$U\cup V$$ of two disjoint open sets $$U,V$$ such that $$0\in U$$ and $$F\subset V$$. For $$i\in\{0,1\}$$ consider the closed subspace $$X_i=\{(x_n)_{n\in\omega}\in X:\forall n\in\omega\; x_{2n+i}=0\}$$ and observe that $$X=X_0+X_1$$ and $$|U\cap X_1|=\mathfrak c$$.

Let $$(e_i)_{i\in\omega}$$ be the standard orthonormal basis of the Hilbert space $$\ell_2$$. For every $$u\in U\cap X_1$$ we have $$\|u\|<1$$ and $$u+e_0\in F\subset V$$. Consequently, there exist rational numbers $$x_0 such that $$u+x_0e_0\in U$$, $$u+x_0'e_0\in V$$, and $$|x_0'-x_0|<1$$. Since $$u+x_0e_0+e_2\in A\subset V$$, there exist $$x_2 such that $$u+x_0e_0+x_2e_2\in U$$, $$u+x_0e_0+x_2'e_2\in V$$ and $$|x_2'-x_2|<1/4$$. Continuing by induction we shall construct two sequences of rational numbers $$(x_{2i})_{i\in\omega}$$ and $$(x_{2i}')_{i\in\omega}$$ such that for every $$n\in\omega$$ $$u+\sum_{i=0}^nx_{2i}e_{2i}\in U,$$ $$u+\sum_{i=0}^nx_{2i}e_{2i}+(x_{2n}'-x_{2n})e_{2n}\in V\mbox{; and }$$$$|x_{2n}'-x_{2n}|<1/2^{2n}.$$ Since $$U\subset\{x\in \ell_2:\|x\|<1\}$$, the series $$\sum_{i=0}^\infty x_{2i}e_{2i}$$ is convergent and the point $$s(u)=u+\sum_{i=0}^\infty x_{2i}e_{2i}$$ belongs to the separatior $$S$$.

Since the map $$s:U\cap X_1\to S$$, $$s:u\mapsto s(u)$$, is injective, the separator $$S$$ has cardinality $$|S|\ge|U\cap X_1|=\mathfrak c$$. $$\;\;\;\;\;$$

• Very nice argument by the way :) Commented Feb 28, 2019 at 2:23
• @D.S.Lipham Thank you. Commented Feb 28, 2019 at 7:52
• @D.S.Lipham Good idea, but the Lelek fan somehow hides this product structure of the Erdos spaces. By the way, it seems that you did not notice my answer to your question mathoverflow.net/questions/297924/embedding-into-c-times-0-1? Commented Mar 2, 2019 at 22:38
• @D.S.Lipham No, because every metrizable separable space contains a dense $G_\delta$-set, which is zero-dimensional. This follows from Lavrentiev Theorem saying that every homeomorphism between subsets of Polish spaces extends to a homeomorphism of $G_\delta$-subsets. Commented Mar 6, 2019 at 5:53
• Yes but I want to assume $X$ is not zero-dimensional (see second sentence in my previous comment). I think it may be true in this case. Commented Mar 6, 2019 at 12:52