Denote by $X := [\mathbb{N}]^\infty$ the set of infinite subsets of $\mathbb{N}$. Recall that the Ellentuck topology is a topology on $X$ generated by sets of the form $\{A\text{ infinite} \mid s\text{ is an initial segment of }A\text{ and }A\setminus s\subset B\}$ for some finite $s\subset \mathbb{N}$ and (infinite) $B\subset \mathbb{N}$.

In the paper On completely Ramsey sets by Szymon Plewik, it was proved that the Ellentuck topology is not normal. The argument essentially reduced to a construction of a closed separable subspace, say $Y\subset X$, containing a discrete closed subset $Z$ of cardinality $2^{\aleph_0}$. Suppose that $X$ is normal. Because $Y$ is a closed subset of $X$, $Y$ is normal as well. On the one hand, since $Y$ is separable, the set of continuous function on $Y$, denoted by $C(Y)$, is of cardinality $\le 2^{\aleph_0}$. On the other hand, by the Tietze extension theorem, every (continuous) function on $Z$ can be extended to a continuous function on $Y$, and so $C(Y)$ is of cardinality at least $2^{2^{\aleph_0}}$. A contradiction.

I am wondering if one can demonstrate an explicit construction of two disjoint closed subsets of $X$ that cannot be separated by two disjoint open neighborhoods?


I’ll use the machinery of Proposition $\mathbf{4}$ and its proof in Szymon Plewik, On completely Ramsey sets.

Let $H=\{V\cup A^*\in U:|A|=\omega\}$ and $K=\{V\cup A^*\in U:|A|<\omega\}$; $H$ and $K$ are disjoint closed discrete sets in $[\omega]^\omega$ in the Ellentuck topology. Suppose that $G$ is an Ellentuck-open set containing $H$; I’ll show that every Ellentuck-open nbhd of $K$ meets $G$.

For each infinite $A\subseteq\omega$ there is a finite $s_A\subseteq V\cup A^*$ such that

$$V\cup A^*\in\langle s_A,V\cup A^*\rangle\subseteq G\;.$$

The set $[\omega]^\omega\setminus\{\omega\}$ is a dense $G_\delta$ in the natural topology on $\wp(\omega)$ (i.e., the Cantor space topology), and there are only countably many finite subsets of $\omega$, so by the Baire category theorem there is a finite $s\subseteq\omega$ such that the closure $C$ of $\mathscr{A}=\left\{A\in[\omega]^\omega:s_A=s\right\}$ in the natural topology has non-empty natural interior. Thus, there are disjoint finite $t,x\subseteq\omega$ such that $\langle t,\omega\setminus x\rangle\subseteq C$.

For $i=0,1$ let $\pi_i:\omega\times\omega\to\omega:\langle n_0,n_1\rangle\mapsto n_i$. Let $t_0=\pi_0\big[h^{-1}[s]\big]$ and $x_0=\pi_1\big[h^{-1}[s]\big]$; if $s_A=s$, then $A\in\langle t_0,\omega\setminus x_0\rangle$. Thus, we may assume that $t_0\subseteq t$ and $x_0\subseteq x$. Let $\langle r,V\cup t^*\rangle$ be a basic Ellentuck nbhd of $V\cup t^*$; we may assume that $r$ is large enough so that $\pi_0\big[h^{-1}[r]\big]=t$ and $\pi_1\big[h^{-1}[r]\big]\supseteq x$. Moreover, $s\setminus V\subseteq h[t\times x]\subseteq t^*$, so $s\subseteq V\cup t^*$, and we may assume that $s\subseteq r$.

Now fix $A\in\mathscr{A}\cap\left\langle t,\omega\setminus\pi_1\big[h^{-1}[r]\big]\right\rangle$; clearly $r\subseteq V\cup A^*$. Let $B=V\cup(A^*\cap t^*)$; then

$$\begin{align*} B&\in\langle r,V\cup(A^*\cap t^*)\rangle\\ &=\langle r,V\cup A^*\rangle\cap\langle r,V\cup t^*\rangle\\ &\subseteq\langle s,V\cup A^*\rangle\cap\langle r,V\cup t^*\rangle\\ &\subseteq G\cap\langle r,V\cup t^*\rangle\;, \end{align*}$$

so $V\cup t^*\in K$ is in the Ellentuck closure of $G$. Thus, $H$ and $K$ cannot be separated by disjoint Ellentuck-open sets.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.