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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?

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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.

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