# A conjecture on closed discrete subset

I am struggling with this old problem, which is also posted here:

Let $X$ satisfy countable chain condition(abbreviated as CCC) and $X$ has a regular $G_\delta$-diagonal. Then the cardinality of $X$ is at most $\mathfrak c$.

$X$ has a regular $G_\delta$-diagonal iff there is a collection of open sets of $X^2$, say $\lbrace U_n: n\in N\rbrace$, such that $\Delta=\bigcap\lbrace \overline{U_n}: n \in N\rbrace$, where $\Delta=\lbrace(x,x): x \in X\rbrace$.

Note that the question is answered; however I hope to get new proof.

By certain effort, If $|X|>\mathfrak c$, we can get an uncountable closed discrete subset $S$ of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point.

I would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a Hausdorff space. If $S \subset X$ is an uncountable closed discrete subset of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point. Then could we obtain an uncountale collection of disjoint open sets in $X$?

Thanks for your any help.

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Assuming you want $U_x$ to be a neighbourhood of $x$, your condition on $S$ is equivalent to $S$ being closed discrete, whenever $X$ is a regular space. That's because $X$ is regular if and only if for every point $x \in X$ and every open neighbourhood $U$ of $x$ there is an open set $V$ such that $x \in V \subset \overline{V} \subset U$. Now, there are plenty of regular ccc spaces with uncountable closed discrete subspaces. For example, let $\kappa$ be any uncountable cardinal and $X=\{x \in 2^\kappa: (\exists \beta < \kappa)(x(\beta)=1 \wedge (\forall \gamma>\beta)(x(\gamma)=0)) \}$ with the topology induced from the usual topology on $2^\kappa$. $X$ has the ccc as a dense subspace of the ccc space $2^\kappa$. For $\alpha < \kappa$ define a point $x_\alpha \in X$ as follows: $x_\alpha(\beta)=1$ for $\beta \leq \alpha$ and $x_\alpha(\beta)=0$, for $\beta > \alpha$. Then $D=\{x_\alpha: \alpha < \kappa \}$ is a closed discrete subspace of $X$ of cardinality $\kappa$. To see that, note that $\{[\sigma] \cap X: \sigma \in Fn(\kappa,2)\}$ is a base for $X$, where $Fn(\kappa,2)$ denotes the set of all finite partial functions from $\kappa$ to $2$ and $[\sigma]:=\{x \in 2^\kappa: \sigma \subset x\}$. We have $[\{(\alpha,1), (\alpha+1,0)\}] \cap D=\{x_\alpha\}$ for every $\alpha < \kappa$ and hence $D$ is discrete. Moreover, $D$ is closed because if $x \in X \setminus D$, then there are ordinals $\alpha, \beta < \kappa$, $\alpha < \beta$ such that $x(\alpha)=0$ and $x(\beta)=1$ and hence $[\{(\alpha,0), (\beta,1)\}]\cap X$ is an open neighbourhood of $x$ which misses $D$.
I don't think the set $D$ is closed. If $\gamma$ is a limit ordinal, $y_{\gamma}(\beta)=1$ for $\beta<\gamma$, and $y_{\gamma}(\beta)=0$ for $\beta\geq\gamma$, then $(x_{\alpha})_{\alpha<\gamma}\rightarrow y_{\gamma}$. – Joseph Van Name May 6 '13 at 0:33
You're right, Joseph. Thanks for pointing that out! I think that to fix that it suffices to redefine $X$ to be the set of all $x \in 2^\kappa$ such that there is $\beta < \kappa$ with $x(\beta)=1$ and $x(\gamma)=0$ for every $\gamma > \beta$. I edited my answer accordingly. – Santi Spadaro May 6 '13 at 7:25
I claim that every discrete space of non-measurable cardinality can be embedded as a closed subspace of some completely regular space satisfying the countable chain condition. Let $S$ be an uncountable discrete space whose cardinality is below the first measurable cardinal. Since discrete spaces of non-measurable cardinality are realcompact, the space $S$ is embeddable as a closed subspace of some product $\mathbb{R}^{I}$ for some $I$. However, the space $\mathbb{R}^{I}$ satisfies the countable chain condition. We can give $\mathbb{R}$ a Borel probability measure. If we take the infinite product measure, then we get a measure $\mu$ on $\mathbb{R}^{I}$ such that $\mu(U)>0$ for each non-empty open set $U$. Therefore there can only be countably many disjoint open sets in $\mathbb{R}^{I}$ since $\mu(\mathbb{R}^{I})=1$.
A non-measurable cardinal is a cardinal $\kappa$ such that there is no non-principal $\sigma$-complete ultrafilter on $\kappa$. Said differently, a non-measurable cardinal is a cardinal below the first measurable cardinal(if it exists). Being something called a large cardinal axiom, the existence of a measurable cardinal is independent from the standard axioms of set theory, and the first measurable cardinal is extremely large if it exists, so the non-measurability of $S$ is not too great of a restriction. – Joseph Van Name May 4 '13 at 12:33