Size of the closure of a set

Why in a Hausdorff sequentially compact space the size of the closure of a countable subset is less or equal than $c$ ? I can see why this is true when the space if first countable but we are not assuming so.

Solution 1:

The statement is consistently false.

In vaughn's article "Countably compact and sequentially compact spaces" in the Handbook of Set-Theoretic Topology, he gives the following equivalences for an infinite cardinal $\kappa$:

• $\{0,1\}^\kappa$ is sequentially compact
• $\kappa < \mathfrak{s}$ (where $\mathfrak{s}$ is the "splitting number")
• every compact space of weight $\leq\kappa$ is sequentially compact

Thus if we are in a universe where $\omega<\omega_1=\kappa<\mathfrak{s}\leq\mathfrak{c}<2^\kappa$ then the space $\{0,1\}^{\omega_1}$ gives a separable space which is sequentially compact but has cardinality greater than $\mathfrak{c}$.

Added: including the forcing construction to get such a model (from van Douwen's article):

Start with ground model $M\models(\mathfrak{c}=\omega_2 \wedge 2^{\omega_1}=\omega_3)$ then obtain an iterated ccc extension $\langle M_\eta : \eta\in\omega_2\rangle$ by adding $X_\eta\in[\omega]^\omega$ at stage $\eta\in\omega_2$ s.t. $\forall Y\in M\cap[\omega]^\omega$ either $X_\eta\subseteq^* Y$ or $X_\eta\subseteq^* (\omega\setminus Y)$. Then in $M_{\omega_2}$ we have $\mathfrak{c}=\omega_2<2^{\omega_1}$, $\mathfrak{t}=\omega_1$, $\mathfrak{s}=\omega_2$, exactly as desired. (Details are left as an exercise).

Solution 2:

Here is some progress in the positive direction.

Theorem. In any Hausdorff space, the closure of a countable set has size at most $2^{\mathfrak{c}}$, where $\mathfrak{c}$ is the continuum.

Proof. Suppose that $X$ is a Hausdorff topological space with a countable set $D$. For any point $a$ in the closure $\bar D$, we may consider the collection of open sets containing $a$, and their trace on $D$. That is, consider $F_a=\{U\cap D\mid a\in U\text{ open }\}$. Since there are only continuum many subsets of $D$, there are therefore at most $2^{\mathfrak{c}}$ many possible such families.

If the closure $\bar D$ had size larger than $2^{\mathfrak{c}}$, then there would be at least two (in fact many) distinct points $a$ and $b$ in $\bar D$ for which $F_a=F_b$. Let $U$ and $V$ be disjoint neighborhoods of $a$ and $b$. Let $U_1$ be another neighborhood of $a$ such that $U_1\cap D=V\cap D$, which must exist since $F_a=F_b$. Thus, $U\cap U_1$ is a neighborhood of $a$ that is disjoint from $D$, contradicting $a\in\bar D$. QED

The bound is sharp, even for compact Hausdorff spaces, in the sense that the Stone–Čech compactification $\beta\mathbb{N}$ is a Hausdorff topological space of size $2^c$ with a countable dense set. But $\beta\mathbb{N}$ is not sequentially compact, so this is not actually a counterexample to your question.

What the argument actually shows is that if $a$ and $b$ are in the closure of the countable set $D$, then $F_b$ is not a subset of $F_a$. If it were, we could find disjoint neighborhoods $U$ and $V$ of $a$ and $b$, respectively, and then $V\cap D\in F_b$, and so there is a neighborhood $U_1$ of $a$ with $U_1\cap D=V\cap D$, making $U\cap U_1$ a neighborhood of $a$ having no points from $D$, a contradiction. By symmetry, we conclude $F_a$ and $F_b$ are incomparable with respect to $\subset$ for all $a,b\in\bar D$.

I suspect that such a line of reasoning could be improved when there is sequential compactness, perhaps by using a cardinal characteristic, such as the splitting number.