# 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.

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@Bogdan: How do you know it's true? Also where is this question from? I can't decide if it's true or not. But none of the counterexamples in 'Counterexamples in Topology' will work. – Nuno Nov 23 '10 at 22:57
I posted an answer pointing out that the Stone–Čech compactification of $\mathbb{N}$ is compact Hausdorff of size $2^{2^{\aleph_0}}$ and has a countable dense set. But it seems not to be sequentially compact, so it doesn't quite make a counterexample. – JDH Nov 24 '10 at 4:51
I found at.yorku.ca/cgi-bin/… but it says that any point in the closure is a sequential limit, which is not obvious to me in the absence of first-countability. It may be that this follows from sequential compactness but I don't see how. – Nate Eldredge Nov 24 '10 at 21:15
@Nate: I found that too and agree with you that it isn't obvious. Also, thanks for asking for a clarification there. – Nuno Nov 24 '10 at 22:19
@Nate Eldredge: Do you think this question is inappropriate to MathOverflow? Maybe it's not true at all, and a counterexample doesn't seem easy to find. – Nuno Nov 27 '10 at 17:52

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).

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Thanks, this is very nice! (I'll wait a while before awarding the bounty. This will hopefully attract more attention to the question and your answer.) – Jonas Meyer Feb 8 '11 at 19:24
Thanks, I stuck in the forcing construction, since it's not a priori obvious how to force such a model. – Apollo Feb 8 '11 at 19:35
I really appreciate that, along with the references, particularly since I lack the background. Is the van Douwen article "The integers and topology" from the same book? – Jonas Meyer Feb 8 '11 at 19:40
Yes. It's a great book if you're studying set-theoretic topology. – Apollo Feb 8 '11 at 19:46
This is the Hewitt-Marczewski-Pondiczery Theorem: let $\kappa\geq\aleph_0$ be a cardinal and $\beta\leq2^\kappa$. Then if $X_\alpha$ are topological spaces with $d(X_\alpha)\leq\kappa$ for $\alpha<\beta$ then $d(\prod_{\alpha<\beta}X_\alpha)\leq\kappa$. In our case, we are using the special case for $\kappa=\aleph_0$: the product of no more than continuum many separable spaces is separable. We are taking the product of $\omega_1<2^\omega$ many finite spaces. ($d(X)$ is the density of a space $X$.) – Apollo Feb 8 '11 at 20:00

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.

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Yes, I wouldn't be surprised if there weren't an independence result here, something like $\text{MA}+2^{\aleph_0}=\aleph_2$ implying that every separable, sequentially-compact, (+compact?) space is $\leq\aleph_2$. For example, there's the result that if $\mathfrak{c}\leq\aleph_2$ then every compact, sequentially-compact space is pseudo-radial, etc. – Apollo Feb 8 '11 at 22:36

It seems like you don't need all the conditions, so maybe I am missing something. There are only continuum many sequences of points from a countable set. So the limit points are at most continuum many.

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I believe that is why he said that he can see why it is true if the space is first countable. In general, elements of the closure need not be limits of sequences. Since first-countability isn't assumed, one needs to know why separability and sequential compactness is enough. I don't off hand. Also, Hausdorff is used to show that limits are unique, because in general a sequence might converge to every point in a set of arbitrary cardinality. – Jonas Meyer Nov 23 '10 at 18:56
Suppose we take an element in the closure of the countable subset, say C, how can we guarantee that this element is the limit of a sequence of elements of C? don't we need first countability for this? – student Nov 23 '10 at 18:57
@Jonas: exactly. I don't see it either. Perhaps you meant "Hausdorff" instead of separability? – student Nov 23 '10 at 18:58
No, I meant separability, because the closure of the countable set is a separable, sequentially compact, Hausdorff space. – Jonas Meyer Nov 23 '10 at 18:59
Yes, the first countability was what I was missing – Ross Millikan Nov 23 '10 at 19:21