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.
Size of the closure of a set
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2@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. – 2010-11-23
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4I 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. – 2010-11-24
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2I found http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist;task=show_msg;msg=1381.0001 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. – 2010-11-24
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1@Nate: I found that too and agree with you that it isn't obvious. Also, thanks for asking for a clarification there. – 2010-11-24
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2@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. – 2010-11-27
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1@Nuno: Seems like a perfectly reasonable MO question to me, and a recent meta.MO thread seemed to agree that it is fine to repost math.SE questions to MO (where appropriate for the latter) after a reasonable time with no answers. – 2010-11-27
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1@Bogdan: I think you should post this question on MathOverflow, since you didn't get any right response here. If you don't mind, I can post it there for you. @Nate Eldredge: Good to know that. Do you have the link for the thread? I search a bit, but didn't find it. Maybe because I'm not familiar with meta.MO. Also, if Bogdan don't answer this comment we can post this question there. – 2010-11-27
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0@Nuno: http://tea.mathoverflow.net/discussion/791/what-is-a-duplicate/#Item_0 – 2010-11-28
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0@Nate Eldredge: Thank you very much. – 2010-11-28
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0One more comment: the claim at that "For pseudoradial Hausdorff spaces X we have |X| <= d(X)^c(X) (<= 2^d(X))" (at the at.yorku.ca link) is not correct (it is the second inequality that is a problem.) Otherwise we would have a fine contradiction coming from the fact that if $\mathfrak{c}\leq\aleph_2$ then all compact, sequentially compact spaces are pseudo-radial, but my (consistent) counterexample is a separable, compact, sequentially-compact space (hence pseudo-radial) and $\mathfrak{c}=\aleph_2$ but $|X|>2^{\aleph_0}$. – 2011-02-08
3 Answers
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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0Thanks, 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.) – 2011-02-08
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0Thanks, I stuck in the forcing construction, since it's not a priori obvious how to force such a model. – 2011-02-08
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0I 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? – 2011-02-08
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0Yes. It's a great book if you're studying set-theoretic topology. – 2011-02-08
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0Why is $\{0,1\}^{\omega_1}$ separable in this model? – 2011-02-08
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1This 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 $X$.) – 2011-02-08
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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0Yes, 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. – 2011-02-08
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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3I 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. – 2010-11-23
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1Suppose 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? – 2010-11-23
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0@Jonas: exactly. I don't see it either. Perhaps you meant "Hausdorff" instead of separability? – 2010-11-23
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0No, I meant separability, because the closure of the countable set is a separable, sequentially compact, Hausdorff space. – 2010-11-23
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0Yes, the first countability was what I was missing – 2010-11-23