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How to prove the statement:

Assuming the continuum hypothesis, the product of any uncountable family of $T_1$ spaces, each having more than one point, is never sequentially compact.

The statement appears in General Topology by Stephen Willard, exercise 17G.6.

My attempt was along the lines of trying to use the sequential compactness (any sequence of points in the space has a convergent subsequence whose limit is in it) of the space defined to build a set that would violate the continuum hypothesis. But I could not find the right construct.

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2 Answers 2

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Here I show that $P:=\{0,1\}^{2^{\Bbb N}}$ is not sequentially compact, using a diagonalisation argument (this holds in plain ZFC, so "absolutely").

If CH holds and $I$ is uncountable and $X_i, i \in I$ is a family of $T_1$ spaces having more than one point, we have that $P$ is homeomorphic to a closed subspace of $\prod_{i \in I} X_i$. (CH is needed to get $|I| \ge \mathfrak{c}= |2^{\aleph_0}|$ from $I$ being merely uncountable) So if the latter were sequentially compact so would $P$ be, quod non.

FYI: there are models of ZFC where $[0,1]^{\aleph_1}$ is sequentially compact. See the Handbook of Set-theoretic Topology for more on this. This shows CH is really needed.

This is the essence of the argument.

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  • $\begingroup$ Out of curiosity, does MA imply that products of $<\mathfrak c$ sequentially compact spaces is sequentially compact? I can sort of see how it would imply that every sequence has a convergent subnet, but I'm not sure whether that is enough to extract an actual sequence. $\endgroup$
    – tomasz
    Jun 21, 2021 at 8:27
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    $\begingroup$ @tomasz IIRC it does. Check the Handbook maybe. It’s related to the tower number cardinal invariant. $\endgroup$
    – Henno Brandsma
    Jun 21, 2021 at 8:30
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    $\begingroup$ @tomasz in Fremlin's very good book "Consequences of Martin's Axiom" 24B says that a product of $< \mathfrak{p}$ sequentially compact spaces is sequentially compact. And MA implies that $\mathfrak{c}=\mathfrak{p}$. $\endgroup$
    – Henno Brandsma
    Jun 21, 2021 at 8:37
  • $\begingroup$ Ah, yes, I should have guessed Fremlin would have said something about that. $\endgroup$
    – tomasz
    Jun 21, 2021 at 8:39
  • $\begingroup$ Would your $P$ have cardinality $\aleph_2$? (In CH) $\endgroup$
    – FShrike
    Jan 30 at 22:06
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Here's the structural overview of a proof, I urge you to fill in all details:

Step 1: Use what is given. So we have $$ X=\prod _{i\in I}X_i$$ where each $X_i$ has at least two distinct points $a_i$ and $b_i$. Also $I$ is uncountable means that $I$ is not smaler than $\Bbb R$, or equivalently not smaller than $\mathcal P(\Bbb N)$. That is, there exists an onto map $f\colon I\to \mathcal P(\Bbb N)$.

Step 2: Construct a tricky sequence $(x^{(n)})_{n\in \Bbb N}$ from the above. That is, use $f$ and the $a_i,b_i$ - but how? Well, we can let $x^{(n)}_i=a_i$ or $=b_i$ depending on whether or not $n\in f(i)$!

Step 3: Finish, i.e., for every subsequence $(x^{(n_k})_{k\in\Bbb N}$ and every point $c\in X$, show that $x^{(n_k}\not \to c$.

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  • $\begingroup$ A subsequence of a product space fails to converge iff. it fails to converge pointwise as a function on one of the components. However, if I let any such failure-component be $i\in I$, then it is entirely possible for a subsequence to converge pointwise since if $f(i)$ is finite, then I can say $x_i\to b_i\because n\notin f(i)$ for large $n$, and if $f(i)$ is infinite then it is entirely possible to say either $x_i\to a_i\because n\in f(i)$ or $x_i\to b_i$. If I then bring in other indices, I cannot reach a contradiction. Could you help me with the argument? $\endgroup$
    – FShrike
    Jan 30 at 21:57

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