Are uncountable products of $\mathbb{R}$ always non-sequential?

Question

Recall that a topological space $X$ is sequential if the closed subsets are exactly the subsets $A \subseteq X$ such that no sequence of points in $A$ can converge to a point in $X \setminus A$.

I have recently stumbled upon this answer by user642796 which at the end lists all uncountable products of $\mathbb{R}$ as examples of non-sequential topological spaces.

In a much more recent comment user Henno Brandsma says the following:

$\mathbb{R}^\kappa$ is non-sequential if $\kappa \ge \mathfrak{c}$. I believe it is consistent that $\mathbb{R}^{\omega_1}$ is sequential, it depends on your set theory. So the last item [that uncountable products of $\mathbb{R}$ are non-sequential] is not fully correct.

This comment has gone unresponded to for almost three weeks, and I have doubts the answerer will ever respond to it. (According to the profile they haven't been seen since Nov 2017.)

So who is right? Are uncountable products of $\mathbb{R}$ always non-sequential, or does it depend "on your set theory"?

Answer 1

I think it might depend on how weak you allow your set theory to be. If you consider set theory to be ZFC (and its extensions), then user642796 is correct that uncountable products of $\mathbb{R}$ are always non-sequential. If you allow set theory to be weaker than ZFC (in particular by removing the Axiom of Choice), then it's possible that Henno Brandsma is correct. I'll speak a bit more (but significantly less authoritatively) about this after showing the ZFC case.

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To show (in ZFC) that $\mathbb{R}^\kappa$ is non-sequential whenever $\kappa$ is an uncountable cardinal, consider the subset $$A = \{ x = (x_\xi)_{\xi < \kappa} \in \mathbb{R}^\kappa : | \{ \xi < \kappa : x_\xi \neq 0 \} | \leq \aleph_0 \}.$$ This can easily be shown to be a non-closed subset of $\mathbb{R}^\kappa$ (it's actually a dense subset of $\mathbb{R}^\kappa$ which is clearly not all of $\mathbb{R}^\kappa$). We now show that no sequence in $A$ can converge to a point outside of $A$.

Let $( a_n = ( a^{(n)}_\xi )_{\xi < \kappa} )_{n \in \mathbb{N}}$ be a convergent sequence of points in $A$ with limit $b = ( b_\xi )_{\xi < \kappa}$. Note that as for each $n \in \mathbb{N}$ the set $\{ \xi < \kappa : a^{(n)}_\xi \neq 0 \}$ is countable, then so is $K = \bigcup_{n \in \mathbb{N}} \{ \xi < \kappa : a^{(n)}_\xi \neq 0 \}$. If $b \notin A$, then as $\{ \xi < \kappa : b_\xi \neq 0 \}$ is uncountable, there must be a $\xi \in \kappa \setminus K$ such that $b_\xi \neq 0$. As $a_n \rightarrow b$ it must be that $a^{(n)}_\xi \rightarrow b_\xi$ (projection mappings are continuous and so preserve sequential limits). But $a^{(n)}_\xi = 0$ for all $n$, and $b_\xi \neq 0$, which is absurd! Therefore it must be that $b \in A$, as required.

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An observant reader will notice that I make an audacious use of the Axiom of Countable Choice (AC_ω) to conclude that $K$ is countable. (I think the statement that "countable unions of countable sets are countable" is strictly weaker than AC_ω, but AC_ω seems to be the weakest "quotable" axiom implying it.) It might be possible that uncountable products of $\mathbb{R}$ are sequential in some models of ZF+¬AC_ω... perhaps $\mathbb{R}^{\omega_1}$ is sequential in a model where $\omega_1$ is a countable union of countable sets.

I am also implicitly using the Axiom of Choice to reduce "all uncountable products of $\mathbb{R}$" to "every $\mathbb{R}^\kappa$ for an uncountable cardinal $\kappa$". If AC_ω fails, then you could have Dedekind finite but infinite sets (which are "uncountable", but are not equipotent to any ordinal). So this is also something to keep in mind. Mind you, considering $\mathbb{R}^I$ for an uncountable set $I$, we don't really run into any problems until trying to show that $\{ i \in I : b_i \neq 0 \} \setminus \bigcup_{n \in \mathbb{N}} \{ i \in I : a^{(n)}_i \neq 0 \}$ is nonempty.

Answer 2

I withdraw the comment because I misremembered: The fact I had in mind is also about a recurring theme on this site: sequential compactness: there is a cardinal $\mathfrak{t}$ of which we can only prove in ZFC that $\aleph_1 \le \mathfrak{t} \le \mathfrak{c} =2^{\aleph_0}$ (so under CH all three are equal to $\aleph_1$ while under Martin's axiom MA we can e.g. have $\mathfrak{t} = \mathfrak{c} = \aleph_2$) such that:

If $X_{i \in I}$ is a family of sequentially compact (Hausdorff) spaces and $|I| < \mathfrak{t}$ then $\prod_{i \in I} X_i $ is sequentially compact and if $|I| \le \mathfrak{t}$, $\prod_{i \in I} X_i$ is countably compact, but e.g. $\{0,1\}^{\mathfrak{t}}$ or $[0,1]^{\mathfrak{t}}$ is not sequentially compact. It generalises the fact that countable products of sequentially compact spaces are sequentially compact, and the well-known fact that $[0,1]^\mathbb{R}$ is not sequentially compact.

In the CH model $[0,1]^{\omega_1}$ is not sequentially compact, while in the MA model it is. Hence the confusion in my mind.

As @palladiumtelemann rightly pointed out, the $\Sigma$-product in the full product shows that a dense subset (so not closed) is sequentially closed in $\mathbb{R}^{\omega_1}$. Sorry to confuse the OP.