If every nonseparable metric space contains a sequence of subsets with no convergent subsequence, does the Continuum Hypothesis hold?


The answer is negative, and in the interests of self-contained items, I write Ashutosh's solution below, recast in terms of the splitting number.


This question was submerged in the discussion of Fedor Petrov's [much-read question](https://mathoverflow.net/questions/181226/if-any-open-set-is-a-countable-union-of-balls-does-it-imply-separability) whether writing open sets as countable unions of balls implies separability in a metric space. The problem was solved by Joel and Ashutosh. As Timothy Chow observed, Sierpinski proved that if the continuum hypothesis is true, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence (Wacław Sierpiński, "Sur l'inversion du théorème de Bolzano-Weierstrass généralisé," Fund. Math. 34 (1947), 155–156). Sierpinski's proof invokes CH strongly in a classic enumeration argument, given in Timothy's comment. Can the appeal to CH be removed?



If every nonseparable metric space contains a sequence of subsets with no convergent subsequence, does the Continuum Hypothesis hold?
Edit: I found a problem with the previous argument. The following seems easier.


Start with a model of $2^{\omega} \geq \omega\_2$ and add $\omega\_1$ Cohen reals. Interpret the Cohen reals $\langle c\_i : i < \omega\_1 \rangle$ as increasing functions in $\omega^{\omega}$. Using $c\_i$'s instead of all of $\omega^{\omega}$ should let you run Sierpinski's argument: The relevant observation is that if $\langle k\_n : n < \omega \rangle$ is strictly increasing, then for some $i < \omega\_1$, both $c\_i$ and its complement meet $\{k\_n : n < \omega\}$ on an infinite set which is enough.


I haven't thought about if you can do this in ZFC alone.


