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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 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?

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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.

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  • $\begingroup$ Is it possible to say: if the splitting number $s$ is $\aleph_{1}$, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence. That looks like the nub of your slick proof. Since $s = \aleph_{1} < c$ is consistent, the question is answered. $\endgroup$ – Avshalom Sep 22 '14 at 11:39
  • $\begingroup$ That's right. I think it would be interesting to know if this can be done in ZFC alone. $\endgroup$ – Ashutosh Sep 22 '14 at 12:01
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This answer is the proof given by Ashutosh, but formulated in terms of the splitting number.

Proposition If the splitting number $s$ is $\aleph_{1}$, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence.

Proof: Following Sierpinski, since the metric space $M$ is non-separable, there exist $d > 0$ and a sequence $\{p_\xi\}_{\xi<\omega_{1}}$ of points in $M$ such that $\varrho(p_\xi,p_\eta)\ge d$ for $\xi<\eta<\omega_{1}$, where $\varrho(x,y)$ is the metric on $M$.

Let $S$ be a splitting family (for $[\omega]^{\omega}$) of size $\aleph_{1}$, $S = \lbrace s^\xi : \xi < \omega_{1} \rbrace$, where $s^\xi = \langle n_1^\xi,n_2^\xi,n_3^\xi,\ldots\rangle$; for a given $k \in \mathbb{N}$, let $E_k$ be the set of all $p_\xi$ such that $k\in \{n_1^\xi,n_2^\xi,\ldots\}$.

The sequence $E_1,E_2,E_3,\ldots$ does not contain any convergent subsequence. For, if $E_{k_1}, E_{k_2},\ldots$ where $k_1<k_2<\cdots$ is an arbitrary subsequence of $E_1,E_2,\ldots$, then there exists $\alpha<\omega_{1}$, such that $s^\alpha$ splits $K = \lbrace k_{n} : n < \omega \rbrace$ into infinite $a = K \cap s^{\alpha}$ and $b = K \setminus s^{\alpha}$ say. Now $p_\alpha\in E_{k_{i}}$ for $k_{i} \in a$ and $p_\alpha \notin E_{k_{j}}$ for $k_{j} \in b.$ The open ball with centre $p_\alpha$ and radius $d$ (which is actually just the singleton $p_\alpha$) intersects each $E_{k_{i}}$ for $k_{i} \in a$ non-trivially but is disjoint from $E_{k_{j}}$ for $k_{j} \in b$. Consequently, $E_{k_1},E_{k_2},E_{k_3},\ldots$ is not convergent. So the sequence $E_1,E_2,\ldots$ contains no convergent subsequence, q.e.d.

Corollary (Sierpinski) CH implies the assertion (*) every nonseparable metric space contains a sequence of subsets with no convergent subsequence.

Proof. CH implies $s = \aleph_{1}$.

Corollary (Ashutosh) The assertion (*) does not imply CH.

Proof. It is relatively consistent that $s = \aleph_{1} < 2^{\aleph_{0}}$. q.e.d.

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