Suppose that $(X,2^X)$ is equipped with a non-atomic probability measure $\mu$ (the existence of such spaces is consistent with ZFC). This induces the $L_1$ pseudometric $\Delta$ on $2^X$, via $\Delta(A,B)=\mu((A\setminus B)\cup(B\setminus A)$.

Question: Assuming ZFC, does the (pseudo)metric space $(2^X,\Delta)$ contain an $\epsilon$-separated subset of cardinality at least $|X|$, for some $\epsilon>0$?

A subset $N$ of a metric space $(\Omega,\rho)$ is $\epsilon$-separated if for all $x,y\in N$ we have $\rho(x,y)>\epsilon$.

  • $\begingroup$ What does "uncountable" modify in "uncountable $\epsilon$-separated subsets"? Does it mean uncountably many pairs of subsets that are $\epsilon$-separated, or that the subsets themselves are uncountable? Additionally, when you say $\epsilon$-separated, do you mean "separated by at most $\epsilon$", or "separated by a nonzero amount that is at most $\epsilon$"? $\endgroup$ – user44191 Jan 20 at 6:50
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    $\begingroup$ Not necessarily - $\mu$ can be $\{0,1\}$-valued, in which case $\Delta$ has only two equivalence classes. $\endgroup$ – Nate Eldredge Jan 20 at 7:03
  • $\begingroup$ I've defined $\epsilon$-separated explicitly now. As for what "uncountable" refers to: it refers to the $\epsilon$-separated set $N$. $\endgroup$ – Aryeh Kontorovich Jan 20 at 7:03
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    $\begingroup$ Yes. See M. Gitik, S. Shelah, Forcings with ideals and simple forcing notions -- Israel J Math 68 (1989) 129-160. $\endgroup$ – Ashutosh Jan 20 at 7:22
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    $\begingroup$ @Ashutosh It would be great if you could turn your comment into an answer -- which I'd be happy to accept. $\endgroup$ – Aryeh Kontorovich Jan 20 at 8:06

The question has a trivial negative answer, as long as an atomlessly measurable cardinal exists (if one doesn't it is vacuously true, of course). Given an atomless probability measure $\mu$ on $(Y, \mathcal{P}(Y))$, let $\lambda$ be the supremum of the cardinalities of the $\epsilon$-separated subsets of $\mathcal{P}(Y)$. We can then define $X = Y + \lambda^{+}$ and extend $\mu$ to $X$ by defining each subset of $\lambda^{+}$ to have measure 0. This is an atomless measure on $\mathcal{P}(X)$, and any $\epsilon$-separated subset of $\mathcal{P}(X)$ defines an $\epsilon$-separated subset of $Y$ (by intersecting with $Y$, which doesn't change the measure). So there is no $\epsilon$-separated subset of $\mathcal{P}(X)$ of cardinality $|X|$, because $|X| \geq \lambda^+ > \lambda$.

So I will reinterpret the question to give a more interesting positive answer, along the lines of what Ashutosh suggested, which is to say, $X$ will have atomlessly measurable cardinality $\kappa$, and $\mu : \mathcal{P}(X) \rightarrow [0,1]$ will be the atomless $\kappa$-additive measure witnessing this fact. The material involved is adequately treated in Fremlin's textbook, though due to its size it is important to know where to look.

For any probability measure space $(X,\Sigma,\mu)$ we can form its measure algebra, $A = \Sigma/\mathcal{N}(\mu)$, where $\mathcal{N}(\mu)$ is the $\sigma$-ideal of sets $S \in \Sigma$ such that $\mu(S) = 0$. That is to say, in $A$ we identify measurable sets up to null sets. For probability spaces, $A$ is a complete Boolean algebra.

The pseudometric $\Delta(A,B)$ in the question then becomes a metric on $A$, which Fremlin says is called the measure metric or Fréchet-Nikodym metric (323A (d)). Then in 521O (a) (ii), Fremlin shows that for a measure algebra whose underlying set is infinite, its density character $d(A)$ (the smallest cardinality of a dense subset) is equal to the Maharam type of $A$, $\tau(A)$. The Maharam type of $A$ is the smallest cardinality of a subset of $A$ that generates $A$ as a complete Boolean algebra (331F).

By the Gitik-Shelah theorem (Theorem 2.6 in the reference provided by Ashutosh), for an atomless probability measure $\mu$ on $\mathcal{P}(X)$, $d(A) > \mathbf{add}(\mu)$, where $\mathbf{add}$ is the additivity. So in the setting described in the second paragraph, $\mathbf{add}(\mu) = \kappa$ by definition, so $d(A) > \kappa$. As $\kappa$ cannot be countable, this shows that $A$ cannot be separable (because separable spaces have $d(A) = \aleph_0$, by definition). This answers the question in the title.

But we still need to deal with the question of the $\epsilon$-separated subset. Given a cardinal $\kappa$, consider the space $2^\kappa$, as the compact product space of the discrete space $2$, and equip it with the Borel $\sigma$-algebra and the unique Radon probability measure $\nu_\kappa$ that treats $2^\kappa$ as a $\kappa$-indexed sequence of independent fair coins. For each $x \in \kappa$, define $$ C_x = \{ f \in 2^\kappa \mid f(x) = 0 \} $$ Then it should be clear that if $x \neq y$ are elements of $\kappa$, $\Delta(C_x,C_y) = \frac{1}{2}$, so $(C_x)_{x \in \kappa}$ is a $\frac{1}{2}$-separated family of Borel subsets of $2^\kappa$ of cardinality $\kappa$. This will be important later. Also, we will use the notation $\mathfrak{B}_\kappa$ for the measure algebra of $(2^\kappa,\nu_\kappa)$, as Fremlin does.

Let's return to the measure algebra $A$ of $(X,\mu)$. By Maharam's classification (332B), $A \cong \prod_{i \in I}\mathfrak{B}_{\kappa_i}$, where $(\kappa_i)_{i \in I}$ is a countable family of cardinals, and under this isomorphism the measure $\mu$ is mapped to a suitable $\sigma$-convex combination of the measures $\nu_{\kappa_i}$, i.e. $\mu = \sum_{i \in I}\alpha_i \nu_{\kappa_i}$, where $\sum_{i \in I}\alpha_i = 1$ and $\alpha_i > 0$ for all $i \in I$. It is easy to show that $$ \tau(A) \leq \prod_{i \in I} \tau(\mathfrak{B}_{\kappa_i}) = \prod_{i \in I}\kappa_i $$ (the fact that $\tau(\mathfrak{B}_{\kappa_i}) = \kappa_i$ is 331K in Fremlin).

We have that each $\kappa_i \leq \tau(A)$, and since the Gitik-Shelah theorem already mentioned (543F for the specific version in Fremlin's book) shows that $\tau(A) > \kappa$, and $\kappa$ is weakly inaccessible, there must be an $i \in I$ such that $\kappa_i > \kappa = |X|$. As $\mathfrak{B}_{\kappa_i}$ embeds (non-unit-preservingly) as a subalgebra of $A$, the image of $(C_x)_{x \in \kappa_i}$ in $A$ is an $\epsilon = \frac{\alpha}{2}$-separated family of cardinality $\kappa_i > |X|$ in $A$, which gives rise to such a family in $\mathcal{P}(X)$ by taking representatives of the equivalence classes.


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