A Sperner type problem on infinite antichains

Question

Let $\mathcal{A} \subset 2^{\mathbb{N}}$ be an antichain (with respect to containment). I want to measure the size of $\mathcal{A}$ in the following way: I create a set, $S$, by flipping a fair coin for each natural number $x \in \mathbb{N}$, if heads I add $x$ to $S$, otherwise I don't. Then the "measure" of $\mathcal{A}$ will be the probability $S \in \mathcal{A}$.

A friend told me about cylindrical sets which seem to capture this notion. Take those collections consisting of $k \in \mathbb{N}$ fixed bits (elements or nonelements) and define their measure to be $\frac{1}{2^k}$. For an example of such a collection take $\mathcal{F} = \{A \subset \mathbb{N}: 3 \in A, 8 \in A, 11 \notin A\}$, here $\mu(\mathcal{F}) = 1/8$. Let $\mu$ be the measure generated from these sets. Unfortunately, $\mu$ seems hard to work with, and I've gotten nowhere.

Question: If $\mathcal{A}$ is an antichain in $2^{\mathbb{N}}$, measurable with respect to $\mu$, does it follow that $\mu(\mathcal{A}) = 0$?

Background: The motivation is trying to extend Sperner's Thoerem to an infinite setting. Other infinite versions of Sperner's Theorem have been considered but on other structures than sets.

Answer 1

$\def\ZZ{\mathbb{Z}}$Assuming the axiom of choice and the continuum hypothesis, this is false! I will "construct" an antichain with outer measure $1$.

Notation It will be convenient to work with $\{ 0,1 \}^{\ZZ}$ rather than $\{ 0,1 \}^{\mathbb{N}}$; this is an isomorphic measure space. We abbreviate $\{0,1 \}^{\ZZ}$ to $\Sigma$. We write an element of $\Sigma$ as $(s_i)_{i \in \ZZ}$. Let $\Sigma_0$ denote the elements of $\Sigma$ for which $s_i$ takes both values infinitely often as $i \to \infty$. For $A$ and $B$ disjoint finite subsets of $\ZZ$, we write $\Omega(A,B)$ for the subset of $\Sigma$ where $s_i = 0$ for $i \in A$ and $s_i = 1$ for $i \in B$. So the $\Omega(A,B)$ are a basis of neighborhoods for $\Sigma$, with $\mu(\Omega(A,B)) = 2^{-|A|-|B|}$. We write $\Omega_0(A,B) = \Sigma_0 \cap \Omega(A,B)$. We write $\tau$ for the map $\Sigma \to \Sigma$ with $\tau(s)_i = s_{i+1}$.

The Strong Mixing Lemma Let $U$ and $V$ be measurable sets in $\Sigma$. Then $\lim_{n \to \infty} \mu(U \cap \tau^n V) = \mu(U) \mu(V)$.

Proof This is very standard ergodic theory. The first online source I could find is page 108 in Descriptive Set Theory and Dynamical Systems. $\square$

Key lemma Let $K$ be a closed set in $\Sigma$ with positive measure and let $s^1$, $s^2$, ... be a countable (or finite) subset of $\Sigma_0$. Then there is an element $t \in K \cap \Sigma_0$ which is incomparable with every $s^i$.

Proof We first replace $K$ by a smaller (but still positive measure) closed set $L$ which is contained in $\Sigma_0$. Define $U$ to be the set of $s$ in $\Sigma$ where $s_0=0$, at least one of $(s_1, s_2)$ are $1$, at least one of $(s_3,s_4, s_5)$ are $0$, at least one of $(s_6, s_7, s_8, s_9)$ are $1$, etcetera. This is an intersection of closed sets, so $U$ is closed. Note that $\mu(U) = \prod_{j \geq 1} (1-2^{-j}) > 0$ so, by the strong mixing lemma, $\mu(K \cap \tau^n U) > \mu(K) \mu(U)/2$ for $n$ sufficiently large. Take $L = K \cap \tau^n U$. As promised, $L$ is a closed set, of positive measure, and contained in $\Sigma_0$.

Since $s^1$ is in $\Sigma_0$, there are infinitely many $n$'s for which $s^1_n=0$ and infinitely many $m$'s for which $s^1_m=1$. Thus, by the strong mixing lemma, we can find $n_1$ so that $s^1_{n_1}=0$ and $\mu(L \cap \Omega(\emptyset, \{ n_1 \})) > (1/3) \mu(L)$. We can repeat this trick to find $m_1$ so that $s^1_{m_1} = 1$ and $\mu(L \cap \Omega(\emptyset, \{ n_1 \}) \cap \Omega(\{ m_1 \}, \emptyset)) > (1/3) \mu(L \cap \Omega(\emptyset, \{ n_1 \})$. Combining the two inequalities, $$\mu(L \cap \Omega(\{m_1 \}, \{ n_1 \})) > (1/9) \mu(L).$$ We have now guaranteed that any element of $L \cap \Omega(\{m_1 \}, \{ n_1 \})$ is incomparable to $s^1$.

Set $L_1 = L \cap \Omega(\{m_1 \}, \{ n_1 \})$. We now induct, finding $(m_2, n_2)$ so that $\mu(L_1 \cap \Omega(\{m_2 \}, \{ n_2 \}) > (1/9) \mu(L_1)$ and every element of $L_2$ is incomparable with $s^2$. Also, since $L_1 \supseteq L_2$, every element of $L_2$ is incomparable with $s^1$. In general, we build $L \supseteq L_1 \supseteq L_2 \supseteq \cdots$ with $\mu(L_k) > 9^{-k} \mu(L)$ and every element of $L_k$ is incomparable with $s^1$, $s^2$, ..., $s^k$.

Since that $L^k$ have positive measure, they are nonempty. Since $\Sigma$ is compact, $\bigcap L_k$ is nonempty. Let $t \in \bigcap L_k$. Then $t \in L \subset \Sigma_0$, and $t \in L \subset K$, and $t$ is incomparable to every $s^k$. $\square$

Lemma $\Sigma$ has continuum many open sets.

Proof There are countably many sets of the form $\Omega(A,B)$, so there are continuum many collections of sets of the form $\Omega(A,B)$. Every open set is the union of such a collection, so there are at most continuum many open sets in $\Sigma$. Also, $\Sigma \setminus \{ x \}$ is open for any $x \in \Sigma$, so there are at least continuum many open sets in $\Sigma$. $\square$

Let $\mathcal{K}$ be the collection of closed subsets of $\Sigma$ with positive measure. Since complementation is a bijeciton between closed subsets and open subsets, the above lemma shows that $\mathcal{K}$ has cardinality the continuum. Now, USING OUR SET THEORY ASSUMPTIONS, put $\mathcal{K}$ in bijection with the first uncouteable ordinal $\omega_1$. For $a \in \omega_1$, we write $K_a$ for the corresponding positive area closed set.

We now use transfinite induction (and the Axiom of Choice) to find elements $t_a$ of $\Sigma$, for each $a \in \omega_1$, so that $t_a \in K_a \cap \Sigma_0$ and each $t_a$ is incomparable to $t_b$ for $b < a$. The key lemma handles the successor case, and the limit case is immediate. (If you haven't seen transfinite induction before, I can write more, but this is long already.)

I claim that $\mathcal{A} := \{ t_a \}_{a \in \omega_1}$ is my desired antichain. It is an antichain, since any two elements are incomparable. Let $U$ be any open set of measure $<1$ and let $K = \Sigma \setminus U$. Then $\mu(K)>0$, so $K$ is equal to some $K_a$. Then $K \cap \mathcal{A}$ contains $t_a$ and is thus nonempty, so $\mathcal{A} \not \subseteq U$. We have shown that $\mathcal{A}$ is not contained in any open set of measure less than $1$, so the outer measure of $A$ is $1$.

Answer 2

This is not an answer, but is definitely too long for the usual format. As Thomas Andrews commented, the probability is probably zero. Since this is a probabilistic problem, it is likely that when $\mu({\cal A})=p>0$, the event $S\subseteq T,S\in{\cal A},T\in{\cal A}$ not only occurs, but occurs with positive probability. It is likely also that there is a uniform bound, i.e. there is a positive $c(p)>0$ depending only on $p$, such that $P(S\subseteq T,S\in{\cal A},T\in{\cal A}) \geq c(p)$ whenever $\mu({\cal A})=p$ and $S,T$ are independent variables with values in $2^{\mathbb N}$, distributed as indicated in the OP.

This last claim is in turn equivalent to the analagous statement on $2^{\lbrace 1,2,\ldots,n \rbrace}$ for large $n$ instead of $2^{\mathbb N}$, and we are then left with a finite counting problem.

One of the proofs of Sperner's theorem uses the well-known fact that $U_n=2^{\lbrace 1,2,\ldots,n \rbrace}$ can be partitioned into $g_n=\binom{n}{\lfloor\frac{n}{2} \rfloor}$ chains. Consider such a partition, and for ${\cal A}\subseteq U_n$ with $|{\cal A}| \geq p|U_n|$, let us denote by $d$ the number of chains whose intersection with $\cal A$ has size at least $2$. Then

$$ p2^n \leq | {\cal A}| \leq dn+(g_n-d) \tag{1} $$

We deduce

$$ P(S\subseteq T,S\in{\cal A},T\in{\cal A})\geq \frac{d}{2^{2n}} \geq \frac{p2^n-g_n}{(n-1)2^{2n}} \tag{2} $$

Note that the RHS is asymptotically equivalent to $\frac{1}{n2^n}$ because $g_n\sim \frac{2^n}{\sqrt{n}}$ by Stirling's formula.This is a crude bound that tends to zero very fast, but perhaps a more careful counting argument along those lines can work.