Let $2^\omega$ be the Cantor cube $\{0,1\}^\omega$, endowed with the standard compact metrizable topology and the standard product measure, called the Haar measure. The Cantor cube is considered as a partially ordered set endowed with the partial order $\le$ defined by ($f\le g$ iff $f(n)\le g(n)$ for all $n\in\omega$).

A subset $A\subset 2^\omega$ is called an antichain if any two distinct elements of $A$ are incomparable in this partial order.

General Problem. How large can a Borel antichain in $2^\omega$ be?

This question can be made more precise using the language of $\sigma$-ideals.

By $\mathcal M$, $\mathcal N$ and $\mathcal E$ we denote the $\sigma$-ideal of meager subsets of $2^\omega$, the $\sigma$-ideal of Haar-null sets in $2^\omega$, and the $\sigma$-ideal generated by closed Haar-null sets, respectively.

It is clear that $\mathcal E\subset\mathcal M\cap\mathcal N$.

Theorem. Each Borel antichain in $2^\omega$ belongs to the $\sigma$-ideal $\mathcal M\cap\mathcal N$.

Proof. Identify $2^\omega$ with the compact topological group $\mathbb Z_2^\omega$ and apply the classical Steinhaus or Piccard-Pettis Theorem, which implies that the difference $A-A$ is a neighborhood of zero and hence contains the characteristic function $\chi$ of some singleton $\{n\}$. Choose any elements $a,b\in A$ with $a-b=\chi$ and conclude that they are comparable in the partial order $\le$ on $2^\omega$.

Now we state more

Precise Problem. Does every Borel antichain $A\subset 2^\omega$ belong to the $\sigma$-ideal $\mathcal E$?

This problem is equivalent to an even

More Precise Question. Assume that a $G_\delta$-set $A$ is an antichain in $2^\omega$. Can its closure $\bar A$ in $2^\omega$ have positive Haar measure?


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