Why can't a Lusin set be sigma-compact?

Question

In A Direct Proof of a Theorem of Telgársky, Scheepers asserts that the Menger game is indeteremined for a Lusin (aka Luzin) subset of the reals. That is, the first player lacks a winning strategy in the Menger game (i.e. the subspace is Menger) but the second player also lacks a winning strategy in the Menger game (i.e. the subspace is not Strategically Menger).

A Lusin subset of the reals has the property that its intersection with every nowhere dense subset of the reals is countable (and therefore its intersection with every meager subset of the reals is countable also).

In Every Lusin set is undetermined in the point-open game Recław provides us the corresponding result for the point-open game, which is dual to the Rothberger game. In particular, since the first player lacks a winning strategy in the Rothberger game, the space is Rothberger and therefore Menger.

Since a Lusin set is metrizable, being Strategically Menger is equivalent to being $\sigma$-compact (due to Scheeper's "Direct Proof"). So, why can't a Lusin set be $\sigma$-compact?

Answer 1

Here's a direct proof that doesn't require any measure theory or descriptive set theory.

A Lusin set cannot contain a compact subspace with non-empty interior in $\mathbb R$. If it did, it would contain an open subinterval of $\mathbb R$, and therefore a copy of the Cantor set within $\mathbb R$, which means its intersection with that nowhere dense Cantor set is uncountable.

It follows that every compact subspace of a Lusin set has empty interior in $\mathbb R$, which means it is nowhere dense in $\mathbb R$. Thus by the definition of a Lusin set, every compact subspace of a Lusin set is countable. Since the Lusin set is uncountable, it is not a countable union of countable sets, and thus cannot be $\sigma$-compact.

Answer 2

As Dave L. Renfro points out in the comments, even more is true - Luzin sets don't have the Baire property; that is, given a Luzin subset of $\mathbb R$, there is no open subset of $\mathbb R$ so that the symmetric difference between that open set and the Luzin set is meager. We argue that here using a big hammer involving category and measure.

If any subset $A$ of a Polish space is residually null (the set of Borel probability measures for which $A$ has zero outer measure is co-meager in the set of all Borel probability measures) and has the Baire property, then it is actually meager - see Theorem 14 in the arXiv version of An Extension of the Baire Property. Topology Proceedings 55 (2020). By results detailed in Miller's chapter, Special Subsets of the Real Line, Luzin sets are actually universally null (outer measure zero relative to all continuous Borel probability measures on $\mathbb R$). Since $\mathbb R$ doesn't have isolated points, this means that Luzin sets are residually null (the set of continuous Borel probability measures on $\mathbb R$ is a dense $G_\delta$ subset of all Borel probability measures). Hence, since Luzin sets are non-meager, they cannot have the Baire property.

For a little more detail on this chain, 3.1 of Miller's chapter asserts that Luzin sets are concentrated on every countable dense subset of $\mathbb R$ and also that any subset of $\mathbb R$ which is concentrated on a countable set has strong measure zero. Then 5.1 asserts that any set which has strong measure zero is actually universally null. So Luzin sets are universally null.