Does every set of reals contain a measure-zero set of the same cardinality? Does it contain a meager set of the same cardinality?

Question

This question arises from an issue in my post on Ashutosh's excellent question on Restrictions of the null/meager ideal.

Question 1. Does every set of reals contain a measure-zero subset of the same cardinality?

In other words, if $A\subset\mathbb{R}$, is there a measure-zero set $B\subset A$ with $|B|=|A|$? Is this assertion at least consistent? Does it follow from the continuum hypothesis? Does it follow from some other cardinal characteristic hypothesis? In the intended application, what is needed is that the assertion is consistent with the additivity number for measure being equal to the continuum. Is this consistent? Can anyone prove the consistency of the failure of the property?

Similarly, in the case of category rather than measure:

Question 2. Does every set of reals contain a meager subset of the same cardinality?

And similarly, is this statement consistent? Does it follow from CH or other cardinal characteristic hypotheses? Is it consistent with the additivity number for the meager ideal being large? Can anyone show the consistency of the failure of the property?

The questions arise in my post on Ashutosh's question, where I had proposed as a solution idea the strategy of a back-and-forth construction of length continuum, where the domain and target remain measure-zero during the course of the construction. But in order for this strategy to succeed, we seem to need to know in the context there that one may extend a given measure-zero set inside another non-measure-zero set to a larger measure-zero set with the same cardinality (and the same with meagerness). I had thought at first that this should be easy, but upon reflection I am less sure about it, and so I ask these questions here.

Answer 1

It is consistent that both of the questions have a negative answer. Indeed, this happens if MA holds.

A set $E$ of reals is called a Luzin set if $E$ has size continuum and for every meager set $X$ the intersection $E\cap X$ has size less than continuum.

A set of reals $E$ is called a Sierpiński set if $E$ has size continuum and for every measure zero set $X$ the intersection $E\cap X$ has size less than continuum.

Theorem: MA implies that there are Luzin and Sierpiński sets.

Proof: To construct a Luzin set, list all Borel nowhere dense sets in order type continuum: $\langle F_\alpha;\alpha<\mathfrak{c}\rangle$. For each $\alpha$ choose some $e_\alpha\notin \bigcup_{\beta<\alpha}F_\beta$; this is possible since MA implies that the union of less than continuum meager sets is meager. $E=\{e_\alpha;\alpha<\mathfrak{c}\}$ has size continuum and its intersection with every closed nowhere dense set has size less than continuum by construction. But since a meager set is contained in a union of countably many closed nowhere dense sets, $E$ must be a Luzin set.

To construct a Sierpiński set, replace "Borel nowhere dense" above by "Borel of measure zero" and "meager" by "measure zero". $\square$

In particular, this shows that assuming cardinal characteristics are large is not helpful for this problem.

The same avoidance idea seems to also show:

Theorem: If $V$ was obtained from $W$ by adding more than $\mathfrak{c}^W$ many Cohen (or random) reals to $W$, then the set of generic reals is Luzin (or Sierpiński) in $V$.

Answer 2

In the paper "Uncountable sets of real numbers with no uncountable subsets of measure zero" it is proved that under the continuum hypothesis, there is an uncountable set of reals which has no uncountable subset of measure zero. The result is in fact due to Sierpiński [Fundam. Math. 5, 177--187 (1924)].

Added remarks: I have decided to give a proof of the fact stated above. Thus assume $CH$, and let $G_\alpha, \alpha<\omega_1$ be an enumeration of all $G_\delta-$sets of measure zero. Note that their union is $\mathbb{R}.$ Let $D_0=G_0$ and for $\alpha>0, D_\alpha=G_\alpha \setminus \bigcup_{\beta<\alpha} G_\beta$. Again the union of $D_\alpha$'s is $\mathbb{R},$ so unboundedly many of them are non-empty. Let $A$ contain one element from each non-empty $D_\alpha.$ It is easily seen that $A$ is as required.

Answer 3

The answer to question 1 is "yes" in the Cohen model.

More precisely, add $\aleph_2$ many Cohen reals to a universe $V_0$ satisfying CH. (Finite support, countable support, or all at once using finite functions -- in this case you can make the continuum arbitrarily large.)

Every set $X$ of size $\aleph_1$ (or more generally: less than continuum) is in some intermediate model, and the next Cohen real will make $X$ of measure zero.

Now let $Y$ be a set of size $\aleph_2$ (or continuum). I claim that there is a Borel measure zero set $B$ with Borel code in $V_0$ which contains $\aleph_2$ many elements of $Y$. Indeed, if every Borel set coded in $V_0$ contains only at most $\aleph_1$ many elements of $Y$, then there is an element $y_0$ of $Y$ not contained in any of these ($\aleph_1$ many) measure zero sets. But then $y_0$ is random over $V_0$. It is well known that there are no random reals over $V_0$ in the Cohen model. (In fact, the reals of $V_0$ are not meager in the extension.)

A similar argument works for the dual question, using random reals. (Countable support, or side-by-side. Again you can make the continuum large.)

In the Mathias model (countable support iterations, continuum becomes $\aleph_2$) a similar argument shows that every set of size continuum has a subset of the same size in the ground model which is meager, and another one which is of measure zero. (The Laver property ensures that no random or Cohen reals appear.) The sets of size $\aleph_1$ appear in an intermediate stage, and the next Mathias real will make them meager and measure zero.

Answer 4

I think that I have a compelete answer (but see PS below).

Let $\mathcal{M}$ (repectively, $\mathcal{N}$) be the ideal of meager (respectively, Lebesgue null) sets in $\mathbb{R}$.

Let $\kappa$ be an infinite cardinal, and $\mathcal{I}$ be an ideal of sets in $\mathbb{R}$.

A set $X\subseteq\mathbb{R}$ is $\kappa$-$\mathcal{I}$-Luzin if $|X|\ge\kappa$ but $|X\cap I|<\kappa$ for every set $I\in\mathcal{I}$. In other words, if it has no subset of cardinality $\kappa$ in $\mathcal{I}$.

A set $X\subseteq\mathbb{R}$ is Luzin (respectively, Sierpi\'nski) if it is $\aleph_1$-$\mathcal{M}$-Luzin (respectively, $\aleph_1$-$\mathcal{N}$-Luzin).

cov$(\mathcal{I})$ is the minimal cardinality of a cover of the real line by elements of the ideal $\mathcal{I}$. The following folklore result is easy.

Theorem: If cov$(\mathcal{I})={}$cof$(\mathcal{I})$ then there is a cov$(\mathcal{I})$-$\mathcal{I}$-Luzin set.

Thus, the answer to Question 1 is "No" if cov$(\mathcal{N})={}$cof$(\mathcal{N})$, and similarly for Question 2. Both hypotheses follow from CH, MA, etc.

Lemma. Assume that there is a $\kappa$-$\mathcal{I}$-Luzin set $X$. Then cf$(\kappa)\le{}$cov$(\mathcal{I})$.

Indeed, by moving to a subset of $X$ we may assume that $|X|=\kappa$. Let $\{I_\alpha : \alpha<\text{cov}(\mathcal{I})\}$ be a family in $\mathcal{I}$ covering the real line. Then $$X=\bigcup_{\alpha<\text{cov}(\mathcal{I})} X\cap I_\alpha.$$ Thus, $\kappa$ is a union of $\text{cov}(\mathcal{I})$ sets of cardinality smaller than $\kappa$.

Theorem (Judah, Shelah): It is consistent that non$(\mathcal{M})={}$non$(\mathcal{N})=\aleph_1$ and there are no Luzin or Sierpi\'nski sets.

Moreover, they prove that a Miller real kills the ground model Luzin and Sierpi\'nski set, and iterate with countable support iteration.

The consistency result you hoped for follows.

Theorem. It is consistent that every set of reals contains a measure-zero subset and a Lebesgue null subset of the same cardinality.

Indeed, in Miller's model, non$(\mathcal{M})={}$non$(\mathcal{N})=\aleph_1$ and the continuum is $\aleph_2$. It follows from the Cihon diagram that $\text{cov}(\mathcal{M})=\text{cov}(\mathcal{N})=\aleph_1$ there. By the above lemma, there are no $\aleph_2$-$\mathcal{M}$-Luzin or $\aleph_2$-$\mathcal{N}$-Luzin sets there. By the Judah--Shelah Theorem, there are also no $\aleph_1$-$\mathcal{M}$-Luzin or $\aleph_1$-$\mathcal{N}$-Luzin sets there.

PS. I am writing this after an all-night work, I hope I do not mess up things.