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

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.

• As a quick comment, MA (and, in particular, CH) imply that there are Luzin sets, which are sets of size continuum whose intersection with every meager set has size less than continuum, showing that Question 2 has a negative answer in this case. You build such sets by diagonalizing against all nowhere dense sets, of which there are continuum many. – Miha Habič Oct 10 '13 at 23:55
• It is consistent without choice (in Solovay's model) that this is the case, although that's not very helpful here I suppose. :-) – Asaf Karagila Oct 10 '13 at 23:58
• Miha, please post your comment as an answer. But I don't agree that there are only continuum many nowhere dense sets, since every subset of the Cantor set is nowhere dense, and there are $2^{2^{\aleph_0}}$ many such subsets. Perhaps you are speaking only of Borel sets? Does this idea still answer the question? – Joel David Hamkins Oct 10 '13 at 23:58
• @Asaf, yes, I am thinking of the ZFC situation, but I would encourage you to post an answer with the $\neg\text{AC}$ situation if it is interesting. – Joel David Hamkins Oct 11 '13 at 0:00
• Joel, thank you for the encouragement, but this is really just an artifact of the perfect set property. So it's not that interesting after all. – Asaf Karagila Oct 11 '13 at 0:03

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 $V$.

• Great! Meanwhile, may we still hope for the positive consistency? – Joel David Hamkins Oct 11 '13 at 0:30
• The argument you give here totally destroys the proof strategy that I gave on Ashutosh's question, since you are not really using MA here, but rather are only using that the additivity numbers are large. So the additivity hypothsis that I needed to run my argument is exactly what prevents the idea from actually succeeding. – Joel David Hamkins Oct 11 '13 at 0:38
• I think the official definition of "Luzin set" is "uncountable, but has no uncountable meager subsets". The version you give ("of size continuum, with no equipotent meager subset") is sometimes called "generalized Luzin set", e.g. in chapter 17 of the book by Just and Weese. Similarly for Sierpiński sets. – Goldstern Oct 13 '13 at 20:49
• @Goldstern Yes, I noticed that when I was later figuring out where I remembered these from. I'll try and get away with this abuse of terminology. – Miha Habič Oct 13 '13 at 21:30

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.

• It still amazes me how every week or so I learn about another remarkable result due to Sierpinski..! – Malik Younsi Oct 12 '13 at 23:32

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.

• More briefly: if cov(measure)=$\aleph_1< 2^{\aleph_0}=\aleph_2$ (or more generally: less than the cofinality of the continuum), and unif(measure)=continuum, then every small set (i.e., of cardinality less than continuum) is of measure zero, and every large set must meet one of the covering null sets in a large set. But the situation "cov(measure)=$\aleph_1<$unif(measure)=$\aleph_2=2^{\aleph_0}$" is well known to be consistent. – Goldstern Oct 12 '13 at 22:02
• I suggest you put this comment into the main text, at the beginning, since it implies a full solution. I didn't notice it when I wrote my answer below. (See my comment at the end of my answer.) – Boaz Tsaban Jul 31 '15 at 14:33

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.

• I now see that Goldstern added a comment to his answer that implies that if cov(M) and cov(N) are both $\aleph_1$ and non(M) and non(N) are both $\aleph_2$ (which is consistent), then one also gets the same theorem. So, my answer only adds that even assuming that all four mentioned cardinals are $\aleph_1$ is consistent with "every set of reals contains a measure-zero subset and a Lebesgue null subset of the same cardinality." – Boaz Tsaban Jul 31 '15 at 14:31