TITLE: 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 [27 upvotes]: 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.

REPLY [4 votes]: 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.