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Is it consistent to have a set of reals $X$ of size $\aleph_3$ such that for every $Y \subseteq X$, $Y$ has measure zero iff $|Y| \leq \aleph_1$?

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    $\begingroup$ This would require that every $Y\subset X$ with $|Y|=\aleph_1$ be a Lebesgue-null set and that every $Y\subset X$ with $|Y|=\aleph_2$ be non-measurable, with inner measure 0 and positive outer measure. I dk whether this is consistent. $\endgroup$
    – DanielWainfleet
    Commented Apr 27, 2017 at 23:15
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    $\begingroup$ (Add to my previous comment): Every uncountable closed set of reals has the cardinal $c$ of the reals. So if $|Y|<c$ then $Y$ has inner measure $0.$ $\endgroup$
    – DanielWainfleet
    Commented Apr 27, 2017 at 23:20

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