If one assumes that the continuum hypothesis is false then there are subsets of the reals of intermediate cardinality, uncountable but smaller than the continuum. What can be said about the necessary properties of such a set e.g. topological properties other than that it cannot be closed?
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1$\begingroup$ It can't be open either. $\endgroup$– hmakholm left over MonicaFeb 3, 2016 at 14:28
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4$\begingroup$ If it's Lebesgue measurable it has measure zero. (And regardless, it has inner measure zero.) Because $m(E)$ is the supremum of $m(K)$ for $K$ a compact subset of $E$. $\endgroup$– David C. UllrichFeb 3, 2016 at 14:29
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2$\begingroup$ Uncountable analytic sets and uncountable Borel sets must have cardinality $\mathfrak c$. $\endgroup$– Martin SleziakFeb 3, 2016 at 14:42
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$\begingroup$ Other comments and replies have given good answers about topology and measure, but there is still MUCH to say here -- I suggest you look at this (PDF), it's quite long but packed with info about "cardinal characteristics," i.e. what can happen with sizes of subsets of the continuum subject to various properties. $\endgroup$– user128390Feb 3, 2016 at 16:08
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Without additional hypotheses, not too much actually.
It cannot be a Borel set, since uncountable Borel sets have a copy of the Cantor set and thus they have cardinality continuum.
It cannot be of positive measure, since if $A$ is of positive measure, $A-A=\{a-b\mid a,b\in A\}$ contains an interval, and thus $A$ had to be of size continuum. Similarly if it isn't meager, then it cannot have the Baire property.
It cannot be a continuous image of a Borel set (or analytic), since those also have copies of the Cantor set.
It might be the complement of an analytic set, but if our set is of cardinality $>\aleph_1$, then it won't be, because uncountable co-analytic sets must have size $\aleph_1$ or $2^{\aleph_0}$.
It might be measurable, since it is consistent that every set of size less than $2^{\aleph_0}$ has measure zero. Similarly it might be meager.
There are many many properties, but the above list should be a reasonable start. Perhaps reading about Martin's Axiom, which have consequences that "most" regularity properties hold for sets of size $<2^{\aleph_0}$ by making them "small" (null sets, meager sets, etc.) and learn what sort of things might happen or might not happen.
answered Feb 3, 2016 at 15:29
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$\begingroup$ This is an interesting result: A CA set (complement of analytic) must be either: countable, $\aleph_1$, or $2^{\aleph_0}$. There exist explicit CA sets whose cardinality can be proved to be $\aleph_1$. $\endgroup$– GEdgarFeb 3, 2016 at 16:05
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$\begingroup$ If I recall correctly, the consistency result is adding a lot of reals over $L$, and the constructible reals make the CA set. But admittedly this is a vague memory. $\endgroup$– Asaf Karagila ♦Feb 3, 2016 at 16:55
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1$\begingroup$ @AsafKaragila I think you do not need any forcing over $L$ to get a thin (no perfect subset) uncountable $\Pi_1^1$ set. $\leq_L \cap \mathbb{R}^L$ is a $\Sigma_2^1$ set. So it is the projection of a certain $\Pi_1^1$ set on $\mathbb{R}^2$. The desired thin set is obtained by applying $\Pi_1^1$-uniformization to this set. The resulting set is thin and is uncountable in $V$ as long as $\omega_1^L = \omega_1^V$. $\endgroup$– WilliamFeb 4, 2016 at 4:15
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1$\begingroup$ @GEdgar There are no CA sets whose cardinality is provably $\aleph_1$. Such a set exists if and only there is a real $x$ so that $\omega_1^{L[x]} = \omega_1^V$. Also if you are curious, the three possibilities you mentioned above follow from the fact that if a $\Pi_1^1(x)$ set has a real not in $L[x]$, then it has a perfect set coded in $L[x]$. $\endgroup$– WilliamFeb 4, 2016 at 4:19