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It's easy to construct a sequence of rational numbers whose set of cluster (or accumulation or limit) points is of any finite cardinality, or is countably infinite, or has the cardinality of the continuum. Thus there appears no obvious barrier to forming a rational sequence with a cluster set of any cardinality up to $\mathfrak{c}$. Now suppose that we reject the continuum hypothesis. Is it consistent with $\mathrm{ZFC}$ that there could be an oracular sequence $(a_0 , a_1 , \dots ) \in \{0, 1\}^\mathbb{N}$ such that the cluster set of the sequence $(n_{2i - 1}/n_{2i} : i = 1, 2, \dots)$, where $n_0$ , $n_1$ , ... are the elements of $\{n \in \mathbb{N} : a_n = 1\}$ in natural order, has cardinality strictly between $\omega$ and $\mathfrak{c}$ ?

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up vote 6 down vote accepted

No, closed sets cannot have intermediate cardinality. One way to see this: suppose a closed subset $K$ of the Cantor set $C$ has greater than countable cardinality. It's easy to see that there must be a partition into two open and closed subsets such that each one has greater than countably many elements of $K$, otherwise you could express $K$ as a countable union of countable sets. Likewise, each of these has at a partition into two closed and open sets with more than countably many elements. Continuing by induction and taking the limit, you get a homeomorphic image of a Cantor set contained in $K$.

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Indeed, the result is due to Cantor himself, who had the strategy of attempting to prove CH by establishing such claims for increasingly complicated sets beyond the closed sets. Although it is consistent with ZFC that there are $\Delta^1_2$ sets of intermediate cardinality, Cantor's original strategy is redeemed by sufficiently large large cardinals, which imply that there can be no projective set that is a counterexample to CH. – Joel David Hamkins Feb 5 '11 at 19:53
An addendum to Joel's comment: The strategy he mentioned succeeds for Borel sets (and slightly farther, $\Sigma^1_1$ sets) even in the absence of large cardinals. ZFC suffices to prove that all uncountable $\Sigma^1_1$ sets have perfect subsets and therefore have the cardinality of the continuum. – Andreas Blass Feb 5 '11 at 22:46
Feeble google: Delta One products are a class of financial derivative that have no optionality and as such have a delta of one (or very close to one) – Bill Thurston Feb 5 '11 at 22:57
For Borel sets there is a nice "modern" argument, first noting that the result holds for closed sets in any Polish space, and then realizing that for any Borel set $A$, one can change the topology slightly so that a Polish space remains Polishable, the collection of Borel sets does not change, and $A$ becomes open. – Andrés E. Caicedo Feb 6 '11 at 0:38
Sorry, I should have said that the $\Delta^1_2$ sets are simultaneously the projection of a co-analytic set (i.e. $\Sigma^1_2$), and the complement of such a set (so also $\Pi^1_2$). A key aspect of this level of complexity is that in Gödel's constructible universe $L$, and elsewhere, there is a well-ordering of $\mathbb{R}$ of complexity $\Delta^1_2$. This is why it is consistent with ZFC that various bad behavior can occur at the level of $\Delta^1_2$, such as non-measurable sets, or sets of intermediate cardinality (in a forcing extension of $L$). – Joel David Hamkins Feb 6 '11 at 2:42

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