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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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 $K$. 

