Past exam of the logic and set theory course of the University of Oxford / 2011 / B / 4 / b / Solution

The Cantor-Schröder-Bernstein theorem states that injections $A\hookrightarrow B$ and $B\hookrightarrow A$ imply a bijection $A\cong B$.

Let $\mathfrak{c}=|\mathbb{R}|$ and $\mathcal{D}=\{S\subseteq \mathbb{R}:S\text{ is dense in }\mathbb{R}\}$. Clearly $|\mathcal{D}|\le 2^{\mathfrak{c}}$. For the reverse bound, let $C$ be the usual Cantor set, of cardinality $\mathfrak{c}$. It is closed with empty interior, so $D=\mathbb{R}\setminus C$ is dense. The mapis injective, because $A=(D\cup A)\cap C$. Applying the Cantor-Schröder-Bernstein theorem givesOne can see $|C|=\mathfrak{c}$ by coding binary sequences as ternary expansions using only digits zero and two, together with the usual $|\{0,1{\}}^{\omega }|=|\mathbb{R}|$. The same fixed $D$ works for all subsets. This is the cardinality of the family of dense subsets of the reals.