The lower density of $A\subseteq\mathbb{N}$ is defined to be $\lambda(A)=\lim\text{inf}\_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$. We set $${\cal C} = \{A\subseteq \mathbb{N}: \lambda(\mathbb{N}\setminus A) = 1 - \lambda(A)\}.$$


Do both ${\cal C}$ and ${\cal P}(\mathbb{N})\setminus {\cal C}$ have cardinality $2^{\aleph\_0}$?



Subsets of $\mathbb{N}$ whose lower density respects complements
The answer is *yes*: Given a set $A$ in $\mathcal{C}$ or in
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$, you can take
any subset $B \subset \mathbb{N} \setminus A$ of lower density $0$,
and $A \cup B$ is still in $\mathcal{C}$ or in
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$, depending on
which of the two classes $A$ itself belongs to.
Now, unless $A$ is cofinite, the cardinality of the set of subsets
of $\mathbb{N} \setminus A$ with lower density $0$ is $2^{\aleph\_0}$,
so you have $2^{\aleph\_0}$ choices for $B$. The result follows since
each of the classes $\mathcal{C}$ and
$\mathcal{P}(\mathbb{N}) \setminus \mathcal{C}$ has at least one
member which is not cofinite, and since the set of cofinite subsets of $\mathbb{N}$
is countable.


