Cardinality of set $ M = \{ A \in \mathcal{P}( \Bbb N ) : |A| = \aleph_0 \land |A^c|=\aleph_0 \} $

Question

Problem:

We have the set $ M = \{ A \in \mathcal{P}( \Bbb N ) : |A| = \aleph_0 \land |A^c|=\aleph_0 \} $ . Show that $ |M| \neq \aleph_0 $

( Here $ A^c $ means the complement of $ A $ with respect to the universe $ \Bbb N $ )

Attempt:
Define $$ E = \{ 2n : n \in \Bbb N \} $$ $$ F = \{ 6n : n \in \Bbb N \} $$ We'll also define $$ T = \{ B \in \mathcal P (\Bbb N) : F \subseteq B \subseteq E \} $$ Note that $ T \subseteq M $ since forall $ \tau \in T $ we have $ F \subseteq \tau \subseteq E $ and $ |\tau| = \aleph_0 , |\tau^c| = \aleph_0 $. We'll show that $ \aleph \leq |T| $ . [ From here I got stuck and couldn't proceed, I couldn't figure what bijection or one-to-one function to define in order to show the last inequality ( from which I'll use the Cantor-Schroeder-Bernstein theorem in order to prove the problem above ) ]

Does anyone know how to proceed? Maybe you have an alternative proof? Thanks in advance!

Answer 1

Let $\beth_1 = |P(\mathbb{N})|$.

Let $B = \{A \in P(\mathbb{N}) \mid |A| < \aleph_0\}$, $C = \{A \in P(\mathbb{N}) \mid |A^c| < \aleph_0\}$, and $D = \{A \in P(\mathbb{N}) \mid |A| = \aleph_0 \land |A^c| = \aleph_0\}$.

Note that $|\{A \in P(\mathbb{N}) \mid |A| < \aleph_0\}| = \aleph_0$. This is because the set of finite sequences of naturals has cardinality $\aleph_0$ (as it's the countable union of $\mathbb{N}^n$ for all $n$ finite), and finite sets can be viewed as finite sequences that are increasing.

This set can be put into bijection with $\{A \in P(\mathbb{N}) \mid |A|^c < \aleph_0\}$ by taking complements, so this set also has cardinality $\aleph_0$.

We see that $B$, $C$, $D$ are pairwise disjoint infinite sets. So we have $\beth_1 = |P(\mathbb{N})| = |B \cup C \cup D| = \max(|B|, |C|, |D|) = \max(\aleph_0, |D|)$.

Therefore, $|D| = \beth_1$.

Answer 2

An easy proof that the cardinality is $2^{\aleph_0}$ is to note that:

1. The cardinality of the collection of finite subsets of $\mathbb{N}$ is $\aleph_0$, hence the cardinality of the collection of infinite subsets must be $2^{\aleph_0}$; and
2. Every infinite subset of $2\mathbb{N}$ lies in $M$, hence $M$ has at least as many elements as there are infinite subsets of $\mathbb{N}$ (since there is an obvious bijection between subsets of $\mathbb{N}$ and subsets of $2\mathbb{N}$).