Consider the following equivalence class:
$${[\mathbb{N}]_s} = \{ A \in P(\mathbb{Z}) : |A| = |A \cup \mathbb{N}| \wedge |\mathbb{N}| = |A \cup \mathbb{N}|\} $$
So, $A$ must be infinite set with the power of $\aleph_0$.
I'm trying to understand the statement below. The set in the middle isn't very clear to me. And apparntly it's power is $\aleph$. $$\{ A \cup {\mathbb{N}_{even}}:A \subseteq {\mathbb{N}_{odd}}\} \subseteq {[\mathbb{N}]_s}$$
I'll appreciate an explanation to make it clear.
Thanks in advance.
First of all, writing that $\aleph$ is equal to that set is false. It is equal to the cardinality of that set, yes. But not to that set specifically.(The remark is related to a previous revision.)Now, the set is the set all of subsets of $\Bbb N$ which include all the even numbers (from $\Bbb N$), to see that it is so note that every subset of $\Bbb N$ including the even numbers can be written as a subset of the odd numbers union the set of all even numbers.
This shows that there is an injection from $\mathcal P(\Bbb N_{odd})$ into $[\Bbb N]_s$, and since the former set has size $2^{\aleph_0}$, or $\aleph$, so much the latter.