how to prove that: $|X| = 2^{\aleph_{0}}$, given $X = \{ A\in P(\mathbb{N}) \vert \ |A^{c}| = \aleph_{0} \}$ Let $X = \{ A\in P(\mathbb{N}) | \ |A^{c}| = \aleph_{0} \}$


Prove\ Disprove that: $|X| = \aleph_{0}$


My attempt:


I'd like to disprove that: $|X|=\aleph_{0}$, By proving that $|X|=2^{\aleph_{0}}$.


For every $A\in X\to A\in P(\mathbb{N})$,then $X\subseteq P(\mathbb{N})\to|X|\leq|P(\mathbb{N})|=2^{\aleph_{0}}$.

Now I'd like to prove that there exists a set, $K$,

such that $|K|=2^{\aleph_{0}}$, and $K \subseteq X$, which will finish the proof.

Suppose by contradiction that there exists no set $K$ as described,

such that: $K\subseteq X$.

Now, let $A=2\mathbb{N}$, then $A^{c}=\mathbb{N}\setminus2\mathbb{N}=\mathbb{N}_{odd}$,

and $|A^{c}|=|\mathbb{N}_{odd}|=\aleph_{0}\to A\in X$.

Notice that for every set $i$, $\alpha_{i}\subseteq2\mathbb{N}:$ $\mathbb{N}_{odd}\subseteq\alpha_{i}^{c}\subseteq\mathbb{N}\to|\alpha_{i}^{c}|=\aleph_{0}$,

therefore for every $i$ and for every $\alpha_{i}$ as descirbed, $\alpha_{i}\in X $.

the collection of all $\alpha_{i}$ exists: $\{\alpha_{i}\vert\ \forall i\in\mathbb{N}:\alpha_{i}\subseteq2\mathbb{N}\}$,hence

$P(2\mathbb{N})=\{\alpha_{i}\vert\ \forall i\in\mathbb{N}:\alpha_{i}\subseteq2\mathbb{N}\}$

hence $P(2\mathbb{N})\subseteq X\Longrightarrow \ 2^{\aleph_{0}}=|P(2\mathbb{N})|\leq|X|$

and this contradicts the assumption,

that there exists no $K\subseteq X:|K|=2^{\aleph_{0}}$.

by Cantor Bernstein we get that $|X|=2^{\aleph_{0}}$.
I'd like to disprove that: $|X|=\aleph_{0}$, By proving that $|X|=2^{\aleph_{0}}$.

For every $A\in X\to A\in P(\mathbb{N})$,then $X\subseteq P(\mathbb{N})\to|X|\leq|P(\mathbb{N})|=2^{\aleph_{0}}$.

Now I'd like to prove that there exists a set, $K$,

such that $|K|=2^{\aleph_{0}}$, and $K \subseteq X$, which will finish the proof.

Suppose by contradiction that there exists no set $K$ as described,

such that: $K\subseteq X$.

Now, let $A=2\mathbb{N}$, then $A^{c}=\mathbb{N}\setminus2\mathbb{N}=\mathbb{N}_{odd}$,

and $|A^{c}|=|\mathbb{N}_{odd}|=\aleph_{0}\to A\in X$.

Notice that for every set $i$, $\alpha_{i}\subseteq2\mathbb{N}:$ $\mathbb{N}_{odd}\subseteq\alpha_{i}^{c}\subseteq\mathbb{N}\to|\alpha_{i}^{c}|=\aleph_{0}$,

therefore for every $i$ and for every $\alpha_{i}$ as descirbed, $\alpha_{i}\in X $.

the collection of all $\alpha_{i}$ exists: $\{\alpha \vert\ \alpha \subseteq2\mathbb{N}\}$,hence

$P(2\mathbb{N})=\{\alpha\vert\ \alpha \subseteq2\mathbb{N}\}$

hence $P(2\mathbb{N})\subseteq X\to \ 2^{\aleph_{0}}=|P(2\mathbb{N})|\leq|X|$

and this contradicts the assumption,

that there exists no $K\subseteq X:|K|=2^{\aleph_{0}}$.

by Cantor Bernstein we get that $|X|=2^{\aleph_{0}}$.

therefore for every $\alpha \subseteq P(2\mathbb{N}),  \alpha \subseteq K$, 

hence $P(2\mathbb{N}) \subseteq K \to  |P(2\mathbb{N})| \leq |K|$, by using Cantor Bernstein theorem we conclude that $|K| = \aleph$.
take $A=\{1,2,3,4\}$

$A^c$={$\mathbb N$ $\setminus ${1,2,3,4}} $

$|A^c|=\aleph_0$
is $|\mathbb N $ $\setminus$ $A^c|$ = $\aleph_0$ ?