I have a set of statements such as:
Proof $\aleph_0+\aleph_0=\aleph_0$
I know that $|\Bbb Z|=\aleph_0$ and that for countable $A,B$ $A\cap B=\emptyset$: $|A\cup B|=|A|+|B|$.
To this I add that if $A=\{-1,-2,...\}$ and $B=\Bbb N$, $|A|=|B|=\aleph_0$ then $|\Bbb Z|=\aleph_0=\aleph_0+\aleph_0=|A|+|B|$. Does this particular case prove the statement?
Also, could you give me any suggestions on:
$\sum_{i=1}^n\aleph_0=\aleph_0, n\in \Bbb N$
$$f(n)=(-1)^{n+1}\left \lfloor \frac{n}{2} \right \rfloor\\f:\mathbb{N} \rightarrow \mathbb{Z}$$ $$f(1)=0 ,1 \to 0 \\f(2)=-1 ,2 \to -1\\f(3)=+1 ,3 \to +1\\f(4)=-2 ,4 \to -2 \\f(5)=+2 ,5 \to +2\\,...\\\left \{ 1,2,3,4,5,6,7,... \right \}\overset{f(n)}{\rightarrow}\left \{ 0,-1,+1,-2,+2,-3,+3,... \right \}$$ f is bijective
so $$card(\mathbb{N})=card( \mathbb{Z})$$ if $$|A|,|B|<\infty \to |A \cup B|=|A|+|B|-|A \cap B|$$