I am new to cardinality proofs and so far I can't even understand exactly what I have to do here. Can anyone enlighten me?
The task is the following: Let $M = \{ n \in \mathbb N^+ \mid n \bmod 3 = 0 \}$. Is it true that $\operatorname{card}(M) = \operatorname{card}(\mathbb N)$?
EDIT:
I just uploaded the original question in German:
It should be an '$=$' sign, not an '_'. With '$=$' it's true, with '_' it makes no sense.
Given that $M$ is infinite (why?), it's enough to show that there is an injection $M\to \Bbb N$. Together with other results, this implies that $card(M) = card(\Bbb N)$. However, it's even easier in this case to show that there is a bijection $\Bbb N\to M$. Consider: $$ f\colon k\mapsto 3(k+1)\colon \Bbb N \to \Bbb N. $$
Every value of $f$ is divisible by $3$, so $\rm{image}(f)\subseteq M$, and any $x\in M$ is of the form $x=3n$ for some $n\in\Bbb N^+$, so $n=k+1$ for some $k\in\Bbb N$. Thus $\rm{image}(f) = M$, so $f$ is a surjection onto $M$: $$ f\colon \Bbb N \stackrel{onto}\longrightarrow M. $$
If $f(j) = f(k)$, then $3(j+1) = 3(k+1)$, so $j+1=k+1$, so $j=k$. Hence, $f$ is an injection too: $$ f\colon \Bbb N \stackrel{1-1}\longrightarrow M. $$ So $f$ is a bijection from $\Bbb N$ to $M$.