Proof about Cardinality/$\aleph_0$

Question

How can I prove that $$\forall n \in \mathbb{N}, n < \aleph_0$$ I was recommended to use a diagonalization argument similar to that used to prove that $|\mathbb{R|} \neq \aleph_0$. Any idea what to do?

Answer 1

The easiest way is to use the pigeonhole principle. Obviously $n\le \aleph_0$ for every $n$, so suppose $\aleph_0\le n$ for some $n$. Then $n+1\le\aleph_0\le n$, which is a contradiction to the pigeonhole principle. Do you see why?

Answer 2

Construct subsets of increasing cardinality of the natural numbers that have cardinality $n$. By induction, they are always less than the aleph null.