Problem: prove that for each $n \in \mathbb{N}$ : $(n,n+1) \cap \mathbb{N} = \varnothing $
attempt:
Indeed, let $S(n)$ be the statement $\{ n : (n,n+1) \cap \mathbb{N} = \varnothing \}$. Clearly, $(1,2) \cap \mathbb{N} = \varnothing$, thus $1 \in S(n)$. Assume now that $S(n)$ is true. Then,
$$ (n+1,n+2) \cap \mathbb{N} = (n,n+1) \cap (\mathbb{N} \setminus [n+1,n+2) ) \cap \mathbb{N} = (n,n+1) \cap \mathbb{N} \cap ( \mathbb{N} \setminus [n+1,n+2) ) = \varnothing $$
and the claim follows by the principle of mathematical induction.
Is this correct proof?