Let $F: \mathbb{N} \to \mathbb{N}$ and for all $n,m \in \mathbb{N}$ if $m>n$ then $F(n) >F(m)$, let $A_n = F^{-1} [\{0,1,\cdots,n-1\}]$, prove that for all $n$ we have that $A_n\subset \mathbb{N}^{< n}$ ?
My attempt : by induction ,
Base case : $A_0 = \emptyset \subset \mathbb{N}^{< 0} = \emptyset$
Assume that $A_n\subset \mathbb{N}^{< n}$, i want to prove that $A_{n+1} \subset \mathbb{N}^{< n+1}$ , otherwise there is $a \in A_{n+1}- \mathbb{N}^{< n+1}$ , meaning $a\geq n+1$ and $F(a)=n$
If $a>n+1$ then $F(a)= n >F(n+1)$ so $F(n+1) \in {0,1,\cdots,n-1}$ meaning $n+1 \in A_n$ contradicting the assumption, but i got stuck in the case $a=n+1$