Is it possible to construct a strictly increasing function $f:\Bbb N\to \Bbb N$ with some property

Question

Let $A$ be a proper infinite subset of $\Bbb N$, say $A=\{n_1<n_2<n_3<\dotsb\}$. Let $B$ be a proper infinite subset of $A$, say $B=\{n_{k_1}<n_{k_2}<n_{k_3}<\dotsb\}$ with the property that $k_1<k_2<k_3<\dotsb$.

Question: Is it possible to construct a strictly increasing function $f:\Bbb N\to \Bbb N$ such that $f(n_{k_j})=k_j, \: \forall j\in \Bbb N$.

I am unable to construct/ product a counter example. Any help/suggestion is appreciated.

Answer 1

No.

If $f$ is strictly increasing, then necessarily $f(n)\ge n$ for all $n$ (by induction). Hence $n_{k_1}>k_1$ would be an immediate show-stopper.