Proof of a Variation on Common Induction

Question

I have been asked to prove the following variation on the common induction induction principle:

Assume $A$ is a set of integers with some $k \in A$, and furthermore whenever $n \in A$, then also $n+1 \in A$. Then A contains all integers above k.

I know that we are using $k$ as the anchor as opposed to $0$ or $1$.

Would the Least Number Principle still apply?

Thank you!

Answer 1

Yep!

What you want to do is take the set $S=\{n\in \Bbb N:n\geq k \text{ and } n\notin A\}$. Then assuming it is not empty, there is a least element $n\in S$. Thus either $n−1\in S$ or it is not. The first case would give $n$ as not a least element, so suppose it is not. Then either $n−1<k$, but then $n=k\in S$, or $n−1\in A$, so $n\in A$.