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I am wondering if the following proof of the well ordering principle is correct by induction.

Well-ordering principle: Every non-empty subset of $\mathbb{N}$ has a least or smallest element.

To prove this, we will prove the following lemma:

Let $n\in\mathbb{N}$ and let S be a nonempty subset of $\mathbb{N}$ such that $n\in S$. Then S has a least element.

Proof by strong induction:

Base: When $n=1$, we have that $1\in S$. Then 1 is the least element of S.

Assume that when $1,2,3,...,k \in S$ then S contains a least element.

Inductive step: Let $n+1\in S$. This can be broken down into two cases.

Case 1: If any of the numbers $1,2,3,...,n\in S$ then S contains a least element by the induction hypothesis.

Case 2: If S doesn't contain any of the numbers $1,2,3,...n$ but $n+1\in S$ then $n+1$ is the least element in S.

We have concluded the lemma.

Now using the lemma to prove the well-ordering principle:

Let $S$ be a nonempty subset of $\mathbb{N}$. This implies $\exists n \in \mathbb{N}$ such that $n\in S$. By the previous lemma, S has a least element.

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Yes, that is essentially how it goes.

Note that what you're doing is really long induction, so strictly speaking your induction step should not go from $n$ to $n+1$, but from "all numbers smaller than $n$" to $n$. Or alternatively the claim you prove by induction should be "if $S$ contains some element that is $\le n$, then $S$ has a smallest element".

If you're working in a very formal setting where you have an actual definition of $<$ on the natural numbers, and axioms for the concepts that go into that definition, then it would be reasonable for you to go into more detail in case 2 to connect your reasoning rigorously to that definition. But if you don't have such a definition, there's not much else you can say but what you're saying already.

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