If k is the least element of S={n∈N∣¬P(n)}, how does P(1) being true mean k is not 1?

Question

I do not understand part of this expression.

Let S={n∈N∣¬P(n)}

I understand (I think) that S is set with elements equal to P(n) such that n is subset of the natural numbers. What does that bent line mean that is right before the P(n)?

Also, there is a second part to this expression.

Let S={n∈N∣¬P(n)}. Assume it is non-empty. Then there is a least element; call it k. Since P(1) is true, we know that k≠1.

How do we know that k does not equal one, even though it is the smallest element of the natural numbers? How does P(1) being true imply this?

Answer 1

You are assuming that $k$ is an element of $S$.

Every element of $S$ by definition is a natural number which does not satisfy $P$.

That is $P(k)$ is not true.

Since $P(1) $ is true and $P(k)$ is not true, $k$ is not equal to $1$