Consider the following $P(n)$ is "Every set of positive integers that contains an integer less than or equal to $n$ must have a smallest element" Prove by ordinary induction that $P(n)$ is true for all $n$.
Would like help with this induction proof. I know the structure of the induction proof but I don't understand how to quantify the question it's asking.
I only have the skeleton of the proof so far
We will prove that $P(n)$ is true for all $n$ greater than or equal to 1, where $P(n)$ is...
Basic Step: Clearly $P(1)$ is true since...
Induction Step: Let $k$ be an arbitrary integer that is greater than or equal to...
Assume that $P(k)$ is true (That is, assume that...)
We want to show that $P(k+1)$ is true.
P(1) is true since every set containing 1 has a smallest element, which is 1.
Assume P(k) is true.
P(k+1): "Every set of positive integers that contains an integer less than or equal to k+1 must have a smallest element"
Every set that contains an element less than or equal to k has a smallest element, by P(k).
This leaves the sets containing k+1 and larger elements. Subtract 1 from each element in these sets. They must still be sets of positive integers, since none of these sets originally contained 1. Also, these sets must each have a smallest element, by P(k). Add 1 back to each elements in the set. Since each set contained an element e <= s for all elements s in the set, e+1 <= s+1 for all elements s in the set, which means that each set still contains a smallest element.