Do we need to rigorously prove why a set is non-empty with non-negative integers?

Question

For proofs using Well-Ordering Principle(WOP), can we prove the set is a non-empty set of nonnegative integers simply by "stating"?

Eg of what I mean by "stating":

Given any integer $n$ and any positive integer $d$, there exist integers $q$ and $r$ such that $n = dq + r$ and $0 ≤ r < d$.

I would start off by saying let set $S={x\underline{\in}N| x=n-dk}$ and then simply "state" that since $x\underline{\in}N$,its a non-empty set of non-negative integers and then proceed to use WOP.

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Meanwhile, some solutions I see prove that its a non-empty set of nonnegative integers rigorously.

Eg:

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Are there any guidelines on when to prove rigorously or when to "state"? Or is it simply up to the reader to accept it?

Answer 1

When you're given a set, you cannot just state it is non-empty by picking an element from the set. In fact, to be able to say $x \in N,$ you need to justify why you can do that.

Many times, it is trivial to show that the set is non-empty. There also are times, when the given assertion trivially holds even for the empty set you can split off the proof into cases when the given set is empty or non-empty. In the latter case, you can pick an element from the set.

So for example, you have two sets $A$ and $B$ and the assertions says $A\subseteq B.$ Then you take two cases.

Case 1: $A = \emptyset.$ Then $A \subseteq B$ is trivial.

Case 2: $A \neq \emptyset.$ Then you pick $x \in A$ and use any known information to conclude $x \in B.$

You cannot say "Let $x \in A$," since $A$ maybe the empty set.

Answer 2

Let $S=\{ x \in N | x^2=-1 \}$. You cannot just claim that since $x\in N$, $S$ is non-empty. You have to actually show that that you can find an element in $S$ at all that satisfies the conditions that define the set. If such elements don't exist, then the set is empty.