Does a sequence require a unique value for a particular natural number? In a question, I was asked to prove the existence of a sequence that converges to $\sup S$,  where every element is an element in a set S.
The solution used defined ($A_n$) for $n$ natural numbers as $|\sup S - 1/n| < A_n < \sup S$.
However for any $N$ there are infinite possible elements that satisfy that inequality. Is the a sufficient definition for a sequence?
 A: Take any element $s\in S$ and build a constant sequence $(a_n)_{n\in\mathbb N}$ on it: $$\forall_{n\in\mathbb N}\ a_n = s$$
The sequence is constant, so it's convergent, hence the existence of a convergent sequence is proven.
A: No, a sequence has to have unique values. In fact a sequence is just another name for a function from $\mathbb N$.
What you know on the other hand is if you have a sequence of non-empty sets $S_n$ there exists a sequence $s_n\in S_n$, that is you can from your ambiguous definition select a unique definition.
In your concrete example you have only to know that there exists an $A_n$ such that $|\sup S - 1/n| < A_n < \sup S$, and by knowing that you know you can select one $A_n$ that fulfils that requirement.
Better would of course be if you have a concrete method to select $A_n$ (this is often possible, but not in your case since you know nothing more about $S$) because if you haven't you might have to rely on the axiom of choice (which could be considered a little bit controversial). 
A: This is a stronger result: it says that all sequences that fulfill the condition do converge to $\sup S$.
The proof is non-constructive (it doesn't give a sequence explicitly).
A: So as I said in the comment, you need the axiom of (countable) choice. For every $n\in\Bbb N$ the set $X_n=[\sup S-\frac1n,\sup S]\cap S$ is non-empty. Then by that axiom there is a function $f:\Bbb N\to\Bbb R$ with $f(n)\in X_n\subseteq S$, one which gives the sequence $A_n=f(n)$. One easily sees that $\lim_{n\to\infty}A_n=\sup S$.
