Proof of uniqueness: first element of a subset $S\subseteq \mathbb{Z}$ 
If a subset $S\subseteq \mathbb{Z}$ has a first element, that element is unique.

This problem frustrates me because I came up with an almost trivial proof, but I feel like it shouldn't be so simple because it is an exercise surrounded by difficult exercises in the textbook.
My attempt:
Suppose the first (least) element is not unique. Then there exist elements $a$ and $b$ such that $a\leq x$ and $b\leq x$ for every $x\in S$. Thus, $a\leq b$ and $b \leq a$, so $a=b$ and the first element is unique.
Is this valid?
Thanks.
 A: Yes, but you don't need to argue towards contradiction here. It just clutters things and you're not really using that.
Let $s\in S$ be a first element, and $t\in S$. If $t$ is a first element, then $s\leq t$ and $t\leq s$. Therefore by antisymmetry of the relation $\leq$, it follows that $s=t$.
(Note that "first" here is taken as minimum, which means that for every $t\in S$, $s\leq t$.)
A: I think it's valid. I guess it depends on what first element means. If first element means an element $a$ so that $a\leq x$ for all $x\in S$ then your response is excelent. This type of "first" element is usually called the minimum.
If by first element you mean an element $a$ such that if $x\leq a$ then $x=a$ then the result is also true since in the natural numbers we always have $a<b, a>b$ or $a=b$ for every $a,b\in \mathbb N$ (the tricotomy law).This kind of "first" element is called a minimal element. So the same thing you said is true. In conclusion good job.
however note that in general a subset can have more than one minimal element in a partially ordered set, while there can be at most one minimum element.
