Proofs with for all statements including uniqueness and divides Let $\mathcal{A}$ be a nonempty finite set of positive integers, with $\forall$ r $\in$ $\mathcal{A}$, $\forall$ s $\in$ $\mathcal{A}$ : r|s or s|r.
(i). Prove $\exists$t $\in$ $\mathcal{A}$: t|a, $\forall$ a $\in$ $\mathcal{A}$
(ii). Prove t is unique
(iii). Suppose $\mathcal{A}$ $\subseteq$ $\mathbb{Z}$, then prove that t need not be unique. 
 A: (i) Let $t$ be the least element in $A$. Such $t$ exist because $A$ is a subset of positive integers hence well ordered. Now for any $a\in A$ and $a\neq t$, since $t<a$ we have $a\not\mid t$ hence $t\mid a$. Also we have $t\mid t$ for completeness.
(ii) There exist only one least element in $A$.
(iii) We can let $A=\{3^x|x\in \mathbb{N}\}\cup\{-3^x|x\in\mathbb{N}\}$ then $3$ and $-3$ are both valid $t$. The reason behind this is because $\mathbb{Z}$ is not a well order unlike $\mathbb{N}$.
A: cr001's answer is correct.  And I upvoted it.  But here's a meta-answer.
I wonder why the specification that $A$ must be finite is stated in the problem.  It isn't necessary.  So I began to wonder if the intuitively obvious fact that every set (infinite or otherwise) of positive integers has a least element hasn't actually been proven.  (It's easy  to prove but it's not trivial. After all not every set of positive rationals have a least element.  {$1/n| n \in \mathbb N$} doesn't.  So why should every set of positive integers have a least element?  [Don't answer; it was a rhetorical question.] 
However every finite set of  real numbers (not just integers, not just rationals, any type of real numbers) has a least element and maybe this had been recently demonstrated and the point of this was to reinforce the idea.
Every finite set of numbers has a least element because:
1) The elements of any finite set can but indexed $x_1, x_2, ...., x_n$.
2) Given any two numbers, $a$ and $b$, one and only one of the following is true:  Either $a < b$; $b < a$; or $a = b$ (although one could quibble that if $a = b$ they are not two numbers but just one.)   One of those must be true and the other two must be false. 
Also for any three numbers $a, b, c$ with $a < b$ and $b < c$ it must also be true that $a < c$.  
In abstract analysis terms we call "<" an "order".
3) Because the elements are finite and indexed we can start at the first and compare every element to every other element and keep a tally.  If $x_1 < x_2$ we let $t := x_1$.  If $x_2 < x_1$ we let $t := x_2$. (As $x_1 \ne x_2$ we know one of these two must be true.)  We then compare $t$ to $x_3$ and reassign if necessary.  And so one.  Once we've compared all all elements of the set, we will have determined that $t < x$ for all $x \ne t$. And we conclude $t$ is the least element of the set and it is unique.
Three important things to note:
A) Condition 1: We can index and access the elements one after another.  This might not be true for an infinite set.  There is no way to do this for the set $U = ${$ 1/n | n \in \mathbb N$}.  That is why we can not conclude $U$ has a least element.
B) The "order" of "being less than" is an abstract one.  We use the idea of being "less than" ($a < b$ means that $b - a$ is a positive amount) as an "order".  But any relationship "~" where exactly one of $a ~ b$ or $b ~ a$ or $a = b$ is true, and where $a ~ b$ and $b ~ c$ means $a~ c$ must be true can be an "order" and any finite set, A,  of anything with such an "order" will have a unique "least" element, t, where $t ~ x$ for all $x \ne t$ in $A$. 
And C) If the "order" weren't "strict" if is was possible to have $a < b$ and $b > a$, or it were possible to have neither $a < b$ nor $b < a$ while $a \ne b$ then we wouldn't be able to make the conclusion.
So if $A$ is a set of positive integers such that for any two $r, s \in A$ either $r|s$ or $s|r$, then "| but not equal" could (for the elements of set A) be considered an "order".
Note: If $r \ne s$ and $r | s$ then $s \not\mid r$. So one and only one of the three apply: ($r \mid s$ and $r \ne s$) or ($s \mid r$ and $r \ne s$) or ($r = s$).
And note that if $r \mid s$ and $s \mid t$ then $r \mid t$ (because if $s = nr; n > 1$ and $t = m s; m > 1$ then $t = mnr; mn > 1$).
So "$\mid$ but not equal" is an "order" that works on finite set $A$ so by the exact same argument, we know there must be a unique "minimal" element" $t$ so that $t \mid x$ for all $x \in A$.
But what if $A$ can have negative elements.  Then it'd be possible, if $s = -t$ for $s \mid t$ and $t \mid s$ BUT $t \ne s$.  So now our argument wouldn't hold.  We could find $t \mid x$ for all $x$ and $s \mid x$ for all $x$, but $t = -s \ne s$ so the elements will not be unique.
On a side note, this is why we can't compare "sizes" of points on a plane or complex numbers.  One point might be "bigger" (further from the origin) than another but multiple different "equal sized" points (equal distance from the origin) exist. 
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POST SCRIPT:  Since I brought it up, how do you prove that an infinite set of positive integers, B, has a unique least element?
Well, Take $x \in B$.  $x > 0$ so $x \ge 1$.  If $1 \in B$ we are done.  If not then $z > 1$ for all $z \in B$.  Compare one after another the numbers {2, .... x}.  As $x \in B$ there must be at least one of these numbers in B.  The first number, n, that we find that is in B will be the unique least number in B.
This can not be done with an infinite set of positive and negative integers because we have no place to start.  We can't start at 0 because we don't know $x > 0$ or that there are no $x < 0$.
