Prove that in any non-empty chain $L$ in $\left\langle A, \le \right\rangle $ exist the smallest element 
In the set $A=\left\{ \left\langle k,l\right\rangle  \in \mathbb Z  \times \mathbb Z: k<l\right\} $ we introduce relation of a partial order $\le$ for which $k' \le k$ and $l \le l'$.Prove that in any non-empty chain $L$ in $\left\langle A, \le \right\rangle $ exist the smallest element.

I have a problem with this task because I think this sentence is false because in $A$ exist an infinite descending sequence: $G=\left\{ \left\langle a,42\right\rangle : a<42\right\}$ and infinite growing sequence $F=\left\{ \left\langle a,b\right\rangle : 0<a<b  \wedge  b=a+1\right\}$ so in $ A $ there is no smallest element or the largest element. That is why also the sentence in the task is false because I can have $L=A$. At the same time, it is an exam task from a few years ago, so it is hard to believe that the thesis would be false.Can you rate this?
 A: There seems to be some confusion with the problem, so let me state things clearly. We consider the set $A=\left\{\langle k,l \rangle\in\mathbb Z\times \mathbb Z\mid k<l\right\}$ with the partial order $\langle k,l \rangle\leq \langle k',l' \rangle$ if $k'\leq k<l\leq l'$. 
Another way to interpret $A$ is as the set of integer intervals (i.e. define $\langle k,l\rangle=\{k,k+1,\ldots, l\}$ ) with non-zero length (i.e. $l-k>0$) ordered by subset inclusion (i.e. $\langle k,l\rangle\leq \langle k',l' \rangle$ if and only if $\langle k,l\rangle\subseteq \langle k',l' \rangle$).
The claim is that any non-empty chain in $A$ has a minimal element, where recall:


*

*A chain $C$ in $A$ is a subset $C\subseteq A$ such that for any $c,c'\in C$,
either $c\leq c'$ or $c'\leq c$; i.e. any two elements are comparable.

*A minimal element of a subset $S\subseteq A$ is an element $m\in S$ such that if $s\in S$, either $m$ and $s$ are not comparable, or $m\leq s$.
Note that in a chain, only the second condition in the definition of a minimal element is relevant.

Now suppose $C$ is a non-empty chain in $A$ and let $L(C)=\{l-k \mid \langle k,l\rangle\in C\}$. Observe that


*

*Since $C$ is nonempty, $L(C)$ is non-empty.

*We have $L\subseteq \mathbb N$: note that by definition, $x\in L$ implies $x=l-k$ for some $\langle k,l\rangle\in A$, hence in particular $k,l\in\mathbb Z$ and $k<l$, thus $x=l-k\in \mathbb N$.
In particular, the Well-Ordering Principle applies, hence $L(C)$ has a least element
$y$. Then by definition, there exists some $k,l\in \mathbb Z$ such that $\langle k,l\rangle \in C$ and $l-k=y$. Moreover, we claim $\langle k,l\rangle$ is minimal. 
To see this, note that since $C$ is a chain, it suffices to show there is no smaller element in $C$. To that end, suppose $\langle k',l'\rangle \leq \langle k,l\rangle$. Then $k\leq k'<l'\leq l$ by definition, which implies $l'-k'\leq l-k$. Since $\langle k',l'\rangle \in C$, we have $l'-k'\in L(C)$. But since $l'-k'\leq l-k=y$ and $y$ is minimal in $L(C)$, it must be that $l'-k'=l-k$. Then $l'=l$ and $k'=k$ (since $(l-l')+(k'-k)=0$ with $(l-l'),(k'-k)\geq 0$ is only possible if the summands are zero), hence $\langle k',l'\rangle=\langle k,l\rangle$.
We have shown that any element of $C$ which is less than or equal to $\langle k,l\rangle$ must be equal to $\langle k,l\rangle$, hence this is the claimed minimal element.

As a final comment, note that the alternate interpretation in terms of intervals makes this proof pretty easy to see. Indeed, the claim is that any non-empty nested collection of integer intervals of nonzero length has an interval which is contained in all others. To see this, note that given any interval $\langle k,l\rangle$, there are only finitely many choices of endpoints $\langle k',l'\rangle$ with $k\leq k'<l'\leq l$, so only finitely many elements of $C$ less than $\langle k,l\rangle$, hence one of them is smallest (as they are totally ordered).
