Totally ordered sets Let T be a totally ordered set that is finite. Does it follow that minimum and maximum of T exist? 
Since T is finite, I believe there exists a minimal of T. From that it maybe able to be shown that the minimal is the minimum but not quite sure whether it is the right approach. 
 A: Yes, so long as $T$ is nonempty. Since $T$ is totally ordered, then minimal is equivalent to minimum (one direction is easy, the other follows by totality/comparability). Similarly for maximal and maximum.
A: Claim: A totally ordered set with at least one minimal element has a minimum element.
Proof sketch: Let $b$ be the minimal element.  If $b$ were not in fact the minimum, then by definition of minimum, "$b \le a$ for all $a$" would be false.  Pick some $a$ for which $b \le a$ is not true, and derive a contradiction using totality.
A: If a totally ordered set is finite and nonempty, then it follows that it has a maximum and a minimum.
A: By induction, you can show that any linearly ordered set $T$ of cardinality $n>0$ has a minimum and a maximum. 
If $n=1$, $T=\{p\}$ and $p$ is the minimum and the maximum of $T$. Then, if $\text{card}(T)=n+1$ take a subset of cardinality $n$ ant its minimum $m$ and its maximum $M$. If $p$ is the last element in $T$, just distinguish the cases $p<m$, $p>M$ and $m<p<M$.
