Let S be an ordered set. Let $A \subset S$ be a nonempty finite subset. Then prove that A is bounded. Furthermore, show that inf A exists and is in A and sup A exists and is in A. Hint : Use Induction.
I tried to construct an inductive argument as suggested by the hint. For the base case, I took a singleton set and it was quite easy to show that it is bounded. Then proceeding along with the argument, I assumed that A (where |A| = k) is bounded. I have no idea how to show that A with |A| = k + 1 is bounded from the inductive hypothesis.
And perhaps I would like to get some hints for the second part of problem.
I assume your set is totally ordered. For the inductive step, assume that every subset of $S$ of cardinality $k$ is bounded. Let $A=\{a_1, \dots, a_k, a_{k+1}\} \subseteq S$ be a subset of cardinality $k+1$. We want to show that $A$ is bounded. Take $\{a_1, \dots,a_k\} \subseteq A$. By hypothesis, it is bounded, so there exist $m,M \in S$ such that $m \leq a_i \leq M$ for all $ i = 1,\dots,k$. Now, if $m \leq a_{k+1}$, $m$ is a lower bound of $A$. If $a_{k+1} \leq m$, then $a_{k+1}$ is a lower bound of $A$. You can argue similarly for the upper bound.