Real analysis - proof concerning infimum and supremum

Question

So I encountered this proof where it asks to prove the infimum is less than or equal to the supremum in a non empty set of real numbers that is bounded.

My approach right now is to prove it by contradiction, where I assume the infimum is greater than the supremum and then I would use the epsilon criterion to point out a contradiction in that if $\inf(S)>\sup(S)$,then the $\inf(S)$ cannot be an $\inf(S)$ at all.

However, the logic seems a bit wrong, as the contradiction lies in supposition not the assumption and I am also trying find a direct proof for the above statement. thanks!

Answer 1

Take $x\in S$; you can do it since it is not empty. Then:

• $\inf(S)\leqslant x$ since $\inf(S)$ is a lower bound of $S$;
• $\sup(S)\geqslant x$, since $\sup(S)$ is an upper bound of $S$.

Therefore, $\inf(S)\leqslant\sup(S)$.

Answer 2

Consider a set $A\subset\mathbb R$ and assume that $A$ isn't empty.

Case 1: If $A$ contains only one point then $\inf A = \sup A$.

Case 2: Suppose $A$ has more than one point. If $x,y\in A$ and $x<y$, then it is clear that $\inf A \le x < y \le \sup A$ by definition.