# Real analysis - proof concerning infimum and supremum

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!

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, $$\inf(S)\leqslant\sup(S)$$.
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 $$x, then it is clear that $$\inf A \le x < y \le \sup A$$ by definition.