Proposition: Given S is non-empty sup(S)=inf(S), prove that the set S has only one element.
What I stated, as this seemed rather trivial.
Proof: Let S be a non-empty subset of R that is bounded. It is given to us that sup(S)=inf(S). The claim is that S, then, has only one element within its set. We proceed by contradiction: Let a,b belong to S where a does not equal b and a is our smallest element, b is our largest element. It follows the inf(S) = a and sup(S) = b. It was given to to us sup(S)=inf(S) which entails that a = b showing uniqueness. Yet, we stated before that a doesn't equal b! We've reached a contradiction. Therefore, S contains only one element if sup(S)=inf(S).
Is this fine or does anyone think this could better stated?
Best Answer
If $\inf S = \sup S = b \in \bar{\mathbb R}$, you know that all elements of $S$ are at the same time $\geq b$ and $\leq b$. The statement follows trivially.
Note that it actually holds for all subsets of $\mathbb R$ if you define $\inf S = \infty$ and $\sup S = -\infty$ when $S$ is empty.