Let $A$ be nonempty and bounded below, and define $B=\{b\in\textbf{R}:b\text{ is a lower bound for }A\}$. Show that $\sup B=\inf A$.
Since $\sup B$ exists by the axiom of completeness, and since $\sup B$ is a lower bound for $A$ by defintion, $\sup B\in B$ and is a maximum of $B$.
Let $i$ be a lower bound for $A$. Then $i\in B$. Since $\sup B$ is a maximum of $B$, $\sup B\geq i$. We now have that $\sup B$ is a lower bound for $A$ and that $\sup B$ is greater than or equal to any other lower bound for $A$, so we may conclude that $\sup B=\inf A$.
Is this proof correct? It seems simpler than most other proofs of this, so I would like to make sure I'm not making a mistake anywhere.
Best Answer
You assumed that $B$ is bounded above. You must prove that.
But that's easy with precise definitions. Is there a number that is as large or larger than all lower bounds of $A$? Well.... if you think this through....(Hint: what about actual elements of $A$? Are they at least as large as every lower bound of $A$?)
Second you assume that $\sup B$ is an lower bound of $A$ by definition. But it is not. $\sup B$ is the least upper bound of the lower bounds of $A$ but there is nothing in the definition that says it is an actual lower bound.
But if it isn't a lower bound of $A$ then there is an $a \in A$ so that $a < \sup B$... which says what about $a$, the set $B$ and upper bounds of $B$? Is $a$ an upper bound of $B$? Are there any elements of $B$ between $a$ and $\sup B$? What does that imply?
With those hints you can indeed prove that $B$ is bounded above and that $\sup B$ is an lower bound of $A$. Once you've done that your argument that that would mean $\sup B = \inf A$ is correct.