Proof that $\inf A = \sup B$

Exercise

Suppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above (I have proven that $B \neq \emptyset$ and that $B$ is bounded above, so I omit the proof here and provide it as an assumption for convenience).

Prove that $\sup B = \inf A$.

I have proven here that $\sup B \in B$, which implied that $\sup B \leq \inf A$. Therefore, my strategy in proving that $\sup B = \inf A$ is to show that $\inf A \leq \sup B$, which I believe I have accomplished in the following demonstration. Of course, we could proceed with proof by contradiction, but I try to avoid indirect proofs when a direct proof is possible.

Proof

Since we know that $\sup B \in B$, then $\sup B$ is the largest element in B. We denote this fact by $M = \sup B$. By definition of $B$, we know that $\inf A \in B$. Because $M$ is the largest element in $B$, we must have the inequality $\inf A \leq M = \sup B$, as desired.

Because $\sup B \leq \inf A$ and $\inf A \leq \sup B$, we conclude by the antisymmetry law that $\inf A = \sup B$.

Solution 1:

Your proof can be greatly simplified. I’ll skip existence of $\sup B$ as that’s trivial.

As $B$ is the set of all lower bounds of $A$, we have that $b \leq a$ for all $a \in A$ and $b \in B$. Note $a \in A$ are all upper bounds of $B$, so we have $\sup B \leq a$ for all $a \in A$ by the very definition of a supremum. By definition of $B$, $\sup B \in B$ and is therefore the maximum of $B$ which means exactly that it is the greatest lower bound of $A$. Therefore, $\inf A = \sup B$.