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.


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 $\inf A = \sup B$.

  • 3
    $\begingroup$ Seems fine to me. $\endgroup$
    – Shahab
    Oct 2 '15 at 23:16

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, $\inf A = \sup B$.


In fact, you don't need to assume $B$ is bounded up, for a in $A$, then any $b$ in $B$ has $b \leq a$.

Then if $\sup(B) > \inf(A)$, by the definition of "$\inf$", then there's a in $A$ such that $a < \sup(B)$. Then by the definition of "$\sup$", there's a $b$ in $B$, such that $b > a$. Then $b$ can not be the lower bound of $A$, contradicts and $\sup(B) \leq \inf(A)$.

If there's a $b$ such that $\sup(B) < b < \inf(A)$, then $b$ is a lower bound of $A$, and $b \leq \sup(B)$, contradicts. Then $\sup(B) \geq \inf(A)$.

So $\sup(B) = \inf(A)$.


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