If the set of all upper bounds of $A$ and $B$ are equal and $\sup A$ exists, then $\sup B$ exists and $\sup A=\sup B$.

Question

Caution: Axiom of Completeness is not assumed here.

Before reading my attempt, I want you to think up a proof of your own.

Here is my attempt:

Let $D$ be the set of all upper bounds of $A$ as well as $B$. Since $\sup A$ exists, $\sup A \le d$ for all $d \in D$. It also means $D$ is bounded below. Now I have to show that $\inf D$ exists and is equal to $\sup A$. (This is the part that I have no idea how to approach.) Then since $\inf D$ exists, $\sup B$ exists and is equal to $\inf D$. Therefore, $\sup A = \sup B$.

This concludes my proof. Two questions:

1. How do I show the part in my proof where I was stuck?
2. Is there any other way to prove it? Preferably much simpler. I want to see some other ways.

Answer 1

Let $U(A)$ be the set of upper bounds for $A.$ The definition of $\sup A$ is $\min U(A),$ which exists iff $U(A)$ has a least member.

Suppose $U(A)=U(B)$ and that $\sup A$ exists. Then $\min U(A)$ exists. So $\min U(B)=\min U(A)$ exists because $U(A)$ and $U(B)$ are the same thing.

Hence $\sup A=\min U(A)=\min U(B)=\sup B.$

Answer 2

You need to at some point invoke the definition of $\sup$ (or something that follows from it), because otherwise you're just talking about a symbol you know nothing about.

One definition of $\sup A$ is a real number $M$ such that:

• For every $a \in A$, $a \le M$.
• For every $\epsilon>0$, there is an $a \in A$ such that $a > M-\epsilon$.

So you want to prove:

1. For every $b \in B$, $b \le M$. This is true because $M$ is an upper bound on $A$, therefore it is an upper bound on $B$.
2. For every $\epsilon>0$, there is a $b \in B$ such that $b > M - \epsilon$. This is easiest to prove by contradiction. Suppose that there is some $\epsilon>0$, such that for all $b \in B$, $b \le M-\epsilon$. Then $M-\epsilon$ is an upper bound on $B$, therefore it is an upper bound on $A$. This means that for all $a \in A$, $a \le M-\epsilon$, which contradicts the definition above. Therefore no such $\epsilon$ can exist.

Obviously this is a bit specific to the definion you're using. Another definiton of $\sup A$, instead of the statement with $\epsilon$, specifies that if $M'$ is another upper bound on $A$, then $M \le M'$. This would lead to a different proof, but the idea is the same.