Cfrgtkky

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.

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

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.