Solving Supremum of Sets Homework Statement

[Lily@pie]

Homework Statement

Let A be a set of real numbers that is bounded above and let B be a subset of real numbers such that A (intersect) B is non-empty.
Show that sup (A(intersect)B) <= sup A

The Attempt at a Solution

I don't know how to start but tried this...
Let C = A (intersect) B
So sup C = sup (A (intersect) B)

Then I thought of trying to prove it by contradiction,
show sup C > sup A leads to a contradiction.

Since for all a in A, a <= sup A.
a < sup C,
can I say that this leads to a contradiction as there exist an a that is larger than c because not all elements in A are in C...

but it seems a bit weak...

 

[HallsofIvy]

The crucial point here is that $C= A\cap B$ is a subset of A. But it does NOT follow that C is a proper subset of A. It might happen that B= A, in which case $A\cap B= A\cap A= A$.

It is not just that a< sup a< sup C but that there must exist a member, x, of C such that $sup A< c\le sup C$

 

[Lily@pie]

Hmm... but is my approach correct?? Because I am totally clueless about this now...

 

[vela]

You can certainly come up with a proof by contradiction, but HallsofIvy's point is that your proposed contradiction isn't really a contradiction. As he noted, if C=A, then all elements of A are in C, so you can't assume there's an element x∈A that's not in C.

Let a = sup(A) and c = sup(C), and assume c>a. Show that a is an upper bound of C and explain why this leads to a contradiction.