# Supremum of sets

1. Aug 13, 2011

### Lily@pie

1. The problem statement, all variables and given/known data
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

3. 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,

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...

2. Aug 13, 2011

### HallsofIvy

Staff Emeritus
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 $sup A< c\le sup C$

3. Aug 13, 2011