- #1
*melinda*
- 86
- 0
Theorem:
For every non empty set E of real numbers that is bounded above there exists a unique real number sup(E) such that
1. sup(E) is an upper bound for E.
2. if y is an upper bound for E then y [itex]\geq[/itex] sup(E).
Prove:
[itex]sup(A\cap B)\leq sup(A)[/itex]
I can show a special case of this,
if [itex]A\cap B=\emptyset [/itex], then [itex]sup(A\cap B)\leq sup(A)[/itex].
Nothing is less than something, right?
Now here's my problem...
Beyond the trivial case, all I have been able to do is draw pictures of sets on a number line. The pictures make the inequality really obvious, but I don't think that pictorial intuition counts as a real proof.
Could anyone give me a pointer on how to set up a real proof?
thanks!
For every non empty set E of real numbers that is bounded above there exists a unique real number sup(E) such that
1. sup(E) is an upper bound for E.
2. if y is an upper bound for E then y [itex]\geq[/itex] sup(E).
Prove:
[itex]sup(A\cap B)\leq sup(A)[/itex]
I can show a special case of this,
if [itex]A\cap B=\emptyset [/itex], then [itex]sup(A\cap B)\leq sup(A)[/itex].
Nothing is less than something, right?
Now here's my problem...
Beyond the trivial case, all I have been able to do is draw pictures of sets on a number line. The pictures make the inequality really obvious, but I don't think that pictorial intuition counts as a real proof.
Could anyone give me a pointer on how to set up a real proof?
thanks!
Last edited: