- #1

*melinda*

- 86

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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!

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