# Inequality of Supremums

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 $\geq$ sup(E).

Prove:

$sup(A\cap B)\leq sup(A)$

I can show a special case of this,

if $A\cap B=\emptyset$, then $sup(A\cap B)\leq sup(A)$.

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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Ok break that inequality into two parts
first prove the equal to part. How would you do that . Assume the su pof A and B are something and that they are equal. That way you can prove the equality part
Then assume that the sup of A is greater than the Sup B. Now what is sup (a inter B) ? What is it less than?

Ok, this is what I have for the equality part if I'm understanding you right.

let $$sup(A)=sup(B)$$

then $$sup(A\cap B)=sup(A)$$

And the inequality would look like this?

let $$sup(B)\leq sup(A)$$

then $$sup(A\cap B)\leq sup(A)$$

Is that right?
It seems too simple...