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Inequality of Supremums

  • Thread starter *melinda*
  • Start date
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!
 
Last edited:
1,444
2
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?
 
86
0
Ok, this is what I have for the equality part if I'm understanding you right.

let [tex]sup(A)=sup(B)[/tex]

then [tex]sup(A\cap B)=sup(A)[/tex]


And the inequality would look like this?

let [tex]sup(B)\leq sup(A)[/tex]

then [tex]sup(A\cap B)\leq sup(A)[/tex]


Is that right?
It seems too simple...
 

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