Can You Prove the Inequality of Supremums Theorem?

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In summary, the theorem states that for any non-empty set E of real numbers that is bounded above, there exists a unique real number sup(E) such that it is an upper bound for E and any other upper bound for E must be greater than or equal to sup(E). The goal is to prove that sup(A∩B)≤sup(A) by first considering the case where A∩B is empty, and then proving the inequality for the general case by assuming the sup of A is greater than the sup of B and finding the relationship between sup(A∩B) and sup(A).
  • #1
*melinda*
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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!
 
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  • #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?
 
  • #3
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...
 

1. What is meant by "inequality of supremums"?

"Inequality of supremums" refers to a mathematical concept that states that the supremum (or least upper bound) of one set is always greater than or equal to the supremum of another set. This means that the maximum value of one set is always greater than or equal to the maximum value of another set.

2. How is "inequality of supremums" used in mathematics?

"Inequality of supremums" is used in various branches of mathematics, such as real analysis, measure theory, and functional analysis. It is a fundamental concept that helps in proving theorems and solving problems related to sets and functions.

3. Can you provide an example of "inequality of supremums"?

Sure, let's say we have two sets: A = {1, 2, 3} and B = {4, 5, 6}. The supremum of set A is 3, while the supremum of set B is 6. According to the inequality of supremums, 6 (the maximum value of set B) is greater than or equal to 3 (the maximum value of set A).

4. What is the relationship between "inequality of supremums" and "inequality of infimums"?

The inequality of supremums and the inequality of infimums are closely related concepts. Just like how the supremum of one set is always greater than or equal to the supremum of another set, the infimum (or greatest lower bound) of one set is always less than or equal to the infimum of another set.

5. Why is understanding "inequality of supremums" important in science?

Understanding "inequality of supremums" is important in science because it allows for a more precise and rigorous approach to analyzing sets and functions. It also plays a crucial role in many mathematical proofs and helps in solving various problems in different fields of science, such as economics, physics, and computer science.

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