Proving Existence of Upper Bound for A if sup A < sup B

[kathrynag]

Homework Statement

If sup A<sup B, then show that there exists an element b\inB that is an upper bound for A.

Homework Equations

The Attempt at a Solution

This one is really frustrating me. I'm having trouble even beginning.
I know sup A implies s\leqb where c is an upper bound for A.
Similarily sup B implies s\leq d where d is an upper bound for B.

 

[╔(σ_σ)╝]

sup B= sup A+ \epsilon

Remember what sup b is and it's closeness to elements of B.

 

[kathrynag]

sup A= sup B+ \epsilon

Remember what sup b is and it's closeness to elements of B.

I have no clue how you got to that.

 

[╔(σ_σ)╝]

Well if x < y that means y= x+b where b= y-x.

 

[kathrynag]

so x<x+b
x<x+y-x
So that statement alone shows that b is an upper bound?

 

[╔(σ_σ)╝]

I made an error while writing the latex. Please review my edit.

 

[╔(σ_σ)╝]

sup B- \epsilon= sup A
Since sup B is the least upper bound what can we say about
sup B- \epsilon_{0} ?
Is it in B?

The point it that if we pick \epsilon_{0} small enough we can find a number that is both in B and is an upper bound of A.

 

[kathrynag]

sup B- \epsilon= sup A
Since sup B is the least upper bound what can we say about
sup B- \epsilon_{0} ?
Is it in B?

The point it that if we pick \epsilon_{0} small enough we can find a number that is both in B and is an upper bound of A.

sup B- \epsilon_{0} must be an upper bound also?

 

[╔(σ_σ)╝]

sup B- \epsilon_{0} must be an upper bound also?

Do you know the definition of sup(B) ?

Given that sup(B) is the least upper bound this means that it is the smallest number that is an upper bound of the set B.

That means given any \epsilon_{0}, no matter how small, there exist b_{0}\inB such that
b_{0} &gt; sup(B) -\epsilon_{0}.

Lets start again.

sup(B) - sup(A) = \alpha &gt;0

If we pick any \epsilon_{0} such that \epsilon_{0} &lt; \alpha we can find an element in B such that
b_{0} &gt; sup(B) -\epsilon. So b_{0} is an upper bound of A.

 

[vela]

You could prove this by contradiction as well. Assume there is no element in B that is an upper bound for A, and then show it leads to the conclusion sup(A)>sup(B), which contradicts the condition sup(A)<sup(B).