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$$\in$$B 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$$\leq$$b 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
$$b_{0} > sup(B) -\epsilon_{0}$$.

Lets start again.

$$sup(B) - sup(A) = \alpha >0$$

If we pick any $$\epsilon_{0}$$ such that $$\epsilon_{0} < \alpha$$ we can find an element in B such that
$$b_{0} > sup(B) -\epsilon$$. 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).