glb counter example

[mat]

Assume that A and B are bounded non-empty sets of the real numbers and that every element of A is less than or equal to every element of B, give a counter example to the following:
glb (A) <= glb (B).

I know the definition of glb is that for all a in A the glb (A) <= a and for all lower bounds, t, of A glb (A) >= t.
I am pretty sure that at least one of the sets will be open on the left, but other than that I have no idea.
Thanks in advance for any help

 

[pka]

Assume that A and B are bounded non-empty sets of the real numbers and that every element of A is less than or equal to every element of B, give a counter example to the following: glb (A) <= glb (B).

Please review the wording of this question.
Every element of A is a lower bound for B That means if \(\displaystyle t\in A\) then \(\displaystyle \text{glb}(A)\le t \le\text{glb}(B)\).

As written it is false. If the conclusion were \(\displaystyle \text{glb}(A)<\text{glb}(B)\) then it is true.

 

[mat]

I am still confused. I thought I was looking for an example where the glb (B) < glb (A)? But if every element of A is less than every element of B and the glb is less than every element of A, then it must be less than every element of B.

 

[pka]

I am still confused. I thought I was looking for an example where the glb (B) < glb (A)? But if every element of A is less than every element of B and the glb is less than every element of A, then it must be less than every element of B.

Once again if if every element of A is less than every element of B then it must be the case that \(\displaystyle \text{glb}(A)\le \text{glb}(B)\)
Take this example \(\displaystyle A=\{0\}\) and \(\displaystyle B=\{n^{-1}:~n\in\mathbb{Z}^+\}\)
Now \(\displaystyle \text{glb}(A)=0= \text{glb}(B)\).

That is why I think that you have copied the question wrongly or the question contains a mistake.

 

[mat]

Okay, I just reread the assumptions on the problem and instead of "every element of A is less than every element of B" it reads "for all x in A there exists a y in B such that x<=y." Sorry! I misread it. I am assuming this changes things because now not every element of A is a lower bound for B.

 

[mat]

okay, it makes more sense now. Would A = [0, 3] and B = [ -1, 4] work?

 

[pka]

Okay, I just reread the assumptions on the problem and instead of "every element of A is less than every element of B" it reads "for all x in A there exists a y in B such that x<=y." Sorry! I misread it. I am assuming this changes things because now not every element of A is a lower bound for B.

It makes a heck of difference.
Consider \(\displaystyle A = \left\{ {{{(n + 1)}^{ - 1}} - 1:n \in {\mathbb{Z}^ + }} \right\}\,\& \;B = \{ 0\} \cup \left\{ {{{(n + 1)}^{ - 1}} - 2:n \in {\mathbb{Z}^ + }} \right\}\).