Can Nested Subsets Prove Equality in Limits?

[mathboy20]

Let F be the label of an non-empty set and let (B_m)_{m \geq 1} be elements in 2^F

Then I need to prove the following:

\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m} = \mathrm{lim}_{m} \ \mathrm{inf} \ \mathrm{B_m} = \cup _{m= 1} ^{\infty} B_{m}

if B_{m} \uparrow which implies that B_{m} \subseteq B_{m+1} for all m \geq 1 and

\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m} = \mathrm{lim}_{m} \ \mathrm{inf} \ \mathrm{B_m} = \cap _{m= 1} ^{\infty} B_{m}

if B_{m} \downarrow which means that B_m \supseteq B_{m+1} for all m \geq 1

how do I go about proving this? Do I need to show the infimum of F first?

Sincerely
mb20

 

[HallsofIvy]

What you have makes no sense. I guess that you really mean 2^F rather than 2^E (or you mean E rather than F in the first line), the power set of F. But then (B_m) is just some collection of subsets of F. Don't you have some conditions on (B_m)? Finally, you say you need to show
\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m}
but that's not a statement you can "show", that's an expression. What is supposed to be true about it?

 

[mathboy20]

My solution

here is my solution

1.
I say B_m "uparrow" if B_m is a subset of B_{m+1} for all m, so they are all nested upward. In this case I want to prove that

limsup B_m = liminf B_m = union of B_m over all m.

Let B be the union over all m of B_m. Since B_m is a subset of F for every m, B is a subset of F. Furthermore, every B_m is contained in B so B is an upper bound for the sequence (B_m). Recall that
liminf <= limsup for abstract reasons, so I show liminf = B then as limsup <= B. liminf {B_m} = sup{inf{B_k: k>= m}: m> 0}.

As B_{k+1} >= B_{k} for every k, so inf{B_k:k>= m}=B_m. Hence
sup{inf{B_k:k>= m}: m>0}=sup{B_m:m>0}. But B_{m+1}>= B_m so
sup{B_m:m>0}=union of all B_m, which is B.

Thus liminf{B_m} = B.

2. I say B_m "downarrow" if B_{m+1} is a subset of B_{m} for all m, so they are all nested downward. In this case I want to prove that

limsup B_m = liminf B_m = intersection of B_m over all m.

As B_{k+1} <= B_k I know that sup{B_k:k>=m}=B_m.

Thus limsup B_m=inf{B_m:m>0}=B -- the intersection of all B_m's.

Does this sound right?

Sincerely Mathman20