Can Nested Subsets Prove Equality in Limits?

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SUMMARY

The discussion focuses on proving the equality of limits for nested subsets in set theory. Specifically, it establishes that if a sequence of sets \( B_m \) is increasing (i.e., \( B_m \uparrow \)), then \( \limsup B_m = \liminf B_m = \bigcup_{m=1}^{\infty} B_m \). Conversely, if the sequence is decreasing (i.e., \( B_m \downarrow \)), then \( \limsup B_m = \liminf B_m = \bigcap_{m=1}^{\infty} B_m \). The proof involves demonstrating that the union or intersection of the sets serves as the limit, depending on the nature of the sequence.

PREREQUISITES
  • Understanding of set theory and the concept of power sets.
  • Familiarity with the definitions of limit superior (limsup) and limit inferior (liminf).
  • Knowledge of nested sequences and their properties in mathematical analysis.
  • Basic proficiency in mathematical notation and expressions.
NEXT STEPS
  • Study the properties of power sets and their applications in set theory.
  • Learn about the concepts of limit superior and limit inferior in more depth.
  • Explore examples of nested sequences and their convergence properties.
  • Investigate related topics such as monotone convergence theorem and its implications.
USEFUL FOR

Mathematicians, students of advanced calculus, and anyone interested in set theory and limits will benefit from this discussion.

mathboy20
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Let F be the label of an non-empty set and let [tex](B_m)_{m \geq 1}[/tex] be elements in [itex]2^F[/itex]

Then I need to prove the following:

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

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

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

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

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

Sincerely
mb20
 
Last edited:
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What you have makes no sense. I guess that you really mean 2F rather than 2E (or you mean E rather than F in the first line), the power set of F. But then (Bm) is just some collection of subsets of F. Don't you have some conditions on (Bm)? Finally, you say you need to show
[tex]\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m}[/tex]
but that's not a statement you can "show", that's an expression. What is supposed to be true about it?
 
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
 
Last edited:

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