Prove the following statements: a) For any sequence of sets (Az)neN!, its Limsup; resp: Liminf set At = lim sup An := ⋃n∈N An ;= lim inf An := â‹‚n∈N An and M = ∞ does not change if one modifies a finite number of sets Ai. Find the Limsup respectively Liminf of the two interval sequences (An) and (Bu) with An = [0, n] respectively Bn = [0, n]. Bn+1 := [−n, 0], n ∈ N b) Show An ∈ At. c) For subsets A, . Az: of a set X holds that: 1A = lim sup Ar and 1A = lim inf A~. Here 1A denotes the indicator function on A, i.e. for x ∈ A, 1A(x) = 1 and for x ∉ A, 1A(x) = 0 d) If An = At holds, we say that the limit of the sequence exists. Show that for ascending sequences (Ap ∈ Az ∈ _) or descending sequences (A1 ≥ A2 ≥ …), the limit always exists
The Correct Answer and Explanation is:
Part a) Limsup and Liminf for the given sequences
We are given two sequences of sets (An)(A_n)(An) and (Bn)(B_n)(Bn), where:
- An=[0,n]A_n = [0, n]An=[0,n], i.e., a sequence of intervals that starts from 0 and extends to nnn.
- Bn=[0,n]B_n = [0, n]Bn=[0,n] and Bn+1=[−n,0]B_{n+1} = [-n, 0]Bn+1=[−n,0], which represent sequences of intervals with the negative side also increasing.
To find the Limsup and Liminf of these sequences, we need to use the definitions of limsup and liminf for sequences of sets:
- lim supAn=⋂n∈N⋃k≥nAk\limsup A_n = \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} A_klimsupAn=⋂n∈N⋃k≥nAk, the set of elements that belong to infinitely many AnA_nAn.
- lim infAn=⋃n∈N⋂k≥nAk\liminf A_n = \bigcup_{n \in \mathbb{N}} \bigcap_{k \geq n} A_kliminfAn=⋃n∈N⋂k≥nAk, the set of elements that belong to all AnA_nAn from some point onward.
For An=[0,n]A_n = [0, n]An=[0,n], we have:
- Limsup of AnA_nAn: The set of all points that appear in infinitely many intervals AnA_nAn. As nnn increases, each interval An=[0,n]A_n = [0, n]An=[0,n] grows larger. Therefore, the limsup of AnA_nAn is the set of all non-negative real numbers [0,∞)[0, \infty)[0,∞).
- Liminf of AnA_nAn: The set of all points that are contained in every AnA_nAn for sufficiently large nnn. Since each set AnA_nAn contains all the points from 000 to nnn, the liminf of AnA_nAn is simply [0,∞)[0, \infty)[0,∞), as all points x∈[0,∞)x \in [0, \infty)x∈[0,∞) will eventually appear in all intervals for large enough nnn.
For Bn=[0,n]B_n = [0, n]Bn=[0,n] and Bn+1=[−n,0]B_{n+1} = [-n, 0]Bn+1=[−n,0], the intervals oscillate between positive and negative values as nnn increases. The limsup and liminf for these sequences are:
- Limsup of BnB_nBn: The limsup will be the union of all intervals from some point onward. As nnn grows, the union will cover the entire set of real numbers, so the limsup is R\mathbb{R}R.
- Liminf of BnB_nBn: The liminf will be the intersection of all intervals from some point onward. As the intervals oscillate between positive and negative values, the liminf will eventually converge to the set [0,0][0, 0][0,0], which is the single point at 0.
Part b) Show that An∈AtA_n \in A_tAn∈At
This part of the problem asks to show that for the sequences (An)(A_n)(An) and (At)(A_t)(At), there is a relationship. To prove that An∈AtA_n \in A_tAn∈At, you would need to use the definitions of limsup and liminf, showing that each interval in the sequence eventually contains all points within the limsup or liminf.
Part c) Indicator function representation
Here, we need to show that:1A=lim supAnand1A=lim infAn1_A = \limsup A_n \quad \text{and} \quad 1_A = \liminf A_n1A=limsupAnand1A=liminfAn
The indicator function 1A(x)1_A(x)1A(x) is defined as:1A(x)={1,if x∈A0,if x∉A1_A(x) = \begin{cases} 1, & \text{if } x \in A \\ 0, & \text{if } x \notin A \end{cases}1A(x)={1,0,if x∈Aif x∈/A
Using this, we can say that the indicator function 1A(x)1_A(x)1A(x) equals 1 if xxx is in the limsup and liminf of the sequence of sets, and 0 otherwise. Therefore, the limsup and liminf can be described in terms of the indicator function, as they represent sets that include certain points at some index nnn (for limsup) or consistently (for liminf).
Part d) Convergence for ascending and descending sequences
For ascending sequences A1⊆A2⊆A3⊆…A_1 \subseteq A_2 \subseteq A_3 \subseteq \dotsA1⊆A2⊆A3⊆… and descending sequences A1⊇A2⊇A3⊇…A_1 \supseteq A_2 \supseteq A_3 \supseteq \dotsA1⊇A2⊇A3⊇…, the limit of the sequence always exists because:
- In an ascending sequence, the sets grow or stay the same, so the limsup and liminf coincide with the union of all sets. The limit exists as the union of all these sets as n→∞n \to \inftyn→∞.
- In a descending sequence, the sets shrink or stay the same, so the limsup and liminf coincide with the intersection of all sets. The limit exists as the intersection of all these sets as n→∞n \to \inftyn→∞.
Therefore, the limit of both ascending and descending sequences of sets always exists.
