limit of sequence of sets∗
CWoo†
2013-03-21 18:40:27
Recall that lim sup and lim inf of a sequence of sets {Ai } denote the limit
superior and the limit inferior of {Ai }, respectively. Please click here to see the
definitions and here to see the specialized definitions when they are applied to
the real numbers.
Theorem. Let {Ai } be a sequence of sets with i ∈ Z+ = {1, 2, . . .}. Then
1. for I ranging over all infinite subsets of Z+ ,
[\
lim sup Ai =
Ai ,
I i∈I
2. for I ranging over all subsets of Z+ with finite compliment,
[\
lim inf Ai =
Ai ,
I i∈I
3. lim inf Ai ⊆ lim sup Ai .
Proof.
1. We need to show, for I ranging over all infinite subsets of Z+ ,
[\
Ai =
∞
∞ [
\
Ak .
(1)
n=1 i=n
I i∈I
Let xTbe an element of the LHS, the left hand side of Equation (1).
S∞Then
+
x ∈ i∈I Ai for some
infinite
subset
I
⊆
Z
.
Certainly,
x
∈
i=1 Ai .
S∞
Now, suppose x ∈ i=k Ai . Since I is infinite, we can findSan l ∈ I such
∞
that l > k. Being a member
x ∈ Al ⊆ i=k+1 Ai . By
S∞ of I, we have that
+
induction, we have x ∈ i=n Ai for all n ∈ Z . Thus x is an element of
the RHS. This proves one side of the inclusion (⊆) in (1).
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1
To show the other inclusion,
S
S∞let x be an element of the RHS. So x ∈
∞
+
A
for
all
n
∈
Z
In
i=n i
i=1 Ai , pick the least element n0 such that
S∞
x ∈ An0 . Next, in i=n0 +1 Ai , pick the least n1 suchTthat x ∈ An1 . Then
the set I = {n0 , n1 , . . .} fulfills the requirement x ∈ i∈I Ai , showing the
other inclusion (⊇).
2. Here we have to show, for I ranging over all subsets of Z+ with Z+ − I
finite,
[\
Ai =
∞ \
∞
[
Ak .
(2)
n=1 i=n
I i∈I
T
Suppose first that x is an element of the LHS so that x ∈ i∈I Ai for some
I with Z+ − I finite. Let n0 be a upper bound of the finite set Z+ − I
such that for any n ∈ Z+ − T
I, n < n0 . This means that any m ≥ n0 , we
∞
have m ∈ I. Therefore, x ∈ i=n0 Ai and x is an element of the RHS.
T∞
Next, suppose x is an element of the RHS so that x ∈ k=n Ak for some n.
Then the set I = {n0 , n0 + 1, . . .} is a subset of Z+ with finite complement
that does the job for the LHS.
3. The set of all subsets (of Z+ ) with finite complement is a subset of the set
of all infinite subsets. The third assertion is now clear from the previous
two propositions. QED
Corollary. If {Ai } is a decreasing sequence of sets, then
\
lim inf Ai = lim sup Ai = lim Ai =
Ai .
Similarly, if {Ai } is an increasing sequence of sets, then
[
lim inf Ai = lim sup Ai = lim Ai =
Ai .
Proof. We shall only show the case when we have a descending chain of
sets, since the other case is completely
T∞ analogous. Let A1 ⊇ A2 ⊇ . . . be a
descending chain of sets. Set A = i=1 Ai . We shall show that
lim sup Ai = lim inf Ai = lim Ai = A.
First, by the definition of limit superior of a sequence of sets:
lim sup Ai =
∞ [
∞
\
n=1 i=n
Ak =
∞
\
An = A.
n=1
Now, by Assertion 3 of the above Theorem, lim inf Ai ⊆ lim sup Ai = A, so we
only need to show that A ⊆ lim inf Ai . But this is immediate from the definition
of A, being the intersection of all Ai with subscripts i taking on all values of
Z+ . Its complement is the empty set, clearly finite. Having shown both the
existence and equality of the limit superior and limit inferior of the Ai ’s, we
conclude that the limit of Ai ’s exist as well and it is equal to A. QED
2