- Source: Set-theoretic limit
In mathematics, the limit of a sequence of sets
A
1
,
A
2
,
…
{\displaystyle A_{1},A_{2},\ldots }
(subsets of a common set
X
{\displaystyle X}
) is a set whose elements are determined by the sequence in either of two equivalent ways: (1) by upper and lower bounds on the sequence that converge monotonically to the same set (analogous to convergence of real-valued sequences) and (2) by convergence of a sequence of indicator functions which are themselves real-valued. As is the case with sequences of other objects, convergence is not necessary or even usual.
More generally, again analogous to real-valued sequences, the less restrictive limit infimum and limit supremum of a set sequence always exist and can be used to determine convergence: the limit exists if the limit infimum and limit supremum are identical. (See below). Such set limits are essential in measure theory and probability.
It is a common misconception that the limits infimum and supremum described here involve sets of accumulation points, that is, sets of
x
=
lim
k
→
∞
x
k
,
{\displaystyle x=\lim _{k\to \infty }x_{k},}
where each
x
k
{\displaystyle x_{k}}
is in some
A
n
k
.
{\displaystyle A_{n_{k}}.}
This is only true if convergence is determined by the discrete metric (that is,
x
n
→
x
{\displaystyle x_{n}\to x}
if there is
N
{\displaystyle N}
such that
x
n
=
x
{\displaystyle x_{n}=x}
for all
n
≥
N
{\displaystyle n\geq N}
). This article is restricted to that situation as it is the only one relevant for measure theory and probability. See the examples below. (On the other hand, there are more general topological notions of set convergence that do involve accumulation points under different metrics or topologies.)
Definitions
= The two definitions
=Suppose that
(
A
n
)
n
=
1
∞
{\displaystyle \left(A_{n}\right)_{n=1}^{\infty }}
is a sequence of sets. The two equivalent definitions are as follows.
Using union and intersection: define
lim inf
n
→
∞
A
n
=
⋃
n
≥
1
⋂
j
≥
n
A
j
{\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n\geq 1}\bigcap _{j\geq n}A_{j}}
and
lim sup
n
→
∞
A
n
=
⋂
n
≥
1
⋃
j
≥
n
A
j
{\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n\geq 1}\bigcup _{j\geq n}A_{j}}
If these two sets are equal, then the set-theoretic limit of the sequence
A
n
{\displaystyle A_{n}}
exists and is equal to that common set. Either set as described above can be used to get the limit, and there may be other means to get the limit as well.
Using indicator functions: let
1
A
n
(
x
)
{\displaystyle \mathbb {1} _{A_{n}}(x)}
equal
1
{\displaystyle 1}
if
x
∈
A
n
,
{\displaystyle x\in A_{n},}
and
0
{\displaystyle 0}
otherwise. Define
lim inf
n
→
∞
A
n
=
{
x
∈
X
:
lim inf
n
→
∞
1
A
n
(
x
)
=
1
}
{\displaystyle \liminf _{n\to \infty }A_{n}={\Bigl \{}x\in X:\liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1{\Bigr \}}}
and
lim sup
n
→
∞
A
n
=
{
x
∈
X
:
lim sup
n
→
∞
1
A
n
(
x
)
=
1
}
,
{\displaystyle \limsup _{n\to \infty }A_{n}={\Bigl \{}x\in X:\limsup _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1{\Bigr \}},}
where the expressions inside the brackets on the right are, respectively, the limit infimum and limit supremum of the real-valued sequence
1
A
n
(
x
)
.
{\displaystyle \mathbb {1} _{A_{n}}(x).}
Again, if these two sets are equal, then the set-theoretic limit of the sequence
A
n
{\displaystyle A_{n}}
exists and is equal to that common set, and either set as described above can be used to get the limit.
To see the equivalence of the definitions, consider the limit infimum. The use of De Morgan's law below explains why this suffices for the limit supremum. Since indicator functions take only values
0
{\displaystyle 0}
and
1
,
{\displaystyle 1,}
lim inf
n
→
∞
1
A
n
(
x
)
=
1
{\displaystyle \liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1}
if and only if
1
A
n
(
x
)
{\displaystyle \mathbb {1} _{A_{n}}(x)}
takes value
0
{\displaystyle 0}
only finitely many times. Equivalently,
x
∈
⋃
n
≥
1
⋂
j
≥
n
A
j
{\textstyle x\in \bigcup _{n\geq 1}\bigcap _{j\geq n}A_{j}}
if and only if there exists
n
{\displaystyle n}
such that the element is in
A
m
{\displaystyle A_{m}}
for every
m
≥
n
,
{\displaystyle m\geq n,}
which is to say if and only if
x
∉
A
n
{\displaystyle x\not \in A_{n}}
for only finitely many
n
.
{\displaystyle n.}
Therefore,
x
{\displaystyle x}
is in the
lim inf
n
→
∞
A
n
{\displaystyle \liminf _{n\to \infty }A_{n}}
if and only if
x
{\displaystyle x}
is in all but finitely many
A
n
.
{\displaystyle A_{n}.}
For this reason, a shorthand phrase for the limit infimum is "
x
{\displaystyle x}
is in
A
n
{\displaystyle A_{n}}
all but finitely often", typically expressed by writing "
A
n
{\displaystyle A_{n}}
a.b.f.o.".
Similarly, an element
x
{\displaystyle x}
is in the limit supremum if, no matter how large
n
{\displaystyle n}
is, there exists
m
≥
n
{\displaystyle m\geq n}
such that the element is in
A
m
.
{\displaystyle A_{m}.}
That is,
x
{\displaystyle x}
is in the limit supremum if and only if
x
{\displaystyle x}
is in infinitely many
A
n
.
{\displaystyle A_{n}.}
For this reason, a shorthand phrase for the limit supremum is "
x
{\displaystyle x}
is in
A
n
{\displaystyle A_{n}}
infinitely often", typically expressed by writing "
A
n
{\displaystyle A_{n}}
i.o.".
To put it another way, the limit infimum consists of elements that "eventually stay forever" (are in each set after some
n
{\displaystyle n}
), while the limit supremum consists of elements that "never leave forever" (are in some set after each
n
{\displaystyle n}
). Or more formally:
= Monotone sequences
=The sequence
(
A
n
)
{\displaystyle \left(A_{n}\right)}
is said to be nonincreasing if
A
n
+
1
⊆
A
n
{\displaystyle A_{n+1}\subseteq A_{n}}
for each
n
,
{\displaystyle n,}
and nondecreasing if
A
n
⊆
A
n
+
1
{\displaystyle A_{n}\subseteq A_{n+1}}
for each
n
.
{\displaystyle n.}
In each of these cases the set limit exists. Consider, for example, a nonincreasing sequence
(
A
n
)
.
{\displaystyle \left(A_{n}\right).}
Then
⋂
j
≥
n
A
j
=
⋂
j
≥
1
A
j
and
⋃
j
≥
n
A
j
=
A
n
.
{\displaystyle \bigcap _{j\geq n}A_{j}=\bigcap _{j\geq 1}A_{j}{\text{ and }}\bigcup _{j\geq n}A_{j}=A_{n}.}
From these it follows that
lim inf
n
→
∞
A
n
=
⋃
n
≥
1
⋂
j
≥
n
A
j
=
⋂
j
≥
1
A
j
=
⋂
n
≥
1
⋃
j
≥
n
A
j
=
lim sup
n
→
∞
A
n
.
{\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n\geq 1}\bigcap _{j\geq n}A_{j}=\bigcap _{j\geq 1}A_{j}=\bigcap _{n\geq 1}\bigcup _{j\geq n}A_{j}=\limsup _{n\to \infty }A_{n}.}
Similarly, if
(
A
n
)
{\displaystyle \left(A_{n}\right)}
is nondecreasing then
lim
n
→
∞
A
n
=
⋃
j
≥
1
A
j
.
{\displaystyle \lim _{n\to \infty }A_{n}=\bigcup _{j\geq 1}A_{j}.}
The Cantor set is defined this way.
Properties
If the limit of
1
A
n
(
x
)
,
{\displaystyle \mathbb {1} _{A_{n}}(x),}
as
n
{\displaystyle n}
goes to infinity, exists for all
x
{\displaystyle x}
then
lim
n
→
∞
A
n
=
{
x
∈
X
:
lim
n
→
∞
1
A
n
(
x
)
=
1
}
.
{\displaystyle \lim _{n\to \infty }A_{n}=\left\{x\in X:\lim _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1\right\}.}
Otherwise, the limit for
(
A
n
)
{\displaystyle \left(A_{n}\right)}
does not exist.
It can be shown that the limit infimum is contained in the limit supremum:
lim inf
n
→
∞
A
n
⊆
lim sup
n
→
∞
A
n
,
{\displaystyle \liminf _{n\to \infty }A_{n}\subseteq \limsup _{n\to \infty }A_{n},}
for example, simply by observing that
x
∈
A
n
{\displaystyle x\in A_{n}}
all but finitely often implies
x
∈
A
n
{\displaystyle x\in A_{n}}
infinitely often.
Using the monotonicity of
B
n
=
⋂
j
≥
n
A
j
{\textstyle B_{n}=\bigcap _{j\geq n}A_{j}}
and of
C
n
=
⋃
j
≥
n
A
j
,
{\textstyle C_{n}=\bigcup _{j\geq n}A_{j},}
lim inf
n
→
∞
A
n
=
lim
n
→
∞
⋂
j
≥
n
A
j
and
lim sup
n
→
∞
A
n
=
lim
n
→
∞
⋃
j
≥
n
A
j
.
{\displaystyle \liminf _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcap _{j\geq n}A_{j}\quad {\text{ and }}\quad \limsup _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcup _{j\geq n}A_{j}.}
By using De Morgan's law twice, with set complement
A
c
:=
X
∖
A
,
{\displaystyle A^{c}:=X\setminus A,}
lim inf
n
→
∞
A
n
=
⋃
n
(
⋃
j
≥
n
A
j
c
)
c
=
(
⋂
n
⋃
j
≥
n
A
j
c
)
c
=
(
lim sup
n
→
∞
A
n
c
)
c
.
{\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\left(\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\bigcap _{n}\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\limsup _{n\to \infty }A_{n}^{c}\right)^{c}.}
That is,
x
∈
A
n
{\displaystyle x\in A_{n}}
all but finitely often is the same as
x
∉
A
n
{\displaystyle x\not \in A_{n}}
finitely often.
From the second definition above and the definitions for limit infimum and limit supremum of a real-valued sequence,
1
lim inf
n
→
∞
A
n
(
x
)
=
lim inf
n
→
∞
1
A
n
(
x
)
=
sup
n
≥
1
inf
j
≥
n
1
A
j
(
x
)
{\displaystyle \mathbb {1} _{\liminf _{n\to \infty }A_{n}}(x)=\liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\sup _{n\geq 1}\inf _{j\geq n}\mathbb {1} _{A_{j}}(x)}
and
1
lim sup
n
→
∞
A
n
(
x
)
=
lim sup
n
→
∞
1
A
n
(
x
)
=
inf
n
≥
1
sup
j
≥
n
1
A
j
(
x
)
.
{\displaystyle \mathbb {1} _{\limsup _{n\to \infty }A_{n}}(x)=\limsup _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\inf _{n\geq 1}\sup _{j\geq n}\mathbb {1} _{A_{j}}(x).}
Suppose
F
{\displaystyle {\mathcal {F}}}
is a 𝜎-algebra of subsets of
X
.
{\displaystyle X.}
That is,
F
{\displaystyle {\mathcal {F}}}
is nonempty and is closed under complement and under unions and intersections of countably many sets. Then, by the first definition above, if each
A
n
∈
F
{\displaystyle A_{n}\in {\mathcal {F}}}
then both
lim inf
n
→
∞
A
n
{\displaystyle \liminf _{n\to \infty }A_{n}}
and
lim sup
n
→
∞
A
n
{\displaystyle \limsup _{n\to \infty }A_{n}}
are elements of
F
.
{\displaystyle {\mathcal {F}}.}
Examples
Let
A
n
=
(
−
1
n
,
1
−
1
n
]
.
{\displaystyle A_{n}=\left(-{\tfrac {1}{n}},1-{\tfrac {1}{n}}\right].}
Then
lim inf
n
→
∞
A
n
=
⋃
n
⋂
j
≥
n
(
−
1
j
,
1
−
1
j
]
=
⋃
n
[
0
,
1
−
1
n
]
=
[
0
,
1
)
{\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\bigcap _{j\geq n}\left(-{\tfrac {1}{j}},1-{\tfrac {1}{j}}\right]=\bigcup _{n}\left[0,1-{\tfrac {1}{n}}\right]=[0,1)}
and
lim sup
n
→
∞
A
n
=
⋂
n
⋃
j
≥
n
(
−
1
j
,
1
−
1
j
]
=
⋂
n
(
−
1
n
,
1
)
=
[
0
,
1
)
{\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n}\bigcup _{j\geq n}\left(-{\tfrac {1}{j}},1-{\tfrac {1}{j}}\right]=\bigcap _{n}\left(-{\tfrac {1}{n}},1\right)=[0,1)}
so
lim
n
→
∞
A
n
=
[
0
,
1
)
{\displaystyle \lim _{n\to \infty }A_{n}=[0,1)}
exists.
Change the previous example to
A
n
=
(
(
−
1
)
n
n
,
1
−
(
−
1
)
n
n
]
.
{\displaystyle A_{n}=\left({\tfrac {(-1)^{n}}{n}},1-{\tfrac {(-1)^{n}}{n}}\right].}
Then
lim inf
n
→
∞
A
n
=
⋃
n
⋂
j
≥
n
(
(
−
1
)
j
j
,
1
−
(
−
1
)
j
j
]
=
⋃
n
(
1
2
n
,
1
−
1
2
n
]
=
(
0
,
1
)
{\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\bigcap _{j\geq n}\left({\tfrac {(-1)^{j}}{j}},1-{\tfrac {(-1)^{j}}{j}}\right]=\bigcup _{n}\left({\tfrac {1}{2n}},1-{\tfrac {1}{2n}}\right]=(0,1)}
and
lim sup
n
→
∞
A
n
=
⋂
n
⋃
j
≥
n
(
(
−
1
)
j
j
,
1
−
(
−
1
)
j
j
]
=
⋂
n
(
−
1
2
n
−
1
,
1
+
1
2
n
−
1
]
=
[
0
,
1
]
,
{\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n}\bigcup _{j\geq n}\left({\tfrac {(-1)^{j}}{j}},1-{\tfrac {(-1)^{j}}{j}}\right]=\bigcap _{n}\left(-{\tfrac {1}{2n-1}},1+{\tfrac {1}{2n-1}}\right]=[0,1],}
so
lim
n
→
∞
A
n
{\displaystyle \lim _{n\to \infty }A_{n}}
does not exist, despite the fact that the left and right endpoints of the intervals converge to 0 and 1, respectively.
Let
A
n
=
{
0
,
1
n
,
2
n
,
…
,
n
−
1
n
,
1
}
.
{\displaystyle A_{n}=\left\{0,{\tfrac {1}{n}},{\tfrac {2}{n}},\ldots ,{\tfrac {n-1}{n}},1\right\}.}
Then
⋃
j
≥
n
A
j
=
Q
∩
[
0
,
1
]
{\displaystyle \bigcup _{j\geq n}A_{j}=\mathbb {Q} \cap [0,1]}
is the set of all rational numbers between 0 and 1 (inclusive), since even for
j
<
n
{\displaystyle j
and
0
≤
k
≤
j
,
{\displaystyle 0\leq k\leq j,}
k
j
=
n
k
n
j
{\displaystyle {\tfrac {k}{j}}={\tfrac {nk}{nj}}}
is an element of the above. Therefore,
lim sup
n
→
∞
A
n
=
Q
∩
[
0
,
1
]
.
{\displaystyle \limsup _{n\to \infty }A_{n}=\mathbb {Q} \cap [0,1].}
On the other hand,
⋂
j
≥
n
A
j
=
{
0
,
1
}
,
{\displaystyle \bigcap _{j\geq n}A_{j}=\{0,1\},}
which implies
lim inf
n
→
∞
A
n
=
{
0
,
1
}
.
{\displaystyle \liminf _{n\to \infty }A_{n}=\{0,1\}.}
In this case, the sequence
A
1
,
A
2
,
…
{\displaystyle A_{1},A_{2},\ldots }
does not have a limit. Note that
lim
n
→
∞
A
n
{\displaystyle \lim _{n\to \infty }A_{n}}
is not the set of accumulation points, which would be the entire interval
[
0
,
1
]
{\displaystyle [0,1]}
(according to the usual Euclidean metric).
Probability uses
Set limits, particularly the limit infimum and the limit supremum, are essential for probability and measure theory. Such limits are used to calculate (or prove) the probabilities and measures of other, more purposeful, sets. For the following,
(
X
,
F
,
P
)
{\displaystyle (X,{\mathcal {F}},\mathbb {P} )}
is a probability space, which means
F
{\displaystyle {\mathcal {F}}}
is a σ-algebra of subsets of
X
{\displaystyle X}
and
P
{\displaystyle \mathbb {P} }
is a probability measure defined on that σ-algebra. Sets in the σ-algebra are known as events.
If
A
1
,
A
2
,
…
{\displaystyle A_{1},A_{2},\ldots }
is a monotone sequence of events in
F
{\displaystyle {\mathcal {F}}}
then
lim
n
→
∞
A
n
{\displaystyle \lim _{n\to \infty }A_{n}}
exists and
P
(
lim
n
→
∞
A
n
)
=
lim
n
→
∞
P
(
A
n
)
.
{\displaystyle \mathbb {P} \left(\lim _{n\to \infty }A_{n}\right)=\lim _{n\to \infty }\mathbb {P} \left(A_{n}\right).}
= Borel–Cantelli lemmas
=In probability, the two Borel–Cantelli lemmas can be useful for showing that the limsup of a sequence of events has probability equal to 1 or to 0. The statement of the first (original) Borel–Cantelli lemma is
The second Borel–Cantelli lemma is a partial converse:
= Almost sure convergence
=One of the most important applications to probability is for demonstrating the almost sure convergence of a sequence of random variables. The event that a sequence of random variables
Y
1
,
Y
2
,
…
{\displaystyle Y_{1},Y_{2},\ldots }
converges to another random variable
Y
{\displaystyle Y}
is formally expressed as
{
lim sup
n
→
∞
|
Y
n
−
Y
|
=
0
}
.
{\textstyle \left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|=0\right\}.}
It would be a mistake, however, to write this simply as a limsup of events. That is, this is not the event
lim sup
n
→
∞
{
|
Y
n
−
Y
|
=
0
}
{\textstyle \limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|=0\right\}}
! Instead, the complement of the event is
{
lim sup
n
→
∞
|
Y
n
−
Y
|
≠
0
}
=
{
lim sup
n
→
∞
|
Y
n
−
Y
|
>
1
k
for some
k
}
=
⋃
k
≥
1
⋂
n
≥
1
⋃
j
≥
n
{
|
Y
j
−
Y
|
>
1
k
}
=
lim
k
→
∞
lim sup
n
→
∞
{
|
Y
n
−
Y
|
>
1
k
}
.
{\displaystyle {\begin{aligned}\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|\neq 0\right\}&=\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|>{\frac {1}{k}}{\text{ for some }}k\right\}\\&=\bigcup _{k\geq 1}\bigcap _{n\geq 1}\bigcup _{j\geq n}\left\{\left|Y_{j}-Y\right|>{\tfrac {1}{k}}\right\}\\&=\lim _{k\to \infty }\limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|>{\tfrac {1}{k}}\right\}.\end{aligned}}}
Therefore,
P
(
{
lim sup
n
→
∞
|
Y
n
−
Y
|
≠
0
}
)
=
lim
k
→
∞
P
(
lim sup
n
→
∞
{
|
Y
n
−
Y
|
>
1
k
}
)
.
{\displaystyle \mathbb {P} \left(\left\{\limsup _{n\to \infty }\left|Y_{n}-Y\right|\neq 0\right\}\right)=\lim _{k\to \infty }\mathbb {P} \left(\limsup _{n\to \infty }\left\{\left|Y_{n}-Y\right|>{\tfrac {1}{k}}\right\}\right).}
See also
List of set identities and relations – Equalities for combinations of sets
Set theory – Branch of mathematics that studies sets
References
Kata Kunci Pencarian:
- Daftar simbol matematika
- Daftar masalah matematika yang belum terpecahkan
- Set-theoretic limit
- Limit inferior and limit superior
- Tree (set theory)
- Io
- Inverse limit
- Greisen–Zatsepin–Kuzmin limit
- Christopher Langan
- General topology
- Category of topological spaces
- Direct limit
