Set-theoretic limit

This article needs additional citations for verification. Please help by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources:  –  ·  ·  · scholar · JSTOR (April 2015) (Learn how and when to remove this template message)

In mathematics, the limit of a sequence of sets ${\displaystyle A_{1},A_{2},\ldots }$ (subsets of a common set ${\displaystyle X}$) 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 ${\displaystyle x=\lim _{k\to \infty }x_{k},}$ where each ${\displaystyle x_{k}}$ is in some ${\displaystyle A_{n_{k}}.}$ This is only true if convergence is determined by the discrete metric (that is, ${\displaystyle x_{n}\to x}$ if there is ${\displaystyle N}$ such that ${\displaystyle x_{n}=x}$ for all ${\displaystyle n\geq N}$). 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 ${\displaystyle \left(A_{n}\right)_{n=1}^{\infty }}$ is a sequence of sets. The two equivalent definitions are as follows.

• Using union and intersection: define^[1]
  $${\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n\geq 1}\bigcap _{j\geq n}A_{j}}$$

  
  and
  $${\displaystyle \limsup _{n\to \infty }A_{n}=\bigcap _{n\geq 1}\bigcup _{j\geq n}A_{j}}$$

  
  If these two sets are equal, then the set-theoretic limit of the sequence ${\displaystyle A_{n}}$ 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 ${\displaystyle \mathbb {1} _{A_{n}}(x)}$ equal ${\displaystyle 1}$ if ${\displaystyle x\in A_{n},}$ and ${\displaystyle 0}$ otherwise. Define^[1]
  $${\displaystyle \liminf _{n\to \infty }A_{n}={\Bigl \{}x\in X:\liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1{\Bigr \}}}$$

  
  and
  $${\displaystyle \limsup _{n\to \infty }A_{n}={\Bigl \{}x\in X:\limsup _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1{\Bigr \}},}$$

  
  where the expressions inside the brackets on the right are, respectively, the limit infimum and limit supremum of the real-valued sequence ${\displaystyle \mathbb {1} _{A_{n}}(x).}$ Again, if these two sets are equal, then the set-theoretic limit of the sequence ${\displaystyle A_{n}}$ 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 ${\displaystyle 0}$ and ${\displaystyle 1,}$ ${\displaystyle \liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1}$ if and only if ${\displaystyle \mathbb {1} _{A_{n}}(x)}$ takes value ${\displaystyle 0}$ only finitely many times. Equivalently, ${\textstyle x\in \bigcup _{n\geq 1}\bigcap _{j\geq n}A_{j}}$ if and only if there exists ${\displaystyle n}$ such that the element is in ${\displaystyle A_{m}}$ for every ${\displaystyle m\geq n,}$ which is to say if and only if ${\displaystyle x\not \in A_{n}}$ for only finitely many ${\displaystyle n.}$ Therefore, ${\displaystyle x}$ is in the ${\displaystyle \liminf _{n\to \infty }A_{n}}$ if and only if ${\displaystyle x}$ is in all but finitely many ${\displaystyle A_{n}.}$ For this reason, a shorthand phrase for the limit infimum is "${\displaystyle x}$ is in ${\displaystyle A_{n}}$ all but finitely often", typically expressed by writing "${\displaystyle A_{n}}$ a.b.f.o.".

Similarly, an element ${\displaystyle x}$ is in the limit supremum if, no matter how large ${\displaystyle n}$ is, there exists ${\displaystyle m\geq n}$ such that the element is in ${\displaystyle A_{m}.}$ That is, ${\displaystyle x}$ is in the limit supremum if and only if ${\displaystyle x}$ is in infinitely many ${\displaystyle A_{n}.}$ For this reason, a shorthand phrase for the limit supremum is "${\displaystyle x}$ is in ${\displaystyle A_{n}}$ infinitely often", typically expressed by writing "${\displaystyle A_{n}}$ i.o.".

To put it another way, the limit infimum consists of elements that "eventually stay forever" (are in each set after some ${\displaystyle n}$), while the limit supremum consists of elements that "never leave forever" (are in some set after each ${\displaystyle n}$). Or more formally:

${\textstyle \lim _{n\in \mathbb {N} }A_{n}=L\quad \Longleftrightarrow }$	for every ${\displaystyle x\in L}$       there is a ${\displaystyle n_{0}\in \mathbb {N} }$ with ${\displaystyle x\in A_{n}}$ for all ${\displaystyle n\geq n_{0}}$ and
	for every ${\displaystyle y\in X\!\setminus \!L}$ there is a ${\displaystyle p_{0}\in \mathbb {N} }$ with ${\displaystyle y\not \in A_{p}}$ for all ${\displaystyle p\geq p_{0}}$.

Monotone sequences[]

The sequence ${\displaystyle \left(A_{n}\right)}$ is said to be nonincreasing if ${\displaystyle A_{n+1}\subseteq A_{n}}$ for each ${\displaystyle n,}$ and nondecreasing if ${\displaystyle A_{n}\subseteq A_{n+1}}$ for each ${\displaystyle n.}$ In each of these cases the set limit exists. Consider, for example, a nonincreasing sequence ${\displaystyle \left(A_{n}\right).}$ Then

From these it follows that Similarly, if ${\displaystyle \left(A_{n}\right)}$ is nondecreasing then

The Cantor set is defined this way.

Properties[]

• If the limit of ${\displaystyle \mathbb {1} _{A_{n}}(x),}$ as ${\displaystyle n}$ goes to infinity, exists for all ${\displaystyle x}$ then
  $${\displaystyle \lim _{n\to \infty }A_{n}=\left\{x\in X:\lim _{n\to \infty }\mathbb {1} _{A_{n}}(x)=1\right\}.}$$

  
  Otherwise, the limit for ${\displaystyle \left(A_{n}\right)}$ does not exist.
• It can be shown that the limit infimum is contained in the limit supremum:
  $${\displaystyle \liminf _{n\to \infty }A_{n}\subseteq \limsup _{n\to \infty }A_{n},}$$

  
  for example, simply by observing that ${\displaystyle x\in A_{n}}$ all but finitely often implies ${\displaystyle x\in A_{n}}$ infinitely often.
• Using the monotonicity of ${\textstyle B_{n}=\bigcap _{j\geq n}A_{j}}$ and of ${\textstyle C_{n}=\bigcup _{j\geq n}A_{j},}$
  $${\displaystyle \liminf _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcap _{j\geq n}A_{j}\quad {\text{ and }}\quad \limsup _{n\to \infty }A_{n}=\lim _{n\to \infty }\bigcup _{j\geq n}A_{j}.}$$

  
• By using De Morgan's law twice, with set complement ${\displaystyle A^{c}:=X\setminus A,}$
  $${\displaystyle \liminf _{n\to \infty }A_{n}=\bigcup _{n}\left(\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\bigcap _{n}\bigcup _{j\geq n}A_{j}^{c}\right)^{c}=\left(\limsup _{n\to \infty }A_{n}^{c}\right)^{c}.}$$

  
  That is, ${\displaystyle x\in A_{n}}$ all but finitely often is the same as ${\displaystyle x\not \in A_{n}}$ finitely often.
• From the second definition above and the definitions for limit infimum and limit supremum of a real-valued sequence,
  $${\displaystyle \mathbb {1} _{\liminf _{n\to \infty }A_{n}}(x)=\liminf _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\sup _{n\geq 1}\inf _{j\geq n}\mathbb {1} _{A_{j}}(x)}$$

  
  and
  $${\displaystyle \mathbb {1} _{\limsup _{n\to \infty }A_{n}}(x)=\limsup _{n\to \infty }\mathbb {1} _{A_{n}}(x)=\inf _{n\geq 1}\sup _{j\geq n}\mathbb {1} _{A_{j}}(x).}$$

  
• Suppose ${\displaystyle {\mathcal {F}}}$ is a