Limsup and Liminf of Sets

Tail Unions and Intersections

For a sequence of sets, a single term may give little information about what happens as the sequence continues. Tail operations collect all terms from a chosen index onward: a tail union records the sets appearing somewhere in that tail, while a tail intersection records what is shared throughout that tail. The limsup and liminf of sets are built by then intersecting or uniting these tails.

Let $X$ be any set, and let $(A_n)_{n\ge1}$ be a sequence of subsets of $X$. No measure, measurability, or topology is assumed. For each positive integer $N$, define the tail union $U_N$ and tail intersection $V_N$ by

U_N=\bigcup_{n\ge N}A_n, \qquad V_N=\bigcap_{n\ge N}A_n.

As $N$ increases, the tails get shorter. Thus $U_{N+1}\subseteq U_N$, while $V_N\subseteq V_{N+1}$. The limsup and liminf are the limits of these nested sequences of tail sets, with the following precise definitions.

Definition: For a sequence $(A_n)_{n\ge1}$ of subsets of $X$, define $$ \limsup_{n\to\infty}A_n=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}A_n, \qquad \liminf_{n\to\infty}A_n=\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}A_n. $$ The limsup is the intersection of the tail unions; the liminf is the union of the tail intersections.

The order of the operations matters. In the limsup, first take a union within each tail, then intersect those unions. In the liminf, first take an intersection within each tail, then unite those intersections. These definitions apply to every sequence of subsets, including sequences with repeated, empty, or equal terms. All complements below are relative to the stated ambient set $X$.

Complement Relations

Complementation exchanges unions and intersections. Applying this fact to each tail gives exact identities between the limits of a sequence and the limits of its termwise complements. These identities are often useful for checking which tail operation belongs in each definition.

Theorem (Complement identities): For any set $X$ and any sequence $(A_n)_{n\ge1}$ of subsets of $X$, let $A_n^c=X\setminus A_n$. Then $$ \left(\limsup_{n\to\infty}A_n\right)^c =\liminf_{n\to\infty}A_n^c, \qquad \left(\liminf_{n\to\infty}A_n\right)^c =\limsup_{n\to\infty}A_n^c. $$

Proof. By the definition of limsup and De Morgan's law for an arbitrary intersection, followed by De Morgan's law for each tail union,

\begin{aligned} \left(\limsup_{n\to\infty}A_n\right)^c &=\left(\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}A_n\right)^c\\ &=\bigcup_{N=1}^{\infty}\left(\bigcup_{n\ge N}A_n\right)^c\\ &=\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}A_n^c\\ &=\liminf_{n\to\infty}A_n^c. \end{aligned}

For the second identity, complement the defining union for the liminf, then complement each tail intersection:

\begin{aligned} \left(\liminf_{n\to\infty}A_n\right)^c &=\left(\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}A_n\right)^c\\ &=\bigcap_{N=1}^{\infty}\left(\bigcap_{n\ge N}A_n\right)^c\\ &=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}A_n^c\\ &=\limsup_{n\to\infty}A_n^c. \end{aligned}

Each use of De Morgan's law is valid for arbitrary families of sets, so the argument does not require $X$ to be nonempty or any of the $A_n$ to be nonempty. $\square$

The Liminf Is Contained in the Limsup

The two limits need not be equal, but they always satisfy an inclusion. The reason is that any tail intersection is contained in every sufficiently compatible tail union: for any two tail indices, a term sufficiently far along belongs to both tails.

Proposition: For every set $X$ and every sequence $(A_n)_{n\ge1}$ of subsets of $X$, $$ \liminf_{n\to\infty}A_n\subseteq\limsup_{n\to\infty}A_n. $$

Proof. Fix positive integers $N$ and $M$, and let $k=\max\{N,M\}$. Since $k\ge N$, the tail intersection $V_N=\bigcap_{n\ge N}A_n$ is contained in $A_k$. Since $k\ge M$, the set $A_k$ is contained in the tail union $U_M=\bigcup_{n\ge M}A_n$. Consequently, $V_N\subseteq U_M$ for every $N$ and $M$. Taking the union over $N$ gives $\bigcup_N V_N\subseteq U_M$ for every $M$. Therefore $\bigcup_N V_N\subseteq\bigcap_M U_M$, which is exactly the stated inclusion by definition. This reasoning also covers empty sets and the empty ambient set. $\square$

Worked Computations

Worked Example: A Periodic Sequence on a Finite Set

Let $X=\{a,b,c,d\}$, and define $A_n=\{a,b\}$ when $n$ is odd and $A_n=\{b,c\}$ when $n$ is even. Compute both set limits.

Every tail contains at least one odd-indexed term and at least one even-indexed term: among any two consecutive positive integers, one is odd and one is even. Therefore every tail union is $U_N=\{a,b\}\cup\{b,c\}=\{a,b,c\}$, and every tail intersection is $V_N=\{a,b\}\cap\{b,c\}=\{b\}$. Intersecting the identical tail unions, and uniting the identical tail intersections, gives

\limsup_{n\to\infty}A_n=\{a,b,c\}, \qquad \liminf_{n\to\infty}A_n=\{b\}.

Check: The complements of the odd and even terms are respectively $\{c,d\}$ and $\{a,d\}$. Their tail unions are all $\{a,c,d\}$, and their tail intersections are all $\{d\}$. Thus the limsup of the complements is $\{a,c,d\}$, which equals $X\setminus\{b\}$, the complement of the original liminf. The liminf of the complements is $\{d\}$, which equals $X\setminus\{a,b,c\}$, the complement of the original limsup. Both complement identities check directly.

Worked Example: A Sequence of Expanding Intervals

On $X=\mathbb{R}$, set $A_n=[0,1+1/n]$ for each $n\ge1$. Compute the tail unions and intersections, including the endpoint behavior.

The right endpoints decrease as $n$ increases. In the tail beginning at $N$, the largest interval is $A_N$, so $U_N=\bigcup_{n\ge N}A_n=[0,1+1/N]$. The intervals in the same tail all contain $[0,1]$. If $x>1$, choose $k\ge N$ so large that $1/k<x-1$; then $1+1/k<x$, so $x\notin A_k$. Hence such an $x$ is excluded from the tail intersection. This proves $V_N=\bigcap_{n\ge N}A_n=[0,1]$ for every $N$. In particular, the endpoint $1$ is included in every interval, while every $x>1$ is eventually excluded. Thus

\limsup_{n\to\infty}A_n =\bigcap_{N\ge1}[0,1+1/N]=[0,1], \qquad \liminf_{n\to\infty}A_n =\bigcup_{N\ge1}[0,1]=[0,1].

Check: Each tail union contains $1$, and for every $x>1$ there is an $N$ with $1/N<x-1$, so $x\notin U_N$; no negative $x$ belongs to any tail union. This verifies their intersection is $[0,1]$. Each tail intersection is already exactly $[0,1]$, so their union is also $[0,1]$. The two calculations agree.

Worked Example: Successive Singletons and Their Complements

Take $X=\mathbb{N}=\{1,2,3,\ldots\}$ and $A_n=\{n\}$. Compute the limits first for $A_n$, then for $B_n=A_n^c$.

For each $N$, the tail union is $U_N=\{N,N+1,N+2,\ldots\}$. The tail intersection is empty: it is contained in $A_N=\{N\}$ and in $A_{N+1}=\{N+1\}$, and these two sets are disjoint. Since the tails $U_N$ shrink as $N$ increases and have no common positive integer, their intersection is empty. The union of the empty tail intersections is also empty. Therefore

\limsup_{n\to\infty}A_n=\varnothing, \qquad \liminf_{n\to\infty}A_n=\varnothing.

For the complements, $B_n=\mathbb{N}\setminus\{n\}$. Every tail union $\bigcup_{n\ge N}B_n$ is $\mathbb{N}$: for any $j\in\mathbb{N}$, there is an $n\ge N$ different from $j$, and then $j\in B_n$. The tail intersection is $\bigcap_{n\ge N}B_n=\{1,\ldots,N-1\}$, with this set understood as empty when $N=1$. Indeed, these are exactly the positive integers not among $N,N+1,\ldots$. It follows that

\limsup_{n\to\infty}B_n=\mathbb{N}, \qquad \liminf_{n\to\infty}B_n =\bigcup_{N\ge1}\{1,\ldots,N-1\} =\mathbb{N}.

Check: The complement of either limit for $A_n$ is $\mathbb{N}$, matching both limits for $B_n$. This agrees with the complement identities. In the tail-intersection calculation, $N=1$ contributes the empty set, while the union over all $N$ still contains every positive integer: $j$ occurs in the term with $N=j+1$.

A Common Tail-Operation Mistake

A frequent error is to take the union of all terms in the sequence and call it the limsup. For the singleton example, $\bigcup_{n\ge1}A_n=\mathbb{N}$, but this is only the first tail union $U_1$. The limsup is the intersection of all tail unions, not just their first member. Since $U_N=\{N,N+1,\ldots\}$, their intersection is empty: a positive integer $j$ is absent from $U_{j+1}$. Thus the correct limsup is $\varnothing$, not $\mathbb{N}$.

The order distinction is equally important for the liminf. It is not the intersection of the tail intersections; it is their union. In the singleton example, every tail intersection is empty, so the liminf is empty. In other sequences, the tail intersections can grow with $N$, and their union collects all of them. Keep the two defining patterns visible when computing:

\text{limsup: intersect the tail unions}, \qquad \text{liminf: unite the tail intersections}.

These constructions need no finiteness or measurability assumptions. If a later setting equips $X$ with a sigma-algebra and measure, additional questions about measurability or measure can be asked, but the set limits themselves are defined before any such structure is chosen.

Check Your Understanding

Use the tail definitions directly in each question. Complements are always taken relative to the stated ambient set.

For $X=\{1,2,3\}$ and $A_n=\{1,2\}$ for odd $n$, $A_n=\{2,3\}$ for even $n$, compute every tail union and tail intersection, then find the two set limits.
Prove that for any sequence of subsets of $X$, the complement of the limsup is the liminf of the termwise complements.
Let $A_n=[0,1-1/n]$ for $n\ge1$. Compute the tail unions and intersections, paying particular attention to whether $1$ belongs to either limit.
For $A_n=\{n\}$ in $\mathbb{N}$, explain why the union of all terms does not equal the limsup.
Using tail unions and intersections, prove that $\liminf A_n\subseteq\limsup A_n$.

What You'll Learn