Algebraic Laws for Set Limsup and Liminf

Set Limits and Their Algebra

Set limsup and liminf record two different kinds of recurring membership. Their algebra resembles ordinary rules for unions, intersections, and complements, but not every familiar-looking equality survives when the operations are applied at each index and then a limit is taken. The key distinction is between membership in infinitely many sets and membership in all sufficiently late sets.

All statements here concern an arbitrary underlying set \(X\) and sequences of subsets of \(X\). They are purely set-theoretic: no sigma-algebra or measure is assumed. The index \(n\) ranges over the positive integers. For a sequence \((A_n)\), define its set limsup and set liminf by

\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.

Thus \(x\in\limsup A_n\) exactly when \(x\) belongs to infinitely many \(A_n\), while \(x\in\liminf A_n\) exactly when there is an \(N\) such that \(x\in A_n\) for every \(n\ge N\). These membership tests will justify each algebraic rule below.

Complement Rules

Theorem (Complement identities): For every sequence of subsets \(A_n\subseteq X\), \[ (\limsup A_n)^c=\liminf A_n^c, \qquad (\liminf A_n)^c=\limsup A_n^c. \] The complements are taken relative to \(X\).

Proof. First, \(x\notin\limsup A_n\) means that \(x\) does not belong to infinitely many \(A_n\). Equivalently, there is an \(N\) such that \(x\notin A_n\) for every \(n\ge N\). This says exactly that \(x\in A_n^c\) for every \(n\ge N\), or \(x\in\liminf A_n^c\). Hence \((\limsup A_n)^c=\liminf A_n^c\).

For the other identity, \(x\notin\liminf A_n\) means that it is not the case that \(x\) belongs to every \(A_n\) from some index onward. Therefore, for every \(N\), there is an \(n\ge N\) for which \(x\notin A_n\). In other words, \(x\) belongs to infinitely many complements \(A_n^c\), so \(x\in\limsup A_n^c\). Conversely, if \(x\in\limsup A_n^c\), then for every \(N\) there is an \(n\ge N\) with \(x\notin A_n\); consequently \(x\) cannot belong to every \(A_n\) from any index onward, and \(x\notin\liminf A_n\). Thus \((\liminf A_n)^c=\limsup A_n^c\). \(\square\)

Both directions matter: the second equality is not established merely by showing that eventual membership in \(A_n\) prevents infinitely many failures. Its converse also requires showing that failure of eventual membership produces infinitely many indices in the complements.

Finite Families: Two Exact Laws

Let \(r\ge1\) be a fixed finite integer, and for each \(n\) let \(A_{n,1},\ldots,A_{n,r}\) be subsets of \(X\). The finite-family identities are

Theorem (Finite-family identities): \[ \limsup_{n\to\infty}\left(\bigcup_{j=1}^{r}A_{n,j}\right) =\bigcup_{j=1}^{r}\limsup_{n\to\infty}A_{n,j}, \qquad \liminf_{n\to\infty}\left(\bigcap_{j=1}^{r}A_{n,j}\right) =\bigcap_{j=1}^{r}\liminf_{n\to\infty}A_{n,j}. \]

Proof. For the first identity, suppose \(x\) belongs to the limsup of the unions. Then \(x\in\bigcup_{j=1}^r A_{n,j}\) for infinitely many \(n\). For each such \(n\), choose an index \(j\in\{1,\ldots,r\}\) with \(x\in A_{n,j}\). Since there are only finitely many possible indices, at least one index \(j_0\) is chosen for infinitely many \(n\); otherwise each index could occur only finitely often, and the union of those finitely many finite collections of indices \(n\) would be finite. Thus \(x\in\limsup_n A_{n,j_0}\), proving inclusion in the right-hand side. Conversely, if \(x\in\limsup_n A_{n,j_0}\) for some \(j_0\), then \(x\in\bigcup_{j=1}^r A_{n,j}\) for infinitely many \(n\). This proves the first equality.

For the second identity, \(x\) belongs to \(\liminf_n\bigcap_{j=1}^r A_{n,j}\) exactly when there is an \(N\) such that for every \(n\ge N\), \(x\in A_{n,j}\) for every \(j\in\{1,\ldots,r\}\). This implies, for each \(j\), eventual membership in \(A_{n,j}\), so \(x\in\bigcap_{j=1}^r\liminf_n A_{n,j}\). Conversely, suppose \(x\in\liminf_n A_{n,j}\) for every \(j\). For each \(j\), choose \(N_j\) such that \(x\in A_{n,j}\) whenever \(n\ge N_j\). The maximum \(N=\max_{1\le j\le r}N_j\) exists because the family is finite. For every \(n\ge N\), \(x\) belongs to every \(A_{n,j}\), hence to their intersection. This proves the reverse inclusion and the second equality. \(\square\)

Worked Example: A Finite Union and Its Limsup

Take \(X=\{a,b\}\). For every positive integer \(n\), let \(A_n=\{a\}\), and let \(B_n=\{b\}\) when \(n\) is odd and \(B_n=\varnothing\) when \(n\) is even. Compute the two limsups and compare them with the limsup of the unions.

The point \(a\) belongs to every \(A_n\), so \(\limsup A_n=\{a\}\). The point \(b\) belongs to \(B_n\) at every odd index, and there are infinitely many odd positive integers, so \(\limsup B_n=\{b\}\). For every \(n\), \(A_n\cup B_n=\{a,b\}\) if \(n\) is odd and \(A_n\cup B_n=\{a\}\) if \(n\) is even. Thus \(a\) belongs to every union, and \(b\) belongs to infinitely many unions. Therefore \(\limsup_n(A_n\cup B_n)=\{a,b\}\).

Check: \((\limsup A_n)\cup(\limsup B_n)=\{a\}\cup\{b\}=\{a,b\}\), agreeing with the computed limsup of the unions.

General Inclusions and Possible Strictness

For intersections and unions in the other two combinations, membership gives inclusions that need not be equalities:

Proposition: For any two sequences of subsets \(A_n,B_n\subseteq X\), \[ \limsup(A_n\cap B_n)\subseteq(\limsup A_n)\cap(\limsup B_n), \qquad (\liminf A_n)\cup(\liminf B_n)\subseteq\liminf(A_n\cup B_n). \] The first inclusion says that membership in both sets at infinitely many common indices implies membership in each set infinitely often. The second says that eventual membership in either one of the two sequences implies eventual membership in their union.

Proof. If \(x\in\limsup(A_n\cap B_n)\), then \(x\in A_n\cap B_n\) for infinitely many \(n\). At each of those indices \(x\in A_n\) and \(x\in B_n\), so \(x\in\limsup A_n\) and \(x\in\limsup B_n\). This proves the first inclusion. If \(x\in\liminf A_n\), then there is an \(N\) such that \(x\in A_n\) for every \(n\ge N\); hence \(x\in A_n\cup B_n\) for every \(n\ge N\), and \(x\in\liminf(A_n\cup B_n)\). The same reasoning applies if \(x\in\liminf B_n\). Taking the union proves the second inclusion. \(\square\)

Worked Example: Strictness for Two Sequences

Let \(X=\{x\}\). For even \(n\), set \(A_n=X\) and \(B_n=\varnothing\); for odd \(n\), set \(A_n=\varnothing\) and \(B_n=X\). Determine the sets in both inclusions.

There are infinitely many even indices, so \(x\in\limsup A_n\), but there are also infinitely many odd indices with \(x\notin A_n\), so \(x\notin\liminf A_n\). Thus \(\limsup A_n=X\) and \(\liminf A_n=\varnothing\). The corresponding odd/even count for \(B_n\) gives \(\limsup B_n=X\) and \(\liminf B_n=\varnothing\). At every index exactly one of \(A_n,B_n\) is \(X\), so \(A_n\cap B_n=\varnothing\) and \(A_n\cup B_n=X\) for every \(n\). Consequently, \(\limsup(A_n\cap B_n)=\varnothing\), while \((\limsup A_n)\cap(\limsup B_n)=X\). Also, \((\liminf A_n)\cup(\liminf B_n)=\varnothing\), while \(\liminf(A_n\cup B_n)=X\).

Check: since \(X=\{x\}\) is nonempty, the first inclusion is strict, \(\varnothing\subsetneq X\), and the second is strict, \(\varnothing\subsetneq X\). The point appears infinitely often in each sequence separately, but not in their intersection at any index; it belongs to the union at every index, although it is not eventually in either individual sequence.

The fixed finite-family limsup-of-unions identity does not extend as an equality to countably many components. The general inclusion from the union of the individual limsups into the limsup of the unions remains valid for any family: if a point belongs to one fixed component at infinitely many indices, it belongs to the union at those indices. The reverse inclusion can fail because the component containing the point can change with the index. Similarly, the liminf-of-unions inclusion proved above remains valid for any family, but it can be strict.

Worked Example: A Countable Union with Changing Components

Let \(X=\{x\}\), and for positive integers \(n,j\), define \(A_{n,j}=X\) if \(n=j\), and \(A_{n,j}=\varnothing\) if \(n\ne j\). Compare the limsup and liminf of the unions over \(j\) with the unions of the individual limsups and liminfs.

For each fixed \(n\), exactly one set in the countable family, namely \(A_{n,n}\), equals \(X\). Therefore \(\bigcup_{j=1}^{\infty}A_{n,j}=X\) for every \(n\), so both \(\limsup_n\bigcup_jA_{n,j}\) and \(\liminf_n\bigcup_jA_{n,j}\) equal \(X\). For each fixed \(j\), \(A_{n,j}=X\) only at the single index \(n=j\) and is empty at every other index. Thus \(\limsup_n A_{n,j}=\varnothing\) and \(\liminf_n A_{n,j}=\varnothing\). Taking the union over \(j\) still gives \(\varnothing\) in both cases.

Check: \(\bigcup_j\limsup_n A_{n,j}=\varnothing\subsetneq X=\limsup_n\bigcup_jA_{n,j}\). The liminf inclusion is also strict: \(\bigcup_j\liminf_n A_{n,j}=\varnothing\subsetneq X=\liminf_n\bigcup_jA_{n,j}\). The changing index \(j=n\) accounts for membership in each union, without any fixed component containing \(x\) infinitely often or eventually.

A Common Mistake

A tempting but invalid argument is: “If \(x\notin\liminf A_n\), then \(x\) fails to belong to \(A_n\) at some index, so \(x\in\limsup A_n^c\).” One failure at a single index is not enough for membership in the limsup, which requires infinitely many failures. The corrected argument uses the negation of eventual membership: if \(x\notin\liminf A_n\), then for every \(N\) there is an \(n\ge N\) such that \(x\notin A_n\). This gives a failure beyond every threshold and therefore infinitely many indices in \(A_n^c\).

For instance, let \(X=\{x\}\), take \(A_1=\varnothing\), and set \(A_n=X\) for every \(n\ge2\). Then \(x\) fails to belong to \(A_1\), but belongs to every \(A_n\) for \(n\ge2\). Hence \(x\in\liminf A_n\), and \(x\notin\limsup A_n^c\): the complement occurs only at the single index \(1\). This illustrates why the quantifiers “for every \(N\)” and “infinitely many indices” cannot be replaced by “at some index.”

How to Choose the Right Rule

When applying a set-limit law, first identify the index-level operation and the type of membership being tested. “Infinitely often” is well matched to a finite union because one of finitely many components must account for infinitely many occurrences. “Eventually” is well matched to a finite intersection because finitely many eventual thresholds can be combined by taking their maximum. If the operation is a limsup of intersections or a liminf of unions, use the general inclusion unless additional hypotheses establish more.

Expression	Valid rule	Equality in general?
\(\limsup_n\bigcup_{j=1}^r A_{n,j}\)	\(\bigcup_{j=1}^r\limsup_n A_{n,j}\)	Yes, for finite \(r\)
\(\liminf_n\bigcap_{j=1}^r A_{n,j}\)	\(\bigcap_{j=1}^r\liminf_n A_{n,j}\)	Yes, for finite \(r\)
\(\limsup_n(A_n\cap B_n)\)	Subset of \((\limsup A_n)\cap(\limsup B_n)\)	No, in general
\((\liminf A_n)\cup(\liminf B_n)\)	Subset of \(\liminf_n(A_n\cup B_n)\)	No, in general
\((\limsup A_n)^c\), \((\liminf A_n)^c\)	\(\liminf A_n^c\), \(\limsup A_n^c\), respectively	Yes

Check Your Understanding

In each question, justify the conclusion using membership tests or one of the proved rules.

Prove that if \(x\in\liminf A_n\), then \(x\in\limsup A_n\).
For three sequences \(A_n,B_n,C_n\), state and prove the finite-union identity for their limsups.
Give two sequences for which \(\limsup(A_n\cap B_n)\) is a proper subset of \((\limsup A_n)\cap(\limsup B_n)\).
Explain why membership in \(\liminf_n\bigcup_j A_{n,j}\) need not imply membership in any one fixed \(\liminf_n A_{n,j}\) when there are countably many indices \(j\).
State the two complement identities and identify which membership condition—infinitely often or eventually—appears on each side.