Membership Is a Question About Indices
For a sequence of sets, the limsup and liminf are built from unions and intersections of tails. Their definitions can therefore be read point by point: fix \(x\), and record exactly which indices \(n\) satisfy \(x\in A_n\). The set limsup asks whether this happens infinitely often; the set liminf asks whether it happens for every sufficiently large index. These are different conditions: infinitely many occurrences need not include an entire tail.
Let \(X\) be a set and let \((A_n)_{n\ge1}\) be a sequence of subsets of \(X\). Write \(\mathbb{N}=\{1,2,\ldots\}\) for the positive integers. The tail-union and tail-intersection definitions are
For each \(x\in X\), define its membership-index set by \(I_x=\{n\in\mathbb{N}:x\in A_n\}\). The membership tests below follow directly from the definitions. In particular, they require no measure or convergence assumptions.
Proof. By the definition of the limsup, \(x\in\limsup_n A_n\) exactly when \(x\in\bigcup_{n\ge N}A_n\) for every \(N\). Membership in this union means that there is at least one \(n\ge N\) with \(x\in A_n\), or equivalently \(I_x\) contains an index at least \(N\), for every \(N\). A subset of the positive integers has an element at least \(N\) for every \(N\) exactly when it is infinite: a finite subset is bounded above, while an infinite subset cannot be bounded above in \(\mathbb{N}\). This proves the limsup test.
By the definition of the liminf, \(x\in\liminf_n A_n\) exactly when there is an \(N\) such that \(x\in\bigcap_{n\ge N}A_n\). This intersection contains \(x\) exactly when \(x\in A_n\) for every \(n\ge N\), which is the stated eventual-membership condition. \(\square\)
Equivalently, limsup membership means that for every \(N\), at least one index \(n\ge N\) contains \(x\); liminf membership means that for some \(N\), every index \(n\ge N\) contains \(x\). The quantifiers “for every \(N\), there exists \(n\)” and “there exists \(N\), for every \(n\)” are not interchangeable.
Proof. Suppose \(x\in\liminf_n A_n\). Then there is an \(N\) such that \(x\in A_n\) for every \(n\ge N\). There are infinitely many positive integers \(n\ge N\), so \(x\in A_n\) for infinitely many indices. By the limsup membership test, \(x\in\limsup_n A_n\). This holds for every \(x\in\liminf_n A_n\), proving the inclusion. \(\square\)
Worked Membership Tests
Worked Example: Alternating Parity on the Integers
Take \(X=\mathbb{Z}\) and \(A_n=\{k\in\mathbb{Z}: k-n\text{ is even}\}\). Thus \(A_n\) consists of the integers with the same parity as \(n\). Determine the limsup and liminf.
Fix \(k\in\mathbb{Z}\). Its membership-index set is \(I_k=\{n\in\mathbb{N}:k-n\text{ is even}\}\). If \(k\) is even, this is the set of positive even indices; if \(k\) is odd, it is the set of positive odd indices. In either case \(I_k\) is infinite: adding \(2\) to any index of the required parity gives another such index. Therefore every \(k\in\mathbb{Z}\) belongs to infinitely many \(A_n\), and the limsup is \(\mathbb{Z}\).
Neither parity class contains a full tail of the positive integers. Indeed, after any \(N\), there are indices of both parities: among \(N\) and \(N+1\), one has parity opposite to \(k\). Thus no \(k\) belongs to every \(A_n\) from some point onward, and the liminf is empty.
Check: for any fixed \(k\), every tail union contains \(k\), since that tail contains an index with the same parity as \(k\). Every tail intersection is empty, since the tail includes an index with the opposite parity, whose set does not contain \(k\). These checks reproduce the two answers from the tail definitions.
Worked Example: A Sequence of Singletons
Let \(X=\mathbb{Z}\) and \(A_n=\{n\}\) for each \(n\in\mathbb{N}\). Determine the two set limits by tracking a fixed integer \(k\).
If \(k\ge1\), then \(I_k=\{k\}\): the point \(k\) belongs to \(A_k\) and to no other \(A_n\). If \(k\le0\), then \(I_k=\varnothing\), since no \(A_n\) contains a nonpositive integer. In both cases \(I_k\) is finite, so no integer belongs to infinitely many of the sets. The limsup is therefore empty. No \(I_k\) contains a full tail either, so the liminf is empty as well.
Check: the union of the tail sets is \(\bigcup_{n\ge N}\{n\}=\{N,N+1,\ldots\}\). No integer belongs to these unions for every \(N\), so their intersection is empty. For every \(N\), the tail intersection \(\bigcap_{n\ge N}\{n\}\) is empty: it includes at least the two distinct singleton sets \(\{N\}\) and \(\{N+1\}\). Hence the union of the tail intersections is also empty, as the membership-index calculation gave.
Worked Example: Divisibility of Positive Integers
Let \(X=\mathbb{N}\subset\mathbb{Z}\), and define \(A_n=\{k\in\mathbb{N}: k\text{ divides }n\}\). Here “\(k\) divides \(n\)” means \(n=qk\) for some \(q\in\mathbb{N}\). Determine the limsup and liminf.
Fix \(k\in\mathbb{N}\). The membership-index set is \(I_k=\{n\in\mathbb{N}:k\text{ divides }n\}=\{k,2k,3k,\ldots\}\). It is infinite, since for every \(N\) one can choose a positive integer \(q\) with \(qk\ge N\), and then \(qk\in I_k\). Consequently every \(k\) belongs to infinitely many sets, so the limsup is \(\mathbb{N}\).
For \(k=1\), every positive integer is divisible by \(1\), so \(I_1=\mathbb{N}\), which contains a full tail. For \(k\ge2\), choose any positive integer \(q\); then \(n=qk+1\) is not divisible by \(k\), because its remainder on division by \(k\) is \(1\). Such indices occur arbitrarily far out: given any \(N\), choose \(q\) large enough that \(qk+1\ge N\). Thus \(I_k\) does not contain a full tail for any \(k\ge2\). Exactly the point \(1\) belongs to the liminf.
Check: for every \(N\) and \(k\in\mathbb{N}\), some multiple \(qk\ge N\) puts \(k\) in the tail union \(\bigcup_{n\ge N}A_n\); hence every \(k\) is in every tail union. For the tail intersection, \(1\) belongs to every \(A_n\). If \(k\ge2\), every tail includes an index \(qk+1\) not divisible by \(k\), so that tail intersection cannot contain \(k\). Taking the union over tails therefore gives exactly \(\{1\}\).
A Common Quantifier Mistake
A frequent error is to treat “infinitely often” as if it meant “eventually,” or to reverse the two tests. The parity example exposes the mistake. Every integer belongs to infinitely many of its sets, so the limsup is all of \(\mathbb{Z}\). But no integer belongs to all sufficiently late sets, so the liminf is empty.
For instance, a wrong calculation might say: “Each \(k\) occurs in one parity class, so it eventually belongs to \(A_n\); hence the liminf is \(\mathbb{Z}\).” The correction is that belonging to one parity class means belonging at every other index, not at every sufficiently large index. Every tail still contains indices of the other parity. The index-set test makes the distinction exact: an infinite \(I_k\) is enough for limsup membership, while liminf membership requires \(I_k\) to contain all indices from some \(N\) onward.
When applying either test, first fix the point \(x\), then write \(I_x\), and only then determine whether it is infinite or contains a tail. This separates the pointwise question from the set operations and prevents an accidental exchange of quantifiers.
Check Your Understanding
For each question, identify the membership-index set for a fixed point before deciding membership in either set limit.
- For \(X=\mathbb{Z}\) and \(A_n=\{k\in\mathbb{Z}: k\le n\}\), determine \(\limsup_n A_n\) and \(\liminf_n A_n\).
- For \(X=\mathbb{Z}\) and \(A_n=\{0\}\) when \(n\) is even and \(A_n=\varnothing\) when \(n\) is odd, determine both set limits.
- State the membership-index condition for \(x\in\limsup_n A_n\), and explain why it is equivalent to having an index \(n\ge N\) for every \(N\).
- State the membership-index condition for \(x\in\liminf_n A_n\). Does having infinitely many membership indices suffice? Give a sequence illustrating your answer.
- For the divisibility example, explain why \(k=1\) is in the liminf but \(k=2\) is not.