When Can a Sequence Have Two Limits?
In the real line with its standard topology, a convergent sequence has only one limit. The reason is not a special property of real-number arithmetic: distinct real numbers have disjoint open neighborhoods. In a general topological space, distinct points need not have disjoint neighborhoods, and a sequence can then converge to more than one point. The neighborhood definition of convergence makes both the phenomenon and the role of separation precise.
Let \((X,\mathcal{T})\) be a topological space. A sequence in \(X\) is a function from \(\mathbb{N}=\{1,2,\dots\}\) to \(X\), written \((x_n)\). It is eventually in a set \(U\) if there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N\).
This definition asks about every neighborhood of a proposed limit, not about distances or a numerical value attached to the sequence. In particular, a sequence may have different limits if the neighborhoods of those points do not force it into incompatible sets.
Indiscrete Spaces: Every Sequence Converges Everywhere
The indiscrete topology on a set \(X\) is \(\mathcal{T}=\{\varnothing,X\}\). If \(X\) is nonempty, the only open neighborhood of any point \(x\in X\) is \(X\) itself. Since every term of every sequence in \(X\) belongs to \(X\), the convergence condition holds for every point.
Worked Example: A Sequence with All Points as Limits
Take \(X=\{a,b,c\}\) with the indiscrete topology \(\{\varnothing,X\}\), and define a sequence by \(x_n=a\) when \(n\) is odd and \(x_n=b\) when \(n\) is even. Determine all its limits.
For the candidate limit \(a\), its only open neighborhood is \(X\). For every \(n\in\mathbb{N}\), the term \(x_n\) is either \(a\) or \(b\), so \(x_n\in X\). Thus the sequence is in this neighborhood for all \(n\), and it converges to \(a\). The same check applies to \(b\) and \(c\): each has \(X\) as its only open neighborhood, and every term lies in \(X\). Therefore the sequence converges to all three points.
Check: for any \(x\in X\), the only neighborhood to test is \(X\), and \(x_n\in X\) for every \(n\). Hence the set of limits is exactly \(X=\{a,b,c\}\).
The empty set causes no exceptional sequence behavior: there is no sequence with terms in an empty set and no candidate limit in it. The assertion above concerns nonempty indiscrete spaces.
Convergence in the Cofinite Topology
For a set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complements in \(X\) are finite. This is a topology: arbitrary unions of open sets are open, since a union containing a nonempty open set has complement contained in that set’s finite complement; and a finite intersection of nonempty open sets has complement equal to a finite union of finite sets, hence finite. Intersections involving \(\varnothing\) are open as well.
Suppose now that \(X\) is infinite. The open neighborhoods of \(x\in X\) are exactly the sets \(X\setminus F\), where \(F\subseteq X\) is finite and \(x\notin F\). The condition \(x\notin F\) ensures \(x\in X\setminus F\), as required for a neighborhood.
Proof. Suppose first that \((x_n)\) converges to \(x\), and fix \(y\in X\) with \(y\ne x\). The set \(U=X\setminus\{y\}\) is open because its complement is finite, and it contains \(x\). Convergence means that \(x_n\in U\) for all sufficiently large \(n\). Thus \(x_n=y\) can occur only among the finitely many indices before that point, so \(\{n\in\mathbb{N}:x_n=y\}\) is finite.
Conversely, suppose every \(y\ne x\) occurs at only finitely many indices. Let \(U\) be any open neighborhood of \(x\). Then \(U=X\setminus F\) for some finite set \(F\) with \(x\notin F\). For each \(y\in F\), the set of indices with \(x_n=y\) is finite by the assumption. Since \(F\) is finite, the union of these index sets is finite. If that union is empty, every term is in \(U\). Otherwise, choose \(N\) larger than every index in the union. For all \(n\ge N\), \(x_n\notin F\), so \(x_n\in U\). This holds for every neighborhood \(U\) of \(x\), and therefore \((x_n)\) converges to \(x\). \(\square\)
Worked Example: A Sequence Converging to Every Point
Let \(X=\mathbb{N}\) with the cofinite topology, and take \(x_n=n\). Find all limits.
Fix any proposed limit \(m\in\mathbb{N}\). For each \(y\in\mathbb{N}\) with \(y\ne m\), the equality \(x_n=y\) holds exactly when \(n=y\). Thus \(y\) occurs at exactly one index, a finite number of times. The proposition shows that \(x_n\to m\). Since the argument holds for every \(m\in\mathbb{N}\), every point of \(X\) is a limit.
Neighborhood check: let \(U=\mathbb{N}\setminus F\) be any open neighborhood of \(m\), where \(F\) is finite and \(m\notin F\). If \(F=\varnothing\), every term is in \(U\). If \(F\ne\varnothing\), it has a largest element \(M\). For every \(n\ge M+1\), \(n\notin F\), so \(x_n=n\in U\). This verifies convergence directly.
Worked Example: A Sequence with Exactly One Limit
Still on \(X=\mathbb{N}\) with the cofinite topology, define \(x_{2k}=1\) and \(x_{2k-1}=k+1\) for \(k\in\mathbb{N}\). Find all limits.
The value \(1\) occurs at every even index \(2k\), so it occurs infinitely often. Every value \(y\ge2\) occurs exactly once: it appears as \(x_{2k-1}\) for the single index choice \(k=y-1\). To test the candidate limit \(1\), every other value therefore occurs finitely often, and the proposition gives \(x_n\to1\).
Now fix any candidate limit \(m\ge2\). The value \(1\ne m\) occurs infinitely often. The open neighborhood \(\mathbb{N}\setminus\{1\}\) contains \(m\), but infinitely many terms of the sequence equal \(1\) and lie outside that neighborhood. The sequence is not eventually in it, so \(x_n\) does not converge to \(m\). Thus \(1\) is its unique limit.
Check: each neighborhood of \(1\) excludes only a finite set of values other than \(1\); each such value occurs once, so only finitely many terms are excluded. In contrast, \(\mathbb{N}\setminus\{1\}\) excludes infinitely many terms, ruling out every \(m\ge2\).
When \(X\) is finite, the cofinite topology is the discrete topology: every subset has finite complement and is open. In particular, the infinite-set assumption matters. In a discrete topology, \(\{x\}\) is an open neighborhood of \(x\), and convergence to \(x\) forces all sufficiently late terms to equal \(x\); a sequence cannot converge to a distinct point as well.
Disjoint Neighborhoods Force Uniqueness
Proof. Suppose \(x\ne y\) have disjoint open neighborhoods \(U\) and \(V\), with \(x\in U\), \(y\in V\), and \(U\cap V=\varnothing\). Assume for contradiction that a sequence \((x_n)\) converges to both points. By convergence to \(x\), there is \(N_1\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N_1\). By convergence to \(y\), there is \(N_2\in\mathbb{N}\) such that \(x_n\in V\) for every \(n\ge N_2\). Let \(N=\max\{N_1,N_2\}\). Then \(x_N\in U\) and \(x_N\in V\), so \(x_N\in U\cap V=\varnothing\), which is impossible. Therefore the sequence cannot converge to both points. \(\square\)
A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods. The theorem therefore shows that every sequence in a Hausdorff space has at most one limit. “At most one” is important: a sequence need not have a limit at all. The hypothesis used in the proof is the existence of disjoint neighborhoods for the particular pair of proposed limits.
A Non-Hausdorff Space Does Not Make Every Limit Ambiguous
Failure to be Hausdorff means that some pair of distinct points cannot be separated by disjoint open neighborhoods. It does not mean that every sequence has multiple limits. The next finite example illustrates both nonuniqueness and uniqueness, depending on the sequence.
Worked Example: Two Sequences in the Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). First consider the constant sequence \(x_n=a\) for every \(n\). It converges to \(a\): the neighborhoods of \(a\) are \(\{a\}\) and \(X\), and every term belongs to both. It also converges to \(b\), because the only open neighborhood of \(b\) is \(X\), which contains every term. Thus this sequence has two distinct limits.
Now consider the constant sequence \(y_n=b\) for every \(n\). It converges to \(b\), since every term lies in the only neighborhood \(X\) of \(b\). It does not converge to \(a\): the open neighborhood \(\{a\}\) of \(a\) contains none of the terms \(y_n=b\), so the sequence is not eventually in \(\{a\}\). Therefore this sequence has exactly one limit, \(b\).
The space is not Hausdorff: the only neighborhood of \(b\) is \(X\), which intersects both possible neighborhoods \(\{a\}\) and \(X\) of \(a\). Hence no pair of open neighborhoods of \(a\) and \(b\) is disjoint.
Common Pitfall: Assuming Non-Hausdorff Points Separate
A tempting but incorrect claim is that distinct points in a cofinite space have disjoint neighborhoods. For distinct \(a,b\) in an infinite set \(X\), consider the open neighborhoods \(U=X\setminus\{b\}\) of \(a\) and \(V=X\setminus\{a\}\) of \(b\). It would be wrong to call these disjoint. Their intersection is
Because \(X\) is infinite, removing the two points \(a,b\) does not leave the empty set, so \(U\cap V\ne\varnothing\). More generally, any two nonempty cofinite open sets intersect: their intersection has complement equal to a finite union of finite sets, so its complement is finite; in an infinite \(X\), that intersection cannot be empty. Thus the uniqueness theorem cannot be applied to distinct points in this space. The sequence \(x_n=n\) on \(\mathbb{N}\) confirms that this lack of separation can matter: it converges to every point.
The safe conclusion is conditional: disjoint open neighborhoods rule out two distinct sequence limits. A space without the Hausdorff property lacks that guarantee for at least one pair, but whether a particular sequence has multiple limits must still be checked from the neighborhoods.
Check Your Understanding
For each question, use the neighborhood definition of convergence and state which topology is being considered.
- In an indiscrete topology on a nonempty set \(X\), prove directly that a sequence converges to every point of \(X\).
- In the cofinite topology on an infinite set \(X\), a sequence takes each value at most once. To which points does it converge?
- In the cofinite topology on an infinite set \(X\), suppose \(y\ne x\) occurs infinitely often in a sequence. Give an open neighborhood of \(x\) that rules out convergence to \(x\).
- Explain why two disjoint open neighborhoods contradict a sequence having both of their points as limits.
- In the stated Sierpinski topology, why does the constant sequence at \(b\) fail to converge to \(a\), even though the space is not Hausdorff?