When a Constant Sequence Has Two Limits
In the real line, a sequence cannot converge to two different points. That familiar fact depends on how neighborhoods separate points. In a general topological space, convergence is defined using open neighborhoods, and those neighborhoods need not distinguish points in both directions. The two-point Sierpinski space gives a particularly small example: one constant sequence converges to both points.
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). A sequence \((x_n)\) in a topological space converges to \(x\) if, for every open neighborhood \(U\) of \(x\), there is an \(N\in\mathbb{N}\) such that \(x_n\in U\) for all \(n\ge N\). The point \(x\) is then called a limit of the sequence. This definition makes no promise that a limit exists or is unique.
All the Limits in the Sierpinski Space
The open neighborhoods of \(a\) are \(\{a\}\) and \(X\). The only open neighborhood of \(b\) is \(X\). Consequently, every sequence in \(X\) converges to \(b\): its terms all belong to \(X\). A sequence converges to \(a\) exactly when it is eventually equal to \(a\), because it must eventually belong to the neighborhood \(\{a\}\).
Worked Example: The Constant Sequence at \(a\)
Set \(x_n=a\) for every \(n\in\mathbb{N}\). To check convergence to \(a\), consider each open neighborhood of \(a\). If \(U=\{a\}\), then \(x_n=a\in U\) for every \(n\), so \(N=1\) works. If \(U=X\), then \(x_n\in U\) for every \(n\), and again \(N=1\) works. Thus \(x_n\to a\).
To check convergence to \(b\), its only open neighborhood is \(X\). Every term \(x_n=a\) belongs to \(X\), so \(N=1\) also works. Thus \(x_n\to b\) as well. The constant sequence at \(a\) has two distinct limits.
Check: The conclusion uses the actual neighborhoods, not the names of the points: \(\{a\}\) is a neighborhood of \(a\), but it is not a neighborhood of \(b\). Since \(b\) has no neighborhood that excludes \(a\), the constant terms at \(a\) satisfy every neighborhood test for convergence to \(b\).
Worked Example: Classifying Every Sequence
Let \((x_n)\) be any sequence with terms in \(X=\{a,b\}\). For convergence to \(b\), the only neighborhood to test is \(X\); each term is in \(X\), so the sequence always converges to \(b\). For convergence to \(a\), the neighborhood \(\{a\}\) requires that there be some \(N\) such that \(x_n=a\) for every \(n\ge N\). Therefore the sequence converges to \(a\) if and only if it is eventually equal to \(a\).
For instance, the constant sequence \(x_n=b\) converges to \(b\), since all its terms lie in \(X\). It does not converge to \(a\): for every \(N\), the term \(x_N=b\) is not in the neighborhood \(\{a\}\). A sequence alternating between \(a\) and \(b\) also converges to \(b\), since every term lies in \(X\), but it does not converge to \(a\), since terms equal to \(b\) occur arbitrarily far along the sequence.
Check: These cases agree with the two tests just derived. Every sequence has \(b\) as a limit; an additional limit \(a\) occurs precisely when all terms from some index onward equal \(a\).
What the Separation Axioms Say Here
The Sierpinski space is \(T_0\). The open set \(\{a\}\) contains \(a\) and excludes \(b\), so it distinguishes the two points in the sense required by \(T_0\). It is not \(T_1\): for the ordered pair \(x=b,y=a\), no open set contains \(b\) while excluding \(a\), because the only open set containing \(b\) is \(X\). It is not Hausdorff either. Every neighborhood of \(b\) is \(X\), which intersects every neighborhood of \(a\); hence there are no disjoint neighborhoods of \(a\) and \(b\).
Proof. Fix \(x\in X\) and consider the constant sequence \(x_n=x\). It converges to \(y\) exactly when every open neighborhood of \(y\) contains \(x\): if each neighborhood contains \(x\), every term lies in it, and if some neighborhood excludes \(x\), no term lies in it. Suppose \(X\) is \(T_1\). For any \(y\ne x\), the \(T_1\) condition gives an open neighborhood of \(y\) that excludes \(x\), so the constant sequence at \(x\) does not converge to \(y\). It does converge to \(x\), because every neighborhood of \(x\) contains \(x\). Conversely, suppose every constant sequence has no limit other than its constant value. Given distinct \(x,y\), the constant sequence at \(x\) does not converge to \(y\). By the convergence definition, some open neighborhood of \(y\) must fail to contain \(x\); that neighborhood contains \(y\) and excludes \(x\). Applying this to every ordered pair of distinct points gives the \(T_1\) condition. \(\square\)
This proposition identifies precisely what \(T_1\) guarantees about constant sequences. It does not say that \(T_1\) guarantees unique limits for every sequence. Hausdorff separation does guarantee that stronger conclusion.
Proof. Suppose a sequence \((x_n)\) converges to both \(x\) and \(y\), and suppose \(x\ne y\). Because the space is Hausdorff, there are disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) gives an index \(N_1\) such that \(x_n\in U\) for all \(n\ge N_1\). Convergence to \(y\) gives an index \(N_2\) such that \(x_n\in V\) for all \(n\ge N_2\). For \(n\ge\max(N_1,N_2)\), the term \(x_n\) must lie in \(U\cap V\), but \(U\cap V=\varnothing\). This is impossible. Thus \(x=y\), so the sequence cannot have two distinct limits. \(\square\)
Comparing \(T_0\), \(T_1\), and Hausdorff
Each step in the separation hierarchy rules out some failures that can occur at the weaker level. \(T_0\) requires a distinction in at least one direction; \(T_1\) requires it in either specified direction; Hausdorff separation requires neighborhoods that are disjoint. The implications \( \text{Hausdorff}\Rightarrow T_1\Rightarrow T_0\) follow directly from the definitions. For the first implication, disjoint neighborhoods of \(x\) and \(y\) give a neighborhood of \(x\) excluding \(y\). For the second, the \(T_1\) condition supplies such a neighborhood for each ordered pair, and therefore supplies a set distinguishing any two distinct points.
Worked Example: Two Points with the Discrete Topology
Give \(X=\{a,b\}\) the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. For distinct points, \(\{a\}\) and \(\{b\}\) are disjoint open neighborhoods. Thus the space is Hausdorff, and hence also \(T_1\) and \(T_0\).
A sequence converges to \(a\) exactly when it is eventually equal to \(a\), because \(\{a\}\) is an open neighborhood of \(a\). It converges to \(b\) exactly when it is eventually equal to \(b\), because \(\{b\}\) is an open neighborhood of \(b\). These two conditions cannot both hold: if both held, then for every sufficiently large index \(n\), the same term \(x_n\) would equal both \(a\) and \(b\), contradicting \(a\ne b\).
Check: The disjoint neighborhoods give the separation property, and the singleton neighborhoods show directly why a sequence cannot converge to both points.
Worked Example: A \(T_1\) Space with a Sequence Having Many Limits
Let \(X=\mathbb{N}\) have the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. Every singleton is closed because its complement is open, so \(X\) is \(T_1\). Consider the sequence \(x_n=n\).
Fix any \(p\in X\) and any open neighborhood \(U\) of \(p\). The complement \(F=X\setminus U\) is finite. If \(F=\varnothing\), every term lies in \(U\), so \(N=1\) works. If \(F\ne\varnothing\), it has a largest element \(m=\max F\). Set \(N=m+1\). For every \(n\ge N\), we have \(n>m\), so \(n\notin F\), and hence \(x_n=n\in U\). This proves \(x_n\to p\). Since \(p\) was arbitrary, this one sequence converges to every point of \(X\).
The space is not Hausdorff. Any two nonempty open sets are cofinite, and their intersection is nonempty: if \(U\) and \(V\) have finite complements, then \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\) is finite, so \(U\cap V\) is infinite. Therefore distinct points cannot have disjoint open neighborhoods.
Check: This example does not contradict the proposition about \(T_1\): it says constant sequences have no other limits in a \(T_1\) space. The sequence \(x_n=n\) is not constant. It also illustrates why the Hausdorff hypothesis in the theorem matters.
A Common Mistake
Common pitfall: Assuming that the limit of a sequence must be its eventual constant value, or applying uniqueness of limits without checking a separation hypothesis. In the Sierpinski space, the constant sequence \(x_n=a\) does not have just the limit \(a\): the neighborhood test shows it also converges to \(b\). The invalid shortcut is to say “the terms are always \(a\), so the only limit is \(a\).” The correction is to test every open neighborhood of each proposed limit. For \(b\), the only such neighborhood is \(X\), and it contains every term \(a\). Thus \(b\) is a limit. In a Hausdorff space the theorem proves uniqueness; without that hypothesis, uniqueness must not be assumed.
Check Your Understanding
For each question, use the open-neighborhood definition of convergence or separation rather than relying on a picture.
- In the Sierpinski space, does the constant sequence at \(b\) converge to \(a\)? Identify the neighborhood that decides the answer.
- Give a sequence in the Sierpinski space that converges to both \(a\) and \(b\), and verify both limits.
- Which separation axiom does the Sierpinski space satisfy among \(T_0\), \(T_1\), and Hausdorff? Justify each answer from the definitions.
- Explain why a Hausdorff space has unique sequence limits, naming the neighborhoods used in the argument.
- In the finite-complement topology on \(\mathbb{N}\), why does the sequence \(x_n=n\) eventually lie in every open neighborhood of any fixed point?