Why Uniqueness Needs Separation
In familiar metric spaces, a sequence does not appear to approach two different points at once. The topological reason is that distinct points in a Hausdorff space have disjoint open neighborhoods. If a sequence converged to both points, it would eventually have to lie in each neighborhood, which is impossible when the neighborhoods are disjoint. We will make this argument precise and compare it with a space where a sequence really does have multiple limits.
Let \((X,\mathcal{T})\) be a topological space. A sequence \((x_n)\) in \(X\) converges to \(x\in X\), written \(x_n\to x\), if for every open neighborhood \(U\) of \(x\), there is an index \(N\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N\). The index \(N\) may depend on \(U\). A space is Hausdorff if for every pair of distinct points \(x,y\in X\), there are open neighborhoods \(U\) of \(x\) and \(V\) of \(y\) such that \(U\cap V=\varnothing\).
Proof. Suppose \(x_n\to x\) and \(x_n\to y\). Assume, for a contradiction, that \(x\ne y\). Since \(X\) is Hausdorff, there are open neighborhoods \(U\) of \(x\) and \(V\) of \(y\) with \(U\cap V=\varnothing\). Convergence to \(x\) gives an index \(N_U\) such that \(x_n\in U\) for all \(n\ge N_U\). Convergence to \(y\) gives an index \(N_V\) such that \(x_n\in V\) for all \(n\ge N_V\). Set \(N=\max\{N_U,N_V\}\). Then \(x_N\in U\) and \(x_N\in V\), so \(x_N\in U\cap V=\varnothing\), a contradiction. Therefore \(x=y\). \(\square\)
The contradiction uses both parts of the definition: convergence forces every sufficiently late term into each chosen neighborhood, and the Hausdorff condition makes those neighborhoods disjoint. No metric, distance, or special formula for the sequence is required.
Using the Argument in Familiar Spaces
Worked Example: A Sequence in the Real Line
Give a direct disjoint-neighborhood argument showing that the sequence \(x_n=1/n\) in \(\mathbb{R}\) with its standard topology cannot converge both to \(0\) and to \(1\).
First verify its convergence to \(0\). Let \(\varepsilon>0\). Choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For every \(n\ge N\), we have \(n\ge N>1/\varepsilon\), hence \(1/n\le 1/N<\varepsilon\). Thus \(|x_n-0|=1/n<\varepsilon\), which proves \(x_n\to0\).
Now choose the open neighborhoods \(U=(-1/3,1/3)\) of \(0\) and \(V=(2/3,4/3)\) of \(1\). They are disjoint because every point of \(U\) is less than \(1/3\), while every point of \(V\) is greater than \(2/3\). For \(n\ge4\), \(0<1/n\le1/4<1/3\), so \(x_n\in U\). On the other hand, for \(n\ge2\), \(1/n\le1/2<2/3\), so \(x_n\notin V\). Therefore the sequence cannot converge to \(1\): convergence to \(1\) would require every sufficiently late term to be in \(V\).
Check: the chosen neighborhoods really are disjoint, and the tail from \(n=4\) lies in \(U\). The sequence has the limit \(0\), but not the distinct proposed limit \(1\), as the theorem predicts.
Worked Example: Disjoint Balls in the Plane
Use the same principle for \(x_n=(1/n,0)\) in \(\mathbb{R}^2\) with the Euclidean topology. Compare the proposed limits \(p=(0,0)\) and \(q=(1,0)\).
The Euclidean distance from \(x_n\) to \(p\) is \(\|x_n-p\|=1/n\). Given \(\varepsilon>0\), choose \(N>1/\varepsilon\). Then \(n\ge N\) implies \(\|x_n-p\|=1/n\le1/N<\varepsilon\), so \(x_n\to p\).
Consider the open balls \(B(p,1/3)\) and \(B(q,1/3)\). Their centers are distance \(\|p-q\|=1\) apart. If a point \(z\) belonged to both balls, the triangle inequality would give \(1=\|p-q\|\le\|p-z\|+\|z-q\|<1/3+1/3=2/3\), which is impossible. Thus the balls are disjoint. For \(n\ge4\), \(\|x_n-p\|=1/n\le1/4<1/3\), so \(x_n\in B(p,1/3)\). For \(n\ge2\), \(\|x_n-q\|=1-1/n\ge1/2>1/3\), so \(x_n\notin B(q,1/3)\). This tail cannot satisfy the neighborhood condition for convergence to \(q\).
Check: the distances to the two centers are \(1/n\) and \(1-1/n\), respectively. For all \(n\ge4\), the first is less than \(1/3\), while the second is greater than \(1/3\); the sequence lies in the first ball and outside the second.
A Supporting Fact About Subsequences
A subsequence records selected terms while preserving their original order. Formally, if \(n_1<n_2<\cdots\) are indices in \(\mathbb{N}\), then \((x_{n_k})\) is a subsequence of \((x_n)\). The following fact explains why selecting terms cannot change a limit.
Proof. Let \(U\) be any open neighborhood of \(x\). Since \(x_n\to x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N\). The indices \(n_k\) increase strictly, so \(n_k\ge k\); in particular, for every \(k\ge N\), \(n_k\ge N\). Therefore \(x_{n_k}\in U\) for every \(k\ge N\). This is the definition of \(x_{n_k}\to x\). \(\square\)
Combining this proposition with uniqueness gives a useful consequence: in a Hausdorff space, a subsequence of a convergent sequence cannot have a different limit. The proposition itself does not require Hausdorffness; the uniqueness conclusion does.
Proof. Suppose \(y\ne x\). The Hausdorff condition provides disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) implies that \(x_n\in U\) for every sufficiently large \(n\). Since \(y\in V\) and \(U\cap V=\varnothing\), none of those terms can equal \(y\). Thus \(y\) can occur only among the finitely many terms before the sequence enters \(U\), contradicting the assumption that it occurs infinitely often. Hence \(y=x\). \(\square\)
Why the Hausdorff Hypothesis Matters
Worked Example: Multiple Limits in the Cofinite Topology
Let \(X=\mathbb{N}\) have the cofinite topology: its open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. Consider the sequence \(x_n=n\). We will show that it converges both to \(1\) and to \(2\), which are distinct points of \(X\).
Fix any point \(p\in X\) and any open neighborhood \(U\) of \(p\). Since \(U\) is open, \(F=X\setminus U\) is finite. If \(F=\varnothing\), every term belongs to \(U\). If \(F\ne\varnothing\), it has a largest element \(M\). For every \(n>M\), \(n\notin F\), so \(x_n=n\in U\). Thus every open neighborhood of \(p\) contains all sufficiently late terms. This proves \(x_n\to p\) for every \(p\in X\), including both \(1\) and \(2\).
This topology is not Hausdorff. Indeed, any two nonempty open sets \(U,V\) have finite complements \(F=X\setminus U\) and \(G=X\setminus V\). Their intersection is \(X\setminus(F\cup G)\). Because \(F\cup G\) is finite and \(X=\mathbb{N}\) is infinite, this intersection is nonempty. Hence no two nonempty open sets are disjoint, so distinct points cannot have disjoint open neighborhoods.
Check: for any chosen neighborhood, only finitely many natural numbers are excluded, and the sequence \(1,2,3,\ldots\) eventually exceeds all of them. The multiple limits do not contradict the theorem: the space fails its Hausdorff hypothesis.
A Common Pitfall
Common pitfall: Distinct points do not automatically have disjoint neighborhoods. For example, in \(\mathbb{R}\), the sets \(U=(-2,2)\) and \(V=(-1,3)\) are neighborhoods of \(0\) and \(1\), respectively, but the calculation \(U\cap V=\varnothing\) would be wrong: in fact \(U\cap V=(-1,2)\), which contains \(0\). A sequence eventually lying in both of these neighborhoods would not produce a contradiction.
The correction is to use the Hausdorff condition to choose neighborhoods that are disjoint. For \(0\) and \(1\), the neighborhoods \((-1/3,1/3)\) and \((2/3,4/3)\) are disjoint. More generally, in a Hausdorff space the definition supplies a disjoint pair for every two distinct points. Without that separation property, as the cofinite example shows, the uniqueness conclusion can fail.
Unique sequence limits matter whenever convergence is used to assign a definite value to a limit. Hausdorffness guarantees that this assignment is unambiguous, but it does not assert that every sequence converges. The theorem says only that if a sequence has a limit, it cannot have two different ones.
Check Your Understanding
Use the neighborhood definition of convergence and the Hausdorff condition in your answers.
- In the uniqueness proof, why is it necessary to take \(N=\max\{N_U,N_V\}\)?
- For \(x_n=1/n\) in the standard topology on \(\mathbb{R}\), give disjoint open neighborhoods of \(0\) and \(2\) and explain why the sequence cannot converge to \(2\).
- Does the proposition about subsequences require the space to be Hausdorff? Justify your answer from its proof.
- Why does the sequence \(x_n=n\) converge to every point in the cofinite topology on \(\mathbb{N}\)?
- Which hypothesis of the uniqueness theorem fails in the cofinite example, and how does that failure allow two distinct limits?