Convergence Is a Neighborhood Condition
A sequence records an ordered list of points, but its limit is determined by what happens to all sufficiently late terms near a proposed point. In a topological space, “near” is described by open neighborhoods rather than by a numerical distance. This definition works whether or not the space has a metric.
Let \((X,\mathcal{T})\) be a topological space. A sequence in \(X\) is a function from \(\mathbb{N}=\{1,2,\dots\}\) to \(X\), written \((x_n)\). An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) with \(x\in U\). To say that a condition holds for all sufficiently large \(n\) means that there is some \(N\in\mathbb{N}\) such that it holds for every \(n\ge N\).
The definition does not require any particular behavior among the first \(N-1\) terms. It requires that, no matter which open neighborhood of \(x\) is chosen, all terms from some point onward lie in that neighborhood. The index \(N\) may depend on the neighborhood.
The Open-Set Form of the Definition
Proof. An open neighborhood of \(x\) is precisely an open set that contains \(x\). Thus the condition in the definition, applied to every open neighborhood, says exactly that for every open \(U\) containing \(x\), some \(N\) satisfies \(x_n\in U\) for every \(n\ge N\). Conversely, every open neighborhood is such an open set, so the stated open-set condition gives the definition. \(\square\)
This formulation is useful when a topology is specified by listing its open sets or by giving a basis. For a basis \(\mathcal{B}\), it is enough to check the basis neighborhoods of \(x\): every open set containing \(x\) contains some basis element that contains \(x\). Conversely, if the condition holds for every open neighborhood, it holds in particular for every basis neighborhood.
Worked Example: A Sequence in the Usual Real Line
Take \(\mathbb{R}\) with its standard topology and let \(x_n=2+\frac{1}{n}\). We verify that \(x_n\to 2\).
Let \(U\) be any open neighborhood of \(2\). Since the standard topology is metric, there is an \(\varepsilon>0\) such that \((2-\varepsilon,2+\varepsilon)\subseteq U\). Choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For every \(n\ge N\), \(1/n\le 1/N<\varepsilon\), and hence \[ 2-\varepsilon<2+\frac{1}{n}<2+\varepsilon. \] Therefore \(x_n\in U\) for all \(n\ge N\), as required.
Numerical check: if \(\varepsilon=1/4\), choose \(N=5\), since \(5>1/(1/4)=4\). For \(n\ge5\), \(0<1/n\le1/5<1/4\), so \(2+1/n\in(7/4,9/4)\). The bound is strict at the neighborhood endpoints, as needed.
Check by distance: \(d(x_n,2)=|(2+1/n)-2|=1/n\), which tends to \(0\). The metric criterion proved below gives the same conclusion.
The Metric Criterion
In a metric space, open balls provide the neighborhoods, so the topological definition can be written using distances. Recall that \(B_d(x,r)=\{y\in X:d(x,y)<r\}\) for \(r>0\).
Proof. Suppose first that \(x_n\to x\). For any \(\varepsilon>0\), the open ball \(B_d(x,\varepsilon)\) is an open neighborhood of \(x\). By convergence, there is \(N\in\mathbb{N}\) such that \(x_n\in B_d(x,\varepsilon)\) for every \(n\ge N\). The definition of the ball gives \(d(x_n,x)<\varepsilon\) for all those \(n\).
Conversely, suppose that the stated distance condition holds for every \(\varepsilon>0\). Let \(U\) be any open neighborhood of \(x\). Because \(U\) is open in the metric topology and contains \(x\), there is \(r>0\) such that \(B_d(x,r)\subseteq U\). Apply the distance condition with \(\varepsilon=r\). For some \(N\), all \(n\ge N\) satisfy \(d(x_n,x)<r\), so \(x_n\in B_d(x,r)\subseteq U\). This holds for every open neighborhood \(U\), proving \(x_n\to x\). \(\square\)
Both the strict inequality in the ball definition and the quantifier “for every \(\varepsilon>0\)” matter. A single distance bound, or a bound that only applies to selected terms, does not establish convergence.
Worked Example: A Sequence That Does Not Converge in the Usual Real Line
Consider \(x_n=(-1)^n\) in \(\mathbb{R}\) with its standard topology. Its terms alternate between \(-1\) and \(1\). We show that no \(x\in\mathbb{R}\) is a limit.
First suppose \(x\ne -1\) and \(x\ne1\). Let \(d=\min\{|x-1|,|x+1|\}\). Both distances are positive, so \(d>0\). The open interval \((x-d/2,x+d/2)\) contains \(x\) but contains neither \(1\) nor \(-1\): each of those points is at distance at least \(d>d/2\) from \(x\). No term of the sequence lies in this neighborhood, so the sequence cannot converge to \(x\).
If \(x=1\), use the open neighborhood \((1/2,3/2)\). Every odd index \(n\) has \(x_n=-1\), which is outside this interval, and odd indices occur arbitrarily far along the sequence. Thus the terms are not all in this neighborhood from any index onward. If \(x=-1\), the neighborhood \((-3/2,-1/2)\) excludes \(1\), and every even index has \(x_n=1\). This also rules out convergence.
These cases include every real candidate \(x\), so the sequence has no limit. In particular, merely noting that its values remain bounded would not establish convergence.
The Topology Can Change Convergence
The same sequence may converge in one topology and fail to converge in another on the same underlying set. The following example uses the lower-limit topology \(\mathbb{R}_\ell\), whose basis consists of the half-open intervals \([a,b)\) with \(a<b\). Thus its open sets are unions of such intervals.
Worked Example: Usual Convergence Need Not Be Lower-Limit Convergence
Let \(x_n=-1/n\), viewed first in \(\mathbb{R}\) with its standard topology. For every \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(n\ge N\), then \[ |x_n-0|=\left|-\frac1n\right|=\frac1n\le\frac1N<\varepsilon. \] The metric criterion shows that \(x_n\to0\) in the usual topology.
Now give the same set \(\mathbb{R}\) the lower-limit topology. The basis element \([0,1)\) is an open neighborhood of \(0\). But \(x_n=-1/n<0\) for every \(n\in\mathbb{N}\), so no term belongs to \([0,1)\). The required neighborhood condition fails, and \(x_n\) does not converge to \(0\) in \(\mathbb{R}_\ell\).
Check: membership in \([0,1)\) requires \(0\le x_n<1\), including the left endpoint and excluding the right endpoint. Since \(x_n<0\) for every \(n\), the failure is at the left endpoint condition, not at the right endpoint.
Worked Example: Convergence in a Discrete Topology
Let \(X=\{a,b\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset of \(X\) is open. Consider a sequence with \(x_n=b\) for \(n=1,2\) and \(x_n=a\) for every \(n\ge3\). We verify its convergence directly.
The set \(\{a\}\) is an open neighborhood of \(a\). For every \(n\ge3\), \(x_n=a\in\{a\}\). If \(U\) is any other open neighborhood of \(a\), then \(a\in U\), so again \(x_n=a\in U\) for every \(n\ge3\). Therefore the sequence converges to \(a\).
It does not converge to \(b\): the open neighborhood \(\{b\}\) of \(b\) contains neither \(x_n\) for \(n\ge3\) nor any term from that point onward. The first two terms do not affect convergence to \(a\), but the requirement about all sufficiently late terms rules out convergence to \(b\).
Check: the open set \(\{a\}\) contains all terms with \(n\ge3\), and the open set \(\{b\}\) does not contain any of them. These two neighborhood tests match the conclusions.
Proof. Suppose \((x_n)\) converges to \(x\). In the discrete topology the singleton \(\{x\}\) is open and contains \(x\). Applying the definition to this neighborhood gives an \(N\) such that \(x_n\in\{x\}\) for all \(n\ge N\), which means \(x_n=x\) for all \(n\ge N\). Conversely, suppose that equality holds for all \(n\ge N\). Every open neighborhood \(U\) of \(x\) contains \(x\), so it contains \(x_n\) for every \(n\ge N\). The sequence therefore converges to \(x\). This argument also covers a one-point discrete space, where every sequence is constant. \(\square\)
A Common Mistake and a Practical Test
Common mistake. When checking \(x_n=1/n\to0\) in \(\mathbb{R}\), one might claim that \(N=1\) works for \(\varepsilon=1/2\), because \(1/n\) gets smaller as \(n\) increases. That choice fails: at \(n=1\), \(|x_1-0|=1\), and \(1\not<1/2\). The convergence condition must hold for every \(n\ge N\), including \(n=N\). A correct choice is \(N=3\): for every \(n\ge3\), \(1/n\le1/3<1/2\). The distinction is between a term eventually becoming close and every term from the chosen index onward satisfying the bound.
For a general topological space, the practical test is to take an arbitrary open neighborhood of the proposed limit and show that all sufficiently late terms lie in it. In a metric space, it is enough to verify the distance inequality for each \(\varepsilon>0\). For a topology given by a basis, it is enough to check basis neighborhoods of the proposed limit. Each method uses the same underlying requirement: the neighborhood is fixed first, and then an index is found that works for every later term.
Check Your Understanding
For each question, specify the topology and use the neighborhood definition or the metric criterion as appropriate.
- In the standard topology on \(\mathbb{R}\), prove that \(x_n=3-2/n\) converges to \(3\).
- In the standard topology on \(\mathbb{R}\), does \(x_n=1+(-1)^n/n\) converge? If so, identify its limit and justify the distance estimate.
- In the lower-limit topology on \(\mathbb{R}\), test whether \(x_n=1/n\) converges to \(0\) by checking an arbitrary basic neighborhood \([0,b)\), where \(b>0\).
- In a discrete space, can a sequence converge to \(x\) without being equal to \(x\) for all sufficiently large indices? Explain using the open neighborhood \(\{x\}\).
- Give an example of a sequence that converges in the usual topology on \(\mathbb{R}\) but not in the lower-limit topology, and identify an open neighborhood witnessing failure.