Convergence of Nets and Neighborhood Tests

Convergence as Eventual Neighborhood Membership

A sequence is indexed by the natural numbers, but a net may be indexed by any directed set. Its indices need not be arranged in a single list: instead, the order specifies which indices count as later than a given index. This makes the phrase “eventually in every neighborhood” meaningful beyond sequences. Here we define convergence in those terms, translate the definition into open-set and metric tests, and examine why limits may fail to be unique in a non-Hausdorff space.

Definition: A directed set is a nonempty set \(D\) with a reflexive, transitive relation \(\leq\) such that for any \(d_1,d_2\in D\), there is \(e\in D\) with \(d_1\leq e\) and \(d_2\leq e\). A net in a space \(X\) is a function \(x:D\to X\), written \((x_d)_{d\in D}\). A property \(P(d)\) holds eventually if there is \(d_0\in D\) such that \(P(d)\) holds for every \(d\geq d_0\).

For each \(d_0\), the set of indices \(d\geq d_0\) is called a tail. Two tails always have a common later tail: if \(d_0,e_0\in D\), directedness gives an index \(f\) with \(d_0\leq f\) and \(e_0\leq f\). This feature is essential when we combine eventual conditions.

Definition: Let \((X,\mathcal{T})\) be a topological space. A net \((x_d)_{d\in D}\) converges to \(x\in X\), written \(x_d\to x\), if for every open neighborhood \(U\) of \(x\), there is \(d_0\in D\) such that \(x_d\in U\) for every \(d\geq d_0\).

The definition says that each neighborhood has its own threshold \(d_0\). There need not be one threshold that works for all neighborhoods at once. Also, the terms before \(d_0\) do not affect whether the net converges to \(x\).

The Open-Set Formulation

Theorem (Open-set test for net convergence): A net \((x_d)_{d\in D}\) in \(X\) converges to \(x\) if and only if for every open set \(U\in\mathcal{T}\) with \(x\in U\), there is \(d_0\in D\) such that \(x_d\in U\) for every \(d\geq d_0\).

Proof. Suppose \(x_d\to x\). Every open set \(U\) containing \(x\) is an open neighborhood of \(x\), so the definition gives an index \(d_0\) such that \(x_d\in U\) whenever \(d\geq d_0\). Conversely, suppose the stated condition holds for every open set containing \(x\). An open neighborhood of \(x\) is exactly such an open set, so the condition applies to every neighborhood in the definition of convergence. Therefore \(x_d\to x\). \(\square\)

This formulation can be used directly: choose an open set around the proposed limit and ask whether the net is eventually inside it. A test involving only one open neighborhood is not enough; convergence requires the condition for every such set.

Worked Example: A Net in the Real Line

Give \(D=\mathbb{N}\times\mathbb{N}\) the coordinatewise order: \((m,n)\leq(k,\ell)\) when \(m\leq k\) and \(n\leq\ell\). It is directed, since \((\max(m,k),\max(n,\ell))\) is an upper bound for any two indices. In \(\mathbb{R}\) with its standard topology, set \(x_{(m,n)}=1/m\). We verify that this net converges to \(0\).

Let \(\varepsilon>0\). Choose \(M\in\mathbb{N}\) with \(1/M<\varepsilon\), and take \(d_0=(M,1)\). If \((m,n)\geq(M,1)\), then \(m\geq M\), so \[ \left|x_{(m,n)}-0\right|=\left|\frac1m\right|=\frac1m\leq\frac1M<\varepsilon. \] Thus the net is eventually in the interval \((-\varepsilon,\varepsilon)\) for every \(\varepsilon>0\).

Check: the metric test below therefore gives \(x_{(m,n)}\to0\). The second coordinate does not affect the values, but it is still part of the directed index set; the chosen threshold \((M,1)\) ensures both coordinate inequalities required for a later index.

The Metric Test

Theorem (Metric test for net convergence): Let \((X,d)\) be a metric space, let \(D\) be directed, and let \((x_d)_{d\in D}\) be a net in \(X\). Then \(x_d\to x\) if and only if for every \(\varepsilon>0\), there is \(d_0\in D\) such that \(d(x_d,x)<\varepsilon\) for every \(d\geq d_0\).

Proof. Suppose \(x_d\to x\), and fix \(\varepsilon>0\). The open ball \(B(x,\varepsilon)\) is an open neighborhood of \(x\). Convergence gives an index \(d_0\) such that \(x_d\in B(x,\varepsilon)\) whenever \(d\geq d_0\). By the definition of an open ball, this means \(d(x_d,x)<\varepsilon\). Conversely, suppose the stated inequality holds for every positive \(\varepsilon\). Let \(U\) be any open neighborhood of \(x\). Since \(U\) is open and contains \(x\), there is \(r>0\) such that \(B(x,r)\subseteq U\). The hypothesis for \(\varepsilon=r\) supplies \(d_0\) for which \(d(x_d,x)<r\) whenever \(d\geq d_0\). Thus \(x_d\in B(x,r)\subseteq U\) for every \(d\geq d_0\). The neighborhood definition gives \(x_d\to x\). \(\square\)

Worked Example: An Oscillating Net Does Not Converge

Use the same directed set \(D=\mathbb{N}\times\mathbb{N}\), and define \(y_{(m,n)}=(-1)^m\) in the standard real line. We show that this net has no limit. Suppose, to the contrary, it converges to some \(L\in\mathbb{R}\). The metric test with \(\varepsilon=1/2\) gives an index \((M,N)\) such that \[ |y_{(m,n)}-L|<\frac12 \] for all \((m,n)\geq(M,N)\).

Choose an even integer \(m_+\geq M\) and an odd integer \(m_-\geq M\), and choose \(n=N\). Both \((m_+,N)\) and \((m_-,N)\) are at least \((M,N)\). Their net values are \(1\) and \(-1\), respectively. Therefore \(|1-L|<1/2\) and \(|-1-L|<1/2\). The triangle inequality would then give \[ 2=|1-(-1)|\leq |1-L|+|L-(-1)|<\frac12+\frac12=1, \] which is impossible.

Check: the argument considered an arbitrary proposed limit \(L\), so it rules out every real limit, not merely \(0\). The issue is that every tail contains indices with both parities in the first coordinate.

Limits in Non-Hausdorff Spaces

Convergence of a net does not always determine a unique limit. Recall that a space is Hausdorff if any two distinct points have disjoint open neighborhoods. The next result explains exactly how that separation condition guarantees uniqueness.

Theorem (Uniqueness of net limits in Hausdorff spaces): If \(X\) is Hausdorff and a net in \(X\) converges to both \(x\) and \(y\), then \(x=y\).

Proof. Suppose \(x_d\to x\) and \(x_d\to y\), and assume for contradiction that \(x\ne y\). Since \(X\) is Hausdorff, there are disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) gives an index \(d_1\) such that \(x_d\in U\) whenever \(d\geq d_1\). Convergence to \(y\) gives an index \(d_2\) such that \(x_d\in V\) whenever \(d\geq d_2\). Directedness gives \(e\in D\) with \(e\geq d_1\) and \(e\geq d_2\). Then \(x_e\in U\cap V\), contradicting \(U\cap V=\varnothing\). Hence \(x=y\). \(\square\)

Worked Example: One Net with Two Limits

Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). These sets form a topology: the empty set and \(X\) are included, and unions and finite intersections of listed sets are again listed. Take \(D=\mathbb{N}\) with its usual order and define the constant net \(x_n=a\).

The open neighborhoods of \(a\) are \(\{a\}\) and \(X\). The net belongs to \(\{a\}\) at every index, so it is eventually in each neighborhood of \(a\), and \(x_n\to a\). The only open neighborhood of \(b\) is \(X\), and every term belongs to \(X\). Thus \(x_n\to b\) as well.

Check: \(a\ne b\), so this space is not Hausdorff: the points have no disjoint open neighborhoods, since every open neighborhood of \(b\) is \(X\). The example confirms that the Hausdorff hypothesis in the uniqueness theorem cannot simply be omitted.

Eventual Membership and a Common Pitfall

Convergence concerns what happens on an entire tail, not merely what happens at arbitrarily large indices. For example, knowing that a net enters a neighborhood at some indices as far along as desired does not show that it stays there eventually. The two statements quantify differently: eventual membership requires one threshold after which every term is in the set.

Common pitfall. For the oscillating net \(y_{(m,n)}=(-1)^m\), one might argue that it “gets close” to \(1\) because there are arbitrarily large even values of \(m\), or gets close to \(-1\) because there are arbitrarily large odd values of \(m\). That does not establish convergence to either point. For any threshold \((M,N)\), there is an even \(m\geq M\) and an odd \(m\geq M\); taking \(n=N\) gives later indices with values \(1\) and \(-1\). In particular, the net is not eventually in \((1/2,3/2)\), a neighborhood of \(1\), and is not eventually in \((-3/2,-1/2)\), a neighborhood of \(-1\). A net must satisfy the neighborhood condition for every neighborhood of its proposed limit.

In a metric space, the metric test makes this requirement quantitative: for each radius \(\varepsilon>0\), all terms on some tail must lie within distance \(\varepsilon\) of the limit. In a general topological space, open neighborhoods provide the test even when no metric is available. In a Hausdorff space the limit, if it exists, is unique; without Hausdorff separation, as the finite example shows, one net may have more than one limit.

Check Your Understanding

Use the definitions and theorems in this tutorial to answer the following.

For the net \(x_{(m,n)}=1/m\) on \(\mathbb{N}\times\mathbb{N}\), which coordinate controls the distance to \(0\), and why must a threshold still be an element of the full index set?
Explain why finding arbitrarily large indices with terms in a neighborhood is weaker than eventual membership.
State the metric test for convergence of a net and identify where openness of metric balls is used in its proof.
In the topology \(\{\varnothing,\{a\},\{a,b\}\}\), which points are limits of the constant net at \(a\)? Verify each using the open neighborhoods.
Where does directedness enter the proof that a net in a Hausdorff space cannot converge to two distinct points?