Subnets and Cluster Points of Nets

Subnets and Cluster Points

A convergent sequence eventually stays in every neighborhood of its limit. A cluster point asks for less: every tail must return to every neighborhood, but the net need not stay there. Subnets formalize a way of selecting indices that move far enough along the original directed set while still finding values near a chosen cluster point.

Recall that a directed set is a set \(I\) with a reflexive, transitive order \(\le\) such that any two indices have a common upper bound. A net in a topological space \((X,\mathcal{T})\) is a family \((x_i)_{i\in I}\) of points of \(X\). The tail starting at \(i_0\) consists of the terms \(x_i\) with \(i\ge i_0\). A net converges to \(x\) if, for every open neighborhood \(U\) of \(x\), there is an index \(i_0\) such that \(x_i\in U\) whenever \(i\ge i_0\).

Definition: A point \(x\in X\) is a cluster point of a net \((x_i)_{i\in I}\) if, for every open neighborhood \(U\) of \(x\) and every \(i_0\in I\), there is some \(i\ge i_0\) such that \(x_i\in U\). A subnet of \((x_i)_{i\in I}\) is a net \((x_{\phi(j)})_{j\in J}\), where \(J\) is directed and \(\phi:J\to I\) is order-preserving and cofinal. Here order-preserving means \(j\le j'\) implies \(\phi(j)\le\phi(j')\), and cofinal means that for every \(i_0\in I\) there is \(j\in J\) with \(\phi(j)\ge i_0\).

With this definition, a subnet eventually has indices beyond any prescribed original index: cofinality gives a \(j_0\) with \(\phi(j_0)\ge i_0\), and order preservation gives \(\phi(j)\ge i_0\) for every \(j\ge j_0\). For a sequence indexed by \(\mathbb{N}\), the familiar subsequence \(x_{n_k}\), with \(n_k\) strictly increasing, is a subnet, using \(\phi(k)=n_k\).

Limits and Cluster Points

Proposition: If a net converges to \(x\), then \(x\) is a cluster point of that net.

Proof. Suppose \(x_i\to x\). Let \(U\) be any open neighborhood of \(x\), and let \(i_0\in I\). Convergence gives an index \(i_1\) such that \(x_i\in U\) for every \(i\ge i_1\). Since \(I\) is directed, there is an index \(i_2\) with \(i_2\ge i_0\) and \(i_2\ge i_1\). Then \(x_{i_2}\in U\) and \(i_2\ge i_0\). This is exactly the cluster-point condition. \(\square\)

The reverse implication does not generally hold. Convergence requires all sufficiently late terms to lie in each neighborhood; clustering requires only that some term in each tail lie there. The distinction is already visible for sequences in the real line.

Worked Example: An Oscillating Sequence

In \(\mathbb{R}\) with its standard topology, let \(x_n=(-1)^n\) for \(n\in\mathbb{N}\). The terms are \(-1,1,-1,1,\ldots\). We determine its cluster points and exhibit subnets converging to them.

Every neighborhood of \(1\) contains all terms equal to \(1\), and there are such terms at arbitrarily large even indices. Thus every tail meets every neighborhood of \(1\), so \(1\) is a cluster point. Every neighborhood of \(-1\) contains all terms equal to \(-1\), and there are such terms at arbitrarily large odd indices. Thus \(-1\) is also a cluster point.

There are no other cluster points. If \(y\notin\{-1,1\}\), let \(r=\min\{|y-1|,|y+1|\}/2\), which is positive. The open interval \((y-r,y+r)\) contains neither \(1\) nor \(-1\), so it contains no term of the net. Hence \(y\) is not a cluster point.

The original sequence does not converge. If it converged to \(y\), Proposition 1 would make \(y\) a cluster point, so \(y\) would have to be \(1\) or \(-1\). The neighborhood \((1/2,3/2)\) of \(1\) misses every odd-indexed term, and odd indices occur arbitrarily late. The neighborhood \((-3/2,-1/2)\) of \(-1\) misses every even-indexed term, and even indices occur arbitrarily late. Neither candidate can be the limit.

Define \(\phi(k)=2k\) for \(k\in\mathbb{N}\). This map is order-preserving and cofinal in \(\mathbb{N}\), and \(x_{\phi(k)}=x_{2k}=1\) for every \(k\). This subnet converges to \(1\). Likewise, \(\psi(k)=2k-1\) is order-preserving and cofinal, and \(x_{\psi(k)}=x_{2k-1}=-1\) for every \(k\), giving a subnet converging to \(-1\).

Check: the even indices satisfy \((-1)^{2k}=1\), and the odd indices satisfy \((-1)^{2k-1}=-1\). Thus the computed subnet values agree with the original formula.

A Cluster Point Is a Subnet Limit

Theorem: Let \((x_i)_{i\in I}\) be a net in a topological space \(X\). A point \(x\in X\) is a cluster point of the net if and only if some subnet of the net converges to \(x\).

Proof. First suppose a subnet \((x_{\phi(j)})_{j\in J}\) converges to \(x\). Let \(U\) be an open neighborhood of \(x\) and \(i_0\in I\). Cofinality gives \(j_0\in J\) with \(\phi(j_0)\ge i_0\). Convergence gives \(j_1\in J\) such that \(x_{\phi(j)}\in U\) for all \(j\ge j_1\). Choose \(j_2\ge j_0,j_1\), using directedness of \(J\). Then order preservation gives \(\phi(j_2)\ge\phi(j_0)\ge i_0\), and convergence gives \(x_{\phi(j_2)}\in U\). Thus every tail of the original net meets every neighborhood of \(x\), so \(x\) is a cluster point.

Conversely, suppose \(x\) is a cluster point. Let \(J\) consist of all pairs \((i,U)\) such that \(i\in I\), \(U\) is an open neighborhood of \(x\), and \(x_i\in U\). Give \(J\) the order \((i,U)\le(i',U')\) when \(i\le i'\) and \(U'\subseteq U\).

This set is nonempty: the cluster-point condition applied to the neighborhood \(X\) and any index provides a suitable pair. It is directed as well. Given \((i,U),(i',U')\in J\), the intersection \(V=U\cap U'\) is an open neighborhood of \(x\). Choose \(k_0\in I\) with \(k_0\ge i,i'\), using directedness of \(I\). By the cluster-point condition, there is \(k\ge k_0\) with \(x_k\in V\). Then \((k,V)\in J\) and it is an upper bound of both given pairs.

Define \(\phi:J\to I\) by \(\phi(i,U)=i\). The order definition makes \(\phi\) order-preserving. It is cofinal: for every \(i_0\in I\), the cluster-point condition applied to \(X\) gives an \(i\ge i_0\), and \((i,X)\in J\). Thus \((x_{\phi(i,U)})_{(i,U)\in J}\) is a subnet.

To check convergence, let \(W\) be any open neighborhood of \(x\). The cluster-point condition gives some \(i\) with \(x_i\in W\), so \((i,W)\in J\). Every \((i',U')\ge(i,W)\) satisfies \(U'\subseteq W\) and \(x_{i'}\in U'\). Therefore \(x_{\phi(i',U')}=x_{i'}\in W\) for every such pair. The subnet is eventually in every open neighborhood of \(x\), so it converges to \(x\). \(\square\)

The construction keeps two pieces of information in each subnet index: the original index \(i\), which ensures progress through the original net, and a neighborhood \(U\) of \(x\), which ensures that later selected terms stay within the chosen neighborhood. The cluster-point condition makes it possible to find an upper bound satisfying both requirements.

Worked Example: The Constant Net in the Sierpinski Space

Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and let \(x_n=a\) for every \(n\in\mathbb{N}\). The open neighborhoods of \(a\) are \(\{a\}\) and \(X\). The open neighborhoods of \(b\) consist only of \(X\).

The net converges to \(a\): every term lies in each neighborhood of \(a\). It also converges to \(b\): the only neighborhood of \(b\) is \(X\), and every term lies in \(X\). In particular, both \(a\) and \(b\) are cluster points. For \(a\), every tail contains \(a\), which lies in each neighborhood of \(a\); for \(b\), every tail contains \(a\in X\), the only neighborhood of \(b\).

The identity map on \(\mathbb{N}\) gives a subnet equal to the original net. That subnet converges to both \(a\) and \(b\), as required by the theorem. This example shows that a subnet limit, like a net limit, need not be unique in every topological space.

Check: the neighborhood list for \(b\) contains \(X\) and no smaller open set, so convergence and clustering at \(b\) follow directly from the definitions.

Reading the Two Conditions Correctly

For a net \((x_i)_{i\in I}\), convergence and clustering differ in the order of their quantifiers. Convergence asks that for each neighborhood there be a tail entirely inside it. Clustering asks that for each neighborhood and each tail, at least one term of that tail lies inside it. A cluster point can therefore be revisited arbitrarily far along the net without being approached in the stronger, eventual sense.

Common pitfall: Do not replace “every tail meets the neighborhood” with “the net is eventually in the neighborhood.” For the sequence \(x_n=(-1)^n\), every tail meets \((1/2,3/2)\), because every tail contains an even index. But the sequence is not eventually in this interval, because every tail also contains an odd index whose term is \(-1\). The neighborhood meets every tail, which establishes clustering at \(1\); it does not establish convergence to \(1\).

Subnets are useful because they extract a genuinely convergent net from the repeated visits that define a cluster point, even when no subsequence is available or convenient. The theorem also gives a test in the other direction: a point reached as a subnet limit must have been met in every tail of the original net. Both directions depend on the cofinal progress of the subnet indices; merely selecting some terms without ensuring that progress would not suffice.

Check Your Understanding

Use the definitions of tail, cluster point, subnet, and convergence to answer these questions.

For \(x_n=1\) when \(n\) is even and \(x_n=0\) when \(n\) is odd in \(\mathbb{R}\), find all cluster points and determine whether the sequence converges.
Explain why a subnet limit must be a cluster point of the original net. Identify where cofinality and order preservation are used.
For \(x_n=1/n\) in \(\mathbb{R}\), verify directly that \(0\) is a cluster point and give a subnet converging to \(0\).
In the Sierpinski space from the example, which open neighborhoods of \(a\) and \(b\) are used to test convergence of the constant net \(x_n=a\)?
State the difference between a net being eventually in a neighborhood and every tail meeting that neighborhood.

Subnets and Cluster Points of Nets

What You'll Learn