Why Nets Detect Hausdorffness
In a Hausdorff space, distinct points have disjoint open neighborhoods. This condition guarantees that a sequence cannot converge to two different points, but sequences do not capture all topological behavior in every space. Nets provide a broader notion of convergence. In fact, they detect Hausdorffness exactly: if a space is not Hausdorff, one can construct a net that converges to two distinct points.
Throughout, \((X,\mathcal{T})\) is a topological space. An open neighborhood of \(x\in X\) means an open set containing \(x\). Recall that \(X\) is Hausdorff if for every pair of distinct points \(x,y\in X\), there are open neighborhoods \(U\) of \(x\) and \(V\) of \(y\) with \(U\cap V=\varnothing\).
Directed Sets, Nets, and Convergence
The direction of the order describes which indices count as later. For example, the natural numbers with their usual order form a directed set, and a sequence is a net indexed by \(\mathbb{N}\). But a net can have a different directed set, allowing it to record convergence patterns that a sequence may not detect.
The phrase “at most one limit” permits a net to have no limit. It says that whenever \(x_d\to x\) and \(x_d\to y\), then \(x=y\).
Hausdorff Spaces Have Unique Net Limits
Proof. Let \((x_d)_{d\in D}\) be a net, and suppose it converges to both \(x\) and \(y\). 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 \(d_1\leq d\). Convergence to \(y\) gives an index \(d_2\) such that \(x_d\in V\) whenever \(d_2\leq d\). Directedness provides \(d_3\) with \(d_1\leq d_3\) and \(d_2\leq d_3\). Then \(x_{d_3}\in U\) and \(x_{d_3}\in V\), contradicting \(U\cap V=\varnothing\). Thus \(x=y\). \(\square\)
The proof uses directedness to find one index that is later than both convergence thresholds. Without that step, the two tails of the net might not be compared at a common index.
Worked Example: Unique Limits in the Real Line
In \(\mathbb{R}\) with its standard topology, verify directly that a net cannot converge to two distinct points. Suppose \(x_d\to a\) and \(x_d\to b\), where \(a<b\). Set \(m=(a+b)/2\). The open intervals \(U=(a-(b-a)/4,m)\) and \(V=(m,b+(b-a)/4)\) contain \(a\) and \(b\), respectively. They are disjoint: every point of \(U\) is less than \(m\), while every point of \(V\) is greater than \(m\).
Convergence to \(a\) makes the net eventually lie in \(U\), and convergence to \(b\) makes it eventually lie in \(V\). At a common later index, directedness would require a net value in both intervals, which is impossible. Hence \(a=b\). This direct calculation illustrates Theorem 1 in the real line.
Check: for \(a=0\) and \(b=1\), the intervals above are \(U=(-1/4,1/2)\) and \(V=(1/2,5/4)\). Each contains its designated point, and the first consists of numbers less than \(1/2\), while the second consists of numbers greater than \(1/2\); they are disjoint.
Building a Net with Two Limits
For the converse, suppose \(x\ne y\) cannot be separated by disjoint open neighborhoods. This means that every open neighborhood \(U\) of \(x\) intersects every open neighborhood \(V\) of \(y\). We use pairs of such neighborhoods to index a net, and put a point from their intersection at each index.
Proof. The forward implication is Theorem 1. For the reverse implication, we prove the contrapositive. Suppose \(X\) is not Hausdorff. Then there are distinct points \(x,y\in X\) such that every open neighborhood of \(x\) intersects every open neighborhood of \(y\).
Let \(D\) consist of triples \((U,V,z)\) such that \(U\) is an open neighborhood of \(x\), \(V\) is an open neighborhood of \(y\), and \(z\in U\cap V\). This set is nonempty: taking \(U=V=X\), any point \(z\in X\) gives a triple, and \(X\) is nonempty because it contains \(x\). Define an order on \(D\) by \((U_1,V_1,z_1)\leq(U_2,V_2,z_2)\) exactly when \(U_2\subseteq U_1\) and \(V_2\subseteq V_1\). The relation is reflexive and transitive. To check directedness, take two indices \((U_1,V_1,z_1)\) and \((U_2,V_2,z_2)\). The sets \(U_1\cap U_2\) and \(V_1\cap V_2\) are open neighborhoods of \(x\) and \(y\). By the assumed failure of separation, their intersection is nonempty. Choose \(z_3\in (U_1\cap U_2)\cap(V_1\cap V_2)\). Then \((U_1\cap U_2,V_1\cap V_2,z_3)\) is an index above both given indices. Thus \(D\) is directed.
Define a net by \(x_{(U,V,z)}=z\). To prove it converges to \(x\), let \(W\) be any open neighborhood of \(x\). The sets \(W\) and \(X\) are neighborhoods of \(x\) and \(y\), so \(W\cap X\ne\varnothing\). Choose \(z_0\in W\); then \(d_0=(W,X,z_0)\in D\). If \(d=(U,V,z)\geq d_0\), the order definition gives \(U\subseteq W\), and \(z\in U\). Hence \(x_d=z\in W\). This proves convergence to \(x\).
To prove convergence to \(y\), let \(W\) be any open neighborhood of \(y\). The neighborhoods \(X\) of \(x\) and \(W\) of \(y\) intersect, so choose \(z_0\in W\) and use \(e_0=(X,W,z_0)\in D\). Whenever \(e=(U,V,z)\geq e_0\), we have \(V\subseteq W\) and \(z\in V\), so \(x_e\in W\). Thus the same net converges to \(y\). Since \(x\ne y\), it has two distinct limits. Therefore, if every net has at most one limit, \(X\) must be Hausdorff. \(\square\)
This construction does not require choosing a point from every neighborhood intersection in advance: the points are included as part of the indices in \(D\). At each directedness check, only one nonempty intersection must be shown to contain a point.
Worked Example: Two Limits in an Indiscrete Space
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). The only open neighborhood of either point is \(X\), so \(a\) and \(b\) have no disjoint open neighborhoods. In particular, \(X\) is not Hausdorff.
Take \(D=\mathbb{N}\) with its usual order and define the constant net \(x_n=a\) for every \(n\). For either \(p=a\) or \(p=b\), the only open neighborhood of \(p\) is \(X\), and \(x_n\in X\) for every \(n\). Thus the net converges both to \(a\) and to \(b\). This is a simple instance of failure of unique net limits.
Check: the two limits are distinct because \(a\ne b\); convergence does not require the net values to equal the proposed limit, only to be eventually in every open neighborhood of it.
Worked Example: A Net with Many Limits in the Cofinite Topology
Give \(X=\mathbb{N}\) the cofinite topology: its open sets are \(\varnothing\) and all subsets whose complements in \(\mathbb{N}\) are finite. Define \(D\) to be the set of finite subsets \(F\subseteq\mathbb{N}\), ordered by inclusion. It is directed because \(F_1\cup F_2\) is finite and contains both \(F_1\) and \(F_2\). Define \(x_F\) to be the least natural number not in \(F\). Such a number exists because a finite set cannot contain all of \(\mathbb{N}\).
Fix any \(p\in\mathbb{N}\) and any open neighborhood \(W\) of \(p\). The complement \(C=\mathbb{N}\setminus W\) is finite. Set \(F_0=C\). Whenever \(F\geq F_0\), we have \(C\subseteq F\), while \(x_F\notin F\). Consequently \(x_F\notin C\), so \(x_F\in W\). This proves \(x_F\to p\). Since \(p\) was arbitrary, this one net converges to every point of \(X\), which confirms that this cofinite space is not Hausdorff.
Check: if \(F=\{1,2,3\}\), then \(x_F=4\); the value is outside \(F\), as required. For the neighborhood \(W=\mathbb{N}\setminus\{2,5\}\), every index \(F\) containing \(\{2,5\}\) has \(x_F\notin\{2,5\}\), hence \(x_F\in W\).
A Common Mistake: Overlapping Neighborhoods
To use Hausdorffness, the neighborhoods must actually be disjoint. Merely choosing two open intervals, one around each point, does not establish separation.
Worked Example: Correcting an Interval Check
Consider the proposed neighborhoods \((-1,1)\) of \(0\) and \((1/2,3/2)\) of \(1\) in the standard real line. A mistaken argument might say that every point of the first is less than \(1\) and every point of the second is greater than \(1/2\), so the intervals are disjoint. These inequalities do not imply disjointness: \(3/4\) satisfies both \(-1<3/4<1\) and \(1/2<3/4<3/2\). In fact their intersection is \((1/2,1)\).
For a valid separation, use \((-1/2,1/2)\) around \(0\) and \((1/2,3/2)\) around \(1\). The first interval contains only numbers less than \(1/2\), while the second contains only numbers greater than \(1/2\); neither contains \(1/2\), so their intersection is empty. These are disjoint open neighborhoods, as required.
Check: \(0\in(-1/2,1/2)\), \(1\in(1/2,3/2)\), and no real number can be both less than and greater than \(1/2\). Thus the corrected intervals meet all parts of the definition.
What the Characterization Gives
The two directions together make net convergence a precise test for Hausdorffness. In one direction, disjoint neighborhoods force two purported limits to coincide. In the other, failure of disjoint-neighborhood separation supplies enough intersecting neighborhoods to build a directed set and a net that eventually lies in every neighborhood of both points.
| Topological condition | Consequence for nets |
|---|---|
| \(X\) is Hausdorff | Every net in \(X\) has at most one limit |
| Some distinct \(x,y\) have no disjoint neighborhoods | There is a net converging to both \(x\) and \(y\) |
| Every net in \(X\) has at most one limit | \(X\) is Hausdorff |
A key distinction is that a net is not necessarily a sequence: its directed index set may encode neighborhood refinements that cannot be arranged as a sequence. The construction above uses pairs of neighborhoods, so it can detect a failure of Hausdorffness without any countability assumption.
Check Your Understanding
For each question, identify the relevant definition before deciding whether the proposed convergence or separation claim holds.
- In the proof that limits are unique, where is directedness used, and what common index does it provide?
- Suppose \(x\ne y\) have no disjoint open neighborhoods. Explain why \(U\cap V\ne\varnothing\) for every open neighborhood \(U\) of \(x\) and \(V\) of \(y\).
- In the cofinite topology on \(\mathbb{N}\), why does the net indexed by finite subsets eventually lie in every open neighborhood of a fixed point?
- Give a net with two distinct limits in a two-point indiscrete space, and verify convergence from the definition.
- Check whether \((-2,1)\) and \((0,2)\) are disjoint neighborhoods of \(0\) and \(1\) in \(\mathbb{R}\). If not, give a point in their intersection.