3.8Nets

Updated 23 Aug 2026Chapter (PDF)

[3.8.1]Definition(Partially Ordered Set, Directed Set)#

A partially ordered set is a set $A$ together with a partial order $\leq$ satisfying the following conditions.

Reflexivity. $a \le a$ for every $a \in A$ .
Symmetry. For every pair $a, b \in A$ , if $a \le b$ and $b \le a$ , then $a = b$ .
Transitivty. For every triple $a, b, c \in A$ , if $a \le b$ and $b \le c$ , then $a \le c$ .

A directed set is some partially ordered set $\Lambda$ such that for any pair $a, b \in \Lambda$ , there is some $c \in \Lambda$ such that $a \le c$ and $b \le c$ . That is, any pair of elements in $\Lambda$ have an upper bound in the set.

[3.8.2]Definition(Cofinal Sets)#

A subset $K$ of a directed set $\Lambda$ is cofinal if, for any $a \in \Lambda$ , there is some $b \in K$ such that $a \le b$ .

[3.8.3]Remark(Intuition behind Cofinality)#

Cofinality is precisely the algebraic translation of “marching off to infinity.” For a standard sequence indexed by $\bZ_+$ , a subset of indices goes to infinity because it eventually exceeds any given integer $N$ . For an arbitrary directed set $\Lambda$ , a subset is cofinal if it eventually exceeds any given element $\alpha \in \Lambda$ .

[3.8.4]Theorem#

Cofinal subsets of a directed set are directed sets.

Proof.

Let $K$ be a cofinal subset of the directed set $\Lambda$ . For any $a, b \in K$ , note there is some $c \in \Lambda$ such that $a \le c$ and $b \le c$ . Since $K$ is cofinal, there is some $k \in K$ such that $c \le k$ . Thus $a, b \le k$ , and so $K$ is a directed set.

[3.8.5]Definition(Net)#

Let $X$ be a topological space. A net in $X$ is a tuple of $X$ indexed by $\Lambda$ . That is, a net $\vb{x}$ is a map $f: \Lambda \to X$ such that $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ , where $x_\lambda = f(\lambda)$ for each $\lambda \in \Lambda$ .

[3.8.6]Definition(Convergence of Net)#

A net $(x_\lambda)_{\lambda \in \Lambda}$ is said to converge to a point $x$ if, for any open neighborhood $U$ of $x$ , there is some $\lambda \in \Lambda$ such that $x_{\lambda'} \in U$ for every $\lambda' \ge \lambda$ .

Many theorems about sequences can be lifted to nets in a completely identical way. We repeat them here anyways.

[3.8.7]Theorem(Hausdorff Guarantees Unique Limit Point)#

Suppose $X$ is a Hausdorff space. Then a convergent net $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ has a unique limit.

Proof.

For the sake of contradiction, suppose $\vb{x} \to x$ and $\vb{x} \to y$ for $x \ne y \in X$ . Construct disjoint open neighborhoods $U_x$ and $U_y$ of $x$ and $y$ respectively. Then note the tail of $\vb{x}$ cannot lie in $U_x$ and $U_y$ simultaneously, since they are disjoint, which is a contradiction. Thus nets converge to, at most, one value.

[3.8.8]Theorem(Nets Probe Elements of Closure)#

Let $X$ be a topological space with $A \subseteq X$ . Then $x \in \bar{A}$ if and only if a net of points from $A$ converges to $x$ .

Proof.

$(\Longrightarrow)$ : Suppose $x \in \bar{A}$ . Let $\Lambda$ be the collection of all open neighborhoods of $x$ , directed by reverse inclusion so that $U_\lambda \subseteq U_{\lambda'}$ for every $\lambda \ge \lambda'$ (where $U_\lambda = \lambda$ ). Since $x \in \bar{A}$ , there is some $x_\lambda \in U_\lambda \cap A$ for each $\lambda \in \Lambda$ . Define the net $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ . For any open neighborhood $U$ of $x$ , note that there is some $\lambda \in \Lambda$ such that $U_\lambda \subseteq U$ (specifically, choosing $\lambda = U$ ). Thus, for each $\lambda' \ge \lambda$ , we have that $x_{\lambda'} \in U_{\lambda'} \subseteq U$ , meaning the net $\vb{x}$ converges to $x$ .

$(\Longleftarrow)$ : Suppose $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ is a net of elements in $A$ that converges to $x \in X$ . For any open neighborhood $U$ of $x$ , note there is some $\lambda \in \Lambda$ such that $x_\lambda \in U$ . Since $x_\lambda \in A$ , $U$ intersects $A$ , meaning $x \in \bar{A}$ .

[3.8.9]Theorem(Nets Probe Continuity)#

Let $f: X \to Y$ . Then $f$ is continuous if and only if, for every convergent net $\vb{x} = (x_\lambda)_{\lambda \in \Lambda} \to x$ , the net $f(x_\lambda) \to f(x)$ .

Proof.

$(\Longrightarrow)$ : Suppose $f$ is continuous. Let $V$ be any open neighborhood of $f(x)$ . By continuity, note that $U = f^{-1}(V)$ is an open neighborhood of $x$ such that $f(U) \subseteq V$ . Then there is some $\lambda' \in \Lambda$ such that $x_\lambda \in U$ for each $\lambda \ge \lambda'$ . Thus $f(x_\lambda) \in f(U) \subseteq V$ for every $\lambda \ge \lambda'$ , meaning $f(x_\lambda) \to f(x)$ .

$(\Longleftarrow)$ : Let $A$ be any subset of $X$ . For any $x \in \bar{A}$ , note there is some convergent net $\vb{x} = (x_\lambda)_{\lambda \in \Lambda}$ drawing values from $A$ such that $\vb{x} \to x$ . By hypothesis, note $f(\vb{x}) \to f(x) \in \bar{f(A)}$ . Thus $f(\bar{A}) \subseteq \bar{f(A)}$ , completing the proof.

Nets seem nice so far, but the true problem is that subsequences don't generalize well. If we generalized subsequences naively by just considering a subset of indices, subnets would be too weak to do anything.

[3.8.10]Remark(The Failure of Naive Subnets)#

Suppose we tried to define a subnet by restricting $f$ to a cofinal subset $J' \subseteq J$ , in exact analogy with passing to a subsequence. Taking $J = \mathbb{Z}_+$ , this recovers precisely the notion of a subsequence, so any theorem we prove about such “subnets” would specialize to a theorem about subsequences. But we already know that subsequences fail to detect compactness: the space $[0,1]^{[0,1]}$ is compact by Tychonoff, and yet it contains a sequence with no convergent subsequence. Under the naive definition, then, this compact space would admit a net with no convergent subnet, and the characterization of compactness we are after would be false.

The obstruction is one of size. To converge in a product, a net must eventually settle in every coordinate at once, and here there are uncountably many coordinates to satisfy simultaneously. A cofinal subset of $\mathbb{Z}_+$ is still countable, so no matter how cleverly we thin the indices, we have only countably many choices with which to meet uncountably many demands. Thinning cannot help us.

The repair, then, is not to select fewer indices but to permit more. By allowing an entirely new directed set $K$ together with a map $g: K \to J$ , we free ourselves from the size of $J$ altogether: $K$ may be far larger than $J$ , and $g$ need not be injective, so the subnet is permitted to revisit indices and to be indexed by a set of much greater cardinality than the original net. What remains is to insist that $K$ still traverses $J$ faithfully, and this is exactly what monotonicity and cofinality accomplish.

[3.8.11]Definition(Subnet)#

Let $f: J \to X$ be a net relative to some directed set $J$ . If $K$ is a directed set together with some map $g: K \to J$ such that

For any $i, j \in K$ such that $i \le j$ , we have $g(i) \le g(j)$ (monotonicity),
$g(K)$ is cofinal in $J$ ,

then we say the map $f \circ g: K \to X$ is a subnet of $f$ .

These properties allow the subnet $f \circ g$ relative to $K$ to lift properties of subsequences to subnets: monotonicity ensures we do not jump backward in the net, and cofinality ensures the subnet marches arbitrarily far out into the original directed set $J$ .

[3.8.12]Theorem(Convergence of Subnets)#

If a net converges to a point, then all subnets must converge to the same point.

[3.8.13]Definition(Accumulation Point)#

Let $(x_\lambda)_{\lambda \in \Lambda}$ be a net in $X$ . We say that $x$ is an accumulation point of the net if, for any open neighborhood $U$ of $x$ , the set of indices $\lambda$ such that $x_\lambda \in U$ is cofinal in $\Lambda$ .

[3.8.14]Remark(Cofinality and Accumulation Points)#

For sequences, an accumulation point is one that the sequence visits “infinitely often.” Cofinality is the exact topological translation of “infinitely often” for nets. If $x$ is an accumulation point, then no matter how far out we march into the directed set $\Lambda$ , we can always find an index further out where the net jumps back into $U$ . Because this set of “good” indices is cofinal, we can build a new directed set $K$ that maps entirely into them, creating a subnet that stays inside $U$ permanently (i.e., a subnet that converges to $x$ ).

[3.8.15]Theorem(Accumulation Points are Limit Points of Subnet)#

A net has an accumulation point if and only if a subnet converges to the accumulation point.

[3.8.16]Theorem(Equivalence of Compactness and Sequential Compactness via Nets)#

A space $X$ is compact if and only if every net in $X$ has a convergent subnet.

The aforementioned theorem says that compact spaces are sequentially compact relative to nets (i.e., if every net has an accumulation point). This closely mirrors the equivalence of compactness and sequential compactness in metric spaces.