Uniform Spaces 1 - Nets

Pre-requisites: point-set topology, metric spaces.

Introduction

I recently became very interested in topological vector spaces, and this interest led me naturally to the concept of uniform spaces, since every topological vector space is a uniform space. Uniform spaces generalize topological vector spaces, metric spaces, topological groups, compact Hausdorff spaces, and more.

This is the first in a series of seven posts about the theory behind uniform spaces. In the next post, I will go deeper into what exactly a uniform space is and what they are used for, but this first post is focused on the idea of nets. At first, I was going to introduce nets in the context of uniform spaces, but the more I read about nets, the more I realized that they are actually a pretty cool tool for point-set topology. So, in this post, I will go over some of the intuition and basic theorems behind nets in a topological space.

The other posts in this series are:

1. Nets
2. Introduction to Uniform Spaces
3. The Completion of a Uniform Space
4. The Bolzano-Weierstrass Theorem
5. Countably Uniform Spaces

Sequences and Topology

The net a generalization of the sequence. Recall that a sequence is a function $a: \mathbb{N} \rightarrow X$, commonly written $a_n$. To emphasize that some function $a$ is a sequence (or net), we will write it as $a_\bullet$. When something is true of $a_n$ all sufficiently large $n$, we say that it holds eventually for $a_\bullet$.

A sequence $a_\bullet$ in a topological space $X$ is said to converge to a point $x \in X$ if for all open sets $U$ such that $x \in U$, the sequence $a_\bullet$ is eventually in $U$. In other words, if $x \in U$ and $U$ is open, then there exists $N \in \mathbb{N}$ such for all $n \geq N$, $a_n \in U$.

If this definition is new to you, it may be worth considering the more common case where $X$ is a metric space. If $(X, d)$ is a metric space, then $a_\bullet$ converges to $x$ (in the above sense) if and only if for all $\epsilon > 0$, $a_\bullet$ is eventually in the open ball $B_\epsilon(x)$. This leads to the following more common characterization of convergence in metric spaces.

Theorem (Topological Convergence of Sequences in Metric Spaces): A sequence $(a_n)$ in a metric space $(X, d)$ converges to $x \in X$ if and only if for all $\epsilon > 0$, $d(a_\bullet, x) < \epsilon$ eventually.

In fact, convergence of sequences completely defines the topology of a metric space, as shown by the following theorems. These theorems are left as an exercise to the interested reader.

Theorem (Closed Subsets of a Metric Space): Let $(X, d)$ be a metric space. A subset $A \subset X$ is closed if and only if whenever $a_\bullet$ is a sequence in $A$ and converges to $x$, then $x \in A$.

Theorem (Open Subsets of a Metric Space): Let $(X, d)$ be a metric space. A subset $A \subset X$ is open if and only if whenever $x_\bullet$ is a convergent sequence in $X$ and converges to $a \in A$, then $x_\bullet$ is eventually in $A$.

These important theorems are no longer true for general topological spaces. For example, consider $X = [0, 1]^{\mathbb{R}}$ with the product topology, viewed as the set of all real-valued functions on $[0, 1]$. Let $A$ be the set of all functions $f \in X$ such that $f(c) = 0$ for all except countably many $c \in [0, 1]$. $A$ is not closed in $X$. To see this, let $h(c) = 1$ for all $c \in [0,1]$. Clearly, $h \not\in A$. If $A$ were closed, then $X - A$ would be open, and would contain an open neighborhood of $h$. Every open neighborhood of $h$ contains on open neighborhood $U_{c_1 \dots c_k}$ of the form:

However, $U_{c_1 \dots c_k}$ contains $g$ where $g(c_i) = 1$ for all $c_i$, and $g(c) = 0$ for all other $c$. Clearly, $g \in A$. So, every open neighborhood of $h$ intersects with $A$. Thus, $A$ is not closed.

On the other hand, if $f_\bullet$ is a sequence in $A$ and $f_\bullet$ converges to $g \in X$, then $g \in A$. To see this, let $C_n = \{c : f(c) \neq 0\}$. Since $(f_n)$ converges to $g$, $g(c) = 0$ for all $c \not\in \bigcup_n C_n$. Since $\bigcup_n C_n$ is countable, $g$ lies in $A$.

Nets and Topology

To salvage these theorems, we introduce the concept of a net.

What is a Net?

A net is simply a function whose domain is a directed set.

Definition: A directed set $(D, \leq)$ is a set $D$ with a binary relation $\leq$ such that the following conditions hold.

1. (Transitive) For all $a, b, c \in D$, if $a \leq b$ and $b \leq c$, then $a \leq c$

2. (Reflexive) For all $a \in D$, $a \leq a$

3. (Upper Bounds) For all $a, b \in D$, there exists $c \in D$ such that $a \leq c$ and $b \leq c$.

4. (Non-Empty) $D \neq \varnothing$

Definition: A net in $X$ is a function $a: D \rightarrow X$ where $D$ is a directed set.

When the order is obvious, we often use the set $D$ to represent the directed set $(D, \leq)$. Some people require that a directed set also satisfies the following axiom, making every directed set into a poset.

5. (Anti-Symmetry) For all $a, b \in D$, if $a \leq b$ and $b \leq a$, then $a = b$.

Realistically, this axiom wouldn't impact the theory of nets at all, since we could replace every non-anti-symmetric net with an anti-symmetic net by identifying $a \sim b$ if $a \leq b$ and $b \leq a$. In my opinion, this axiom just makes the theory of nets slightly more complicated.

Most importantly, $\mathbb{N}$ is a directed set, so every sequence is also a net. In general, though, nets can be a lot stranger than sequences -- nets could be indexed by $\mathbb{R}$, $\mathbb{N}^3$, $\mathcal{P}(\mathbb{R})$, and more.

Convergence of Nets

Similar to sequences, a statement concerning a net $a_\bullet: D \rightarrow X$ is said to hold eventually if there is some $i \in D$ such that for all $j \geq i$, the statement holds for $a_j$. Importantly, if two statements ($T_1$, $T_2$) concerning $a_\bullet$ are both true eventually, then their conjunction ("$T_1$ and $T_2$") is also true eventually. All relevant properties of nets are properties that hold "eventually".

Definition: A net $a_\bullet$ in a topological space $X$ is said to converge to $x \in X$ if for all open sets $U \subset X$ such that $x \in U$, $a_\bullet \in U$ eventually.

Happily, our characterizations of closed and open sets in metric spaces also hold for general topological spaces if sequences are replaced with nets.

Theorem (Closed Subsets of a Topological Space): Let $X$ be a topological space. A subset $A \subset X$ is closed if and only if whenever $a_\bullet$ is a net in $A$ and converges to $x$, then $x \in A$.

Proof: Suppose that $A$ is closed, and let $a_\bullet$ be a net in $A$ which converges to some $x \in X - A$. As $X - A$ is an open set, $a_\bullet \in X - A$ eventually. This is a contradiction. Hence, every time a net in $A$ converges to $x$, we have $x \in A$.

Now suppose that $A$ is some subset of $X$ such that whenever a net in $A$ converges to $y$, $y \in A$. Let $x \in \overline{A}$.

Let $D = \{(U, b) : x \in U \text{ and } b \in A \cap U\}$ where $U$ is open. Define an order on $D$ by: $(U, b) \leq (U', b')$ if and only if $U \subset U'$. Note that if $U$ is open and $x \in U$, then there always exists some $b \in U \cap A$. So, if $(U, b), (U', b') \in D$, then there exists $c \in U \cap U' \cap A$, and $(U \cap U', c)$ is an upper bound on both $(U, b)$ and $(U', b')$. Thus, $D$ is a directed set.

Define the net $a_\bullet: D \rightarrow X$ by: $a_{(U, b)} = b$. We now prove that $a_\bullet$ converges to $x$. Whenever $(U, b) \leq (U', b')$, we have $a_{(U', b')} \in U' \subset U$. So, $a_\bullet$ is eventually in $U$ for any open $U$ containing $x$. Thus, $a_\bullet$ does converge to $x$.

As $x$ is the limit of a net in $A$, we have $x \in A$. Since $x$ was arbitrary, $A$ is closed.

$\square$

Theorem (Open Subsets of a Metric Space): Let $X$ be a topological space. A subset $A \subset X$ is open if and only if whenever $x_\bullet$ is a net in $X$ and converges to $a \in A$, $(x_\bullet)$ is eventually in $A$.

Proof: If $A$ is an open set and $x_\bullet$ converges to $a \in A$, then $x_\bullet$ is eventually in $A$ by the definition of convergence.

Now suppose that $A$ is a set such that whenever $x_\bullet$ is a net which converges to $a \in A$, $x_\bullet$ is eventually in $A$. Thus, whenever $y_\bullet \not\in A$ eventually and $y_\bullet$ converges to $x$, then $x \not\in A$. By the previous theorem, this implies that $X -A$ is closed. Thus, $A$ is open.

$\square$

Nets and Hausdorffness

It is common to say that $a$ is "the limit" of a net $x_\bullet$ if $x_\bullet$ converges to $a$. This terminology can be confusing because a given net may have more than one limit. For example, consider $X = \{x, y\}$ with the indiscrete topology. Consider the constant sequence $a_n = x$. Both $x$ and $y$ are limits of $a_\bullet$. This happens because $x$ and $y$ are indistinguishable under the topology of $X$.

As a slightly more complicated example, give $X$ the new topology

Although $x$ and $y$ are now topologically distinguishable, they are still both limits of $a_\bullet$. This is because $x$ and $y$ are not very strongly separated. Thankfully, everything works if the underlying topological space is Hausdorff. And, at the risk of angering some point-set topologists, all important topological spaces are Hausdorff.

Theorem (Unique Limits iff Hausdorff): A topological space $X$ is Hausdorff if and only if whenever $x_\bullet$ is a net, and $x_\bullet$ converges to both $a$ and $b$, then $a = b$.

Proof: Suppose that $X$ is Hausdorff, $x_\bullet$ is a net, and $x_\bullet$ converges to both $a$ and $b$. If $a \neq b$, then let $a \in U$ and $b \in V$ where $U, V$ are disjoint open subsets of $X$. $x_\bullet \in U$ eventually and $x_\bullet \in V$ eventually. Thus, $x_\bullet \in U \cap V$ eventually. This is impossible, so $X$ has unique limits.

Suppose that $X$ has unique limits, but is not Hausdorff. Let $a, b \in X$ such that $a$ and $b$ can not be separated by disjoint neighborhoods. Let

for $U$ and $V$ open sets. Give $D$ the order: $(U, V, q) \leq (U', V', q')$ if and only if $U \subset U'$ and $V \subset V'$. Note that whenever $a \in U$ and $b \in V$, there exists some $q \in U \cap V$. From this observation, it can be seen that $D$ is a directed set.

Define the net $x_\bullet: D \rightarrow X$ by $x_{(U, V, q)} = q$. If $U$ is open and $a \in U$, then $x_\bullet \in U$ eventually. Thus, $x_\bullet$ converges to $a$. Similarly, $x_\bullet$ converges to $b$. This is a contradiction. Thus, $X$ must be Hausdorff.

$\square$

Nets and Continuity

In metric spaces, there is a strong relationship between continuous functions and convergent sequences. This relationship is codified in the following theorem.

Theorem: Let $X$ anad $Y$ be metric spaces, and let $f: X \rightarrow Y$ be any function. $f$ is continuous if and only if whenever a sequence $a_\bullet$ in $X$ converges to $b$, the sequence $f(a_\bullet)$ in $Y$ converges to $f(b)$.

More generally, nets can be used to determine when a function between topological spaces is continuous. This comes in handy, for example, when proving that functions involving the completion of a uniform space are continuous. For now, it serves as another example of how nets generalize intuitive ideas from metric spaces to more general topological spaces.

Theroem (Continuous iff Preserves Limits): Let $X$ and $Y$ be topological spaces, and let $f: X \rightarrow Y$ be any function. $f$ is continuous if and only if whenever a net $a_\bullet$ in $X$ converges to $x$, the net $f(a_\bullet)$ in $Y$ converges to $f(x)$.

Proof: Let $f$ be a continuous function. Let $a_\bullet$ be a net in $X$ which converges to $x \in X$. For all open subsets $V \subset Y$ containing $f(x)$, $f^{-1}(V)$ is an open subset of $X$ containing $x$. Thus, $a_\bullet \in f^{-1}(V)$ eventually. Hence, $f(a_\bullet) \in V$ eventually. Since $V$ was arbitrary, $f(a_\bullet)$ converges to $f(a)$.

Now, suppose that $f$ preserves the limits of nets. Let $V \subset Y$ be an arbitrary open set. To prove that $f$ is continuous, it suffices to prove that $f^{-1}(V)$ is open. Hence, it suffices to prove that for all $x \in X$ and all nets $a_\bullet$ which converge to $x$, eventually $a_\bullet \in f^{-1}(V)$. Note that $f(a_\bullet)$ converges to $f(x)$, so eventually $f(a_\bullet) \in V$. Thus, eventually $a_\bullet \in f^{-1}(V)$.

$\square$

Nets and Compactness

There is a well-known characterization of when a metric spaces is compact that relies only upon the topology of the metric space.

Theorem: A metric space $(X, d)$ is compact if and only if every arbitrary sequence in $X$ has a convergent subsequence.

Unfortunately, but on-theme for this post, this theorem does not generalize to all topological spaces. The property that every sequence has a convergent subsequence is called sequential compactness, and every compact space is sequentially compact.

But there exist sequentially compact spaces that are not compact. For example, once again take $X = [0,1]^{\mathbb{R}}$ viewed as the real-valued functions on $[0,1]$, and let $A \subset X$ to be the set of all functions that are zero except on a countable subset of $[0,1]$. Since $X$ is Hausdorff and $A$ is not a closed subset of $X$, $A$ is not compact. Since $X$ is compact, every sequence in $X$ has a convergent subsequence. So, every sequence in $A$ has a subsequence which converges in $X$. Its limit is in $A$. Thus, $A$ is sequentially compact but not compact.

To characterize compactness for all topological space, we need to consider nets instead of sequences. It turns out that a topological space is compact if and only if every net has a convergent subnet. To properly state and prove this theorem, we first have to get acquainted with subnets.

Definition: Given a net $a_\bullet: D \rightarrow X$, a subnet is a net $b_\bullet: E \rightarrow X$ function $f: E \rightarrow D$ such that there exists a function $f: E \rightarrow D$ satisfying:

1. For all $i \in E$, $b_i = a_{f(i)}$

2. For all $i, j \in E$, if $i \leq j$, then $f(i) \leq f(j)$.

3. For all $k \in D$, there exists $i \in E$ such that $d \leq f(a)$.

Every subsequence is actually a subnet in disguise. A subsequence of a sequence $a_\bullet$ is can be described as a sequence $b_\bullet$ for which there exists a strictly increasing function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $b_n = a_{f(m)}$. This satisfies the definition of a subnet.

However, subnets can be much more general. For one thing, the domain of a net may not be the same (or even contain) the domains of its subnets. But, subnets and their parent nets are related by one very important property: if some statement is true eventually for $a_\bullet$, and $b_\bullet$ is a subnet of $a_\bullet$, then it is also true eventually for $b_\bullet$.

Lemma: If $a_\bullet$ converges to $x$ and $b_\bullet$ is a subnet of $a_\bullet$, then $b_\bullet$ also converges to $x$.

Proof: If $a_\bullet$ converges to $x$, then for all open subset $U$ containing $x$, $a_\bullet \in U$ eventually. Thus, $b_\bullet \in U$ eventually. Thus, $b_\bullet$ converges to $x$.

$\square$

Subnets are a more complicated idea then subsequences, and can be hard to get a handle on at first. Thankfully, there is an easy criterion that tells us when convergent subnets exist.

Definition: A cluster point of a net $a_\bullet: D \rightarrow X$ is a point $x \in X$ such that for all open sets $U$ such that $x \in U$, and all $i \in D$, there exists $j \geq i$ such that $a_j \in U$.

To simplify this definition, say that some statement about a net $a_\bullet$ is true infinitely often if for all $i$, there exists $j \geq i$ such that the statement is true for $a_j$. This (slightly confusing) terminology is borrowed from sequences, and is not the same as being true for infinitely many $a_j$ terms. Using this terminology, $x$ is a cluster point of $a_\bullet$ if and only if for all neighborhoods $U$ of $x$, $a_\bullet \in U$ infinite often. Importantly, if $b_\bullet$ is a subnet of $a_\bullet$ and some statement is true infinitely often for $b_\bullet$, then it is also true infinite often for $a_\bullet$.

Lemma (Cluster Point iff Convergent Subnet): Let $a_\bullet$ be a net in $X$. $x \in X$ is a cluster point of $U$ if and only if there exists a subnet of $a_\bullet$ that converges to $x$.

Proof: Suppose that $b_\bullet$ is a subnet of $a_\bullet$ that converges to $x$. Let $a_\bullet$ be a net in $X$. If $U$ is an open set containing $x$, then $b_\bullet \in U$ eventually. Thus, $b_\bullet \in U$ infinitely often. Thus, $a_\bullet \in U$ infinitely often. Thus, $x$ is a cluster point of $a_\bullet$.

On the other hand, suppose that $x$ is a cluster point of $a_\bullet: D \rightarrow X$. Define $E = \{(U, j) : x\in U, j \in D, a_j \in U\}$ where $U$ is an open neighborhood of $x$. Give $E$ the ordering $(U, j) \leq (U', j')$ if $U' \subset U$. It is straightforward to show that $E$ is a directed set. Define the net $b_\bullet: E \rightarrow X$ by $b_{(U, j)} = a_j$.

It is again straightforward to show that $b_\bullet$ converges to $x$. Also, $b_\bullet$ is a subnet of $a_\bullet$ with the function $f: E \rightarrow D$ defined by $f((U, j)) = j$. Thus, a subnet of $a_\bullet$ converges to $b_\bullet$.

$\square$

Now that we know a little bit about subnets and cluster points, we can successfully characterize compactness in the language of nets.

Theorem (Convergent Subnets iff Compact): Let $X$ be a topological space. $X$ is compact if and only if every net in $X$ has a convergent subnet if and only if every net in $X$ has a cluster point.

Proof: It suffices to prove that $X$ is compact if and only if every net in $X$ has a cluster point.

Let $X$ be a compact topological space. Suppose for contradiction that $a_\bullet$ is a net in $X$ that has no cluster point. Then, for every $x \in X$, there exists an open neighborhood $U$ of $x$ such that $a_\bullet \not\in U$ eventually. Let $\mathcal{C}$ be the collection of all open sets $U$ such that $a_\bullet$ is eventually not in $U$. We have just shown that $\mathcal{C}$ is an open cover of $X$. Since $X$ is compact, $\mathcal{C}$ has a finite subcover $(U_1, U_2 \dots U_k)$. $a_\bullet$ is eventually not in any of these open sets, so $a_\bullet$ is eventually not in $X$. This is absurd. Thus, every net in $X$ must have a cluster point.

Let $X$ be topological space such that every net has a convergent subnet. Suppose for contradiction that $X$ is not compact. Let $\mathcal{C} = (U_\alpha)_{\alpha \in I}$ be an open cover of $X$ with no finite subcover. Let $F$ be the set of all finite subsets of $I$. By assumption, for all $J \in F$, there exists $q \not\in \bigcup_{\alpha \in J} U_\alpha$. Define $D = \{(J, q)\}$ where $J \in F$ and $q \not\in \bigcup_{\alpha \in J} U_\alpha$. Give $D$ the ordering $(A, q) \leq (A', q')$ if $A \subset A'$. It is straightforward to show that $D$. Define the net $a_\bullet: D \rightarrow X$ by $a_{(A, q)} = q$.

The net $a_\bullet$ has a cluster point $x$. Let $\alpha \in I$ such that $x \in U_\alpha$. Eventually, $a_\bullet$ is not in $U_\alpha$. This is a contradiction. Thus, $X$ must be compact.

$\square$

I'll end with a true application of nets to topology: a short proof of Tychonoff's Theorem. This is a true gem of a proof. As Tychonoff's Theorem is actually equivalent to the Axiom of Choice, it is impossible to get around invoking the axiom of choice.

Theorem (Tychonoff's Theorem): [Assuming choice] Let $(X_\alpha)_{\alpha \in I}$ be a collection of compact topological spaces. Then, the product $\prod_{\alpha \in I} X_\alpha$ is also compact.

Proof: Let $X = \prod_{\alpha \in I} X_\alpha$ and let $\pi_\alpha: X \rightarrow X_\alpha$ be the projection onto $X_\alpha$. To prove that $X$ is compact, it suffices to prove that an arbitrary net $x_\bullet: D \rightarrow X$ has a cluster point. For each $\alpha$, choose a cluster point $a_\alpha$ of $\pi_\alpha(x_\bullet)$. Let $a \in X$ such that $\pi_\alpha(a) = a_\alpha$.

To prove that $a$ is a cluster point of $x_\bullet$, it suffices to prove that $x_\bullet \in \pi_\alpha^{-1}(W)$ infinitely often for each $\alpha$ and each neighborhood $W$ of $a_\alpha$. This is true because $\pi_\alpha(x_\bullet) \in W$ infinite often.

$\square$

Extensions: A sequential space is precisely a space whose topology is defined by its notion of convergent sequences (such as metric spaces and any first-countable space). There is also another tool, filters, which can be used instead of nets when studying uniform spaces. I like nets better because nets are more like sequences.