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LIAM AXON

Uniform Spaces 1 - Nets

Dec 6, 2021
tags: math, topology

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:NXa: \mathbb{N} \rightarrow X, commonly written ana_n. To emphasize that some function aa is a sequence (or net), we will write it as aa_\bullet. When something is true of ana_n all sufficiently large nn, we say that it holds eventually for aa_\bullet.

A sequence aa_\bullet in a topological space XX is said to converge to a point xXx \in X if for all open sets UU such that xUx \in U, the sequence aa_\bullet is eventually in UU. In other words, if xUx \in U and UU is open, then there exists NNN \in \mathbb{N} such for all nNn \geq N, anUa_n \in U.

If this definition is new to you, it may be worth considering the more common case where XX is a metric space. If (X,d)(X, d) is a metric space, then aa_\bullet converges to xx (in the above sense) if and only if for all ϵ>0\epsilon > 0, aa_\bullet is eventually in the open ball Bϵ(x)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 (an)(a_n) in a metric space (X,d)(X, d) converges to xXx \in X if and only if for all ϵ>0\epsilon > 0, d(a,x)<ϵd(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)(X, d) be a metric space. A subset AXA \subset X is closed if and only if whenever aa_\bullet is a sequence in AA and converges to xx, then xAx \in A.

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

These important theorems are no longer true for general topological spaces. For example, consider X=[0,1]RX = [0, 1]^{\mathbb{R}} with the product topology, viewed as the set of all real-valued functions on [0,1][0, 1]. Let AA be the set of all functions fXf \in X such that f(c)=0f(c) = 0 for all except countably many c[0,1]c \in [0, 1]. AA is not closed in XX. To see this, let h(c)=1h(c) = 1 for all c[0,1]c \in [0,1]. Clearly, h∉Ah \not\in A. If AA were closed, then XAX - A would be open, and would contain an open neighborhood of hh. Every open neighborhood of hh contains on open neighborhood Uc1ckU_{c_1 \dots c_k} of the form:

Uc1ck={fX:f(ci)=h(ci),i=1k}U_{c_1 \dots c_k} = \{f \in X : f(c_i) = h(c_i), i = 1 \dots k \}

However, Uc1ckU_{c_1 \dots c_k} contains gg where g(ci)=1g(c_i) = 1 for all cic_i, and g(c)=0g(c) = 0 for all other cc. Clearly, gAg \in A. So, every open neighborhood of hh intersects with AA. Thus, AA is not closed.

On the other hand, if ff_\bullet is a sequence in AA and ff_\bullet converges to gXg \in X, then gAg \in A. To see this, let Cn={c:f(c)0}C_n = \{c : f(c) \neq 0\}. Since (fn)(f_n) converges to gg, g(c)=0g(c) = 0 for all c∉nCnc \not\in \bigcup_n C_n. Since nCn\bigcup_n C_n is countable, gg lies in AA.

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,)(D, \leq) is a set DD with a binary relation \leq such that the following conditions hold.

  1. (Transitive) For all a,b,cDa, b, c \in D, if aba \leq b and bcb \leq c, then aca \leq c

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

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

  4. (Non-Empty) DD \neq \varnothing

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

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

  1. (Anti-Symmetry) For all a,bDa, b \in D, if aba \leq b and bab \leq a, then a=ba = 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 aba \sim b if aba \leq b and bab \leq a. In my opinion, this axiom just makes the theory of nets slightly more complicated.

Most importantly, N\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 R\mathbb{R}, N3\mathbb{N}^3, P(R)\mathcal{P}(\mathbb{R}), and more.

Convergence of Nets

Similar to sequences, a statement concerning a net a:DXa_\bullet: D \rightarrow X is said to hold eventually if there is some iDi \in D such that for all jij \geq i, the statement holds for aja_j. Importantly, if two statements (T1T_1, T2T_2) concerning aa_\bullet are both true eventually, then their conjunction ("T1T_1 and T2T_2") is also true eventually. All relevant properties of nets are properties that hold "eventually".

Definition: A net aa_\bullet in a topological space XX is said to converge to xXx \in X if for all open sets UXU \subset X such that xUx \in U, aUa_\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 XX be a topological space. A subset AXA \subset X is closed if and only if whenever aa_\bullet is a net in AA and converges to xx, then xAx \in A.

Proof: Suppose that AA is closed, and let aa_\bullet be a net in AA which converges to some xXAx \in X - A. As XAX - A is an open set, aXAa_\bullet \in X - A eventually. This is a contradiction. Hence, every time a net in AA converges to xx, we have xAx \in A.

Now suppose that AA is some subset of XX such that whenever a net in AA converges to yy, yAy \in A. Let xAx \in \overline{A}.

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

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

As xx is the limit of a net in AA, we have xAx \in A. Since xx was arbitrary, AA is closed.

\square

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

Proof: If AA is an open set and xx_\bullet converges to aAa \in A, then xx_\bullet is eventually in AA by the definition of convergence.

Now suppose that AA is a set such that whenever xx_\bullet is a net which converges to aAa \in A, xx_\bullet is eventually in AA. Thus, whenever y∉Ay_\bullet \not\in A eventually and yy_\bullet converges to xx, then x∉Ax \not\in A. By the previous theorem, this implies that XAX -A is closed. Thus, AA is open.

\square

Nets and Hausdorffness

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

As a slightly more complicated example, give XX the new topology

{,{y},{x,y}}\{\varnothing, \{y\}, \{x, y\}\}

Although xx and yy are now topologically distinguishable, they are still both limits of aa_\bullet. This is because xx and yy 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 XX is Hausdorff if and only if whenever xx_\bullet is a net, and xx_\bullet converges to both aa and bb, then a=ba = b.

Proof: Suppose that XX is Hausdorff, xx_\bullet is a net, and xx_\bullet converges to both aa and bb. If aba \neq b, then let aUa \in U and bVb \in V where U,VU, V are disjoint open subsets of XX. xUx_\bullet \in U eventually and xVx_\bullet \in V eventually. Thus, xUVx_\bullet \in U \cap V eventually. This is impossible, so XX has unique limits.

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

D={(U,V,q):aUbV, and qUV}D = \{(U, V, q) : a \in U \text{, } b \in V \text{, and } q \in U \cap V \}

for UU and VV open sets. Give DD the order: (U,V,q)(U,V,q)(U, V, q) \leq (U', V', q') if and only if UUU \subset U' and VVV \subset V'. Note that whenever aUa \in U and bVb \in V, there exists some qUVq \in U \cap V. From this observation, it can be seen that DD is a directed set.

Define the net x:DXx_\bullet: D \rightarrow X by x(U,V,q)=qx_{(U, V, q)} = q. If UU is open and aUa \in U, then xUx_\bullet \in U eventually. Thus, xx_\bullet converges to aa. Similarly, xx_\bullet converges to bb. This is a contradiction. Thus, XX 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 XX anad YY be metric spaces, and let f:XYf: X \rightarrow Y be any function. ff is continuous if and only if whenever a sequence aa_\bullet in XX converges to bb, the sequence f(a)f(a_\bullet) in YY converges to f(b)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 XX and YY be topological spaces, and let f:XYf: X \rightarrow Y be any function. ff is continuous if and only if whenever a net aa_\bullet in XX converges to xx, the net f(a)f(a_\bullet) in YY converges to f(x)f(x).

Proof: Let ff be a continuous function. Let aa_\bullet be a net in XX which converges to xXx \in X. For all open subsets VYV \subset Y containing f(x)f(x), f1(V)f^{-1}(V) is an open subset of XX containing xx. Thus, af1(V)a_\bullet \in f^{-1}(V) eventually. Hence, f(a)Vf(a_\bullet) \in V eventually. Since VV was arbitrary, f(a)f(a_\bullet) converges to f(a)f(a).

Now, suppose that ff preserves the limits of nets. Let VYV \subset Y be an arbitrary open set. To prove that ff is continuous, it suffices to prove that f1(V)f^{-1}(V) is open. Hence, it suffices to prove that for all xXx \in X and all nets aa_\bullet which converge to xx, eventually af1(V)a_\bullet \in f^{-1}(V). Note that f(a)f(a_\bullet) converges to f(x)f(x), so eventually f(a)Vf(a_\bullet) \in V. Thus, eventually af1(V)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)(X, d) is compact if and only if every arbitrary sequence in XX 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]RX = [0,1]^{\mathbb{R}} viewed as the real-valued functions on [0,1][0,1], and let AXA \subset X to be the set of all functions that are zero except on a countable subset of [0,1][0,1]. Since XX is Hausdorff and AA is not a closed subset of XX, AA is not compact. Since XX is compact, every sequence in XX has a convergent subsequence. So, every sequence in AA has a subsequence which converges in XX. Its limit is in AA. Thus, AA 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:DXa_\bullet: D \rightarrow X, a subnet is a net b:EXb_\bullet: E \rightarrow X function f:EDf: E \rightarrow D such that there exists a function f:EDf: E \rightarrow D satisfying:

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

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

  3. For all kDk \in D, there exists iEi \in E such that df(a)d \leq f(a).

Every subsequence is actually a subnet in disguise. A subsequence of a sequence aa_\bullet is can be described as a sequence bb_\bullet for which there exists a strictly increasing function f:NNf: \mathbb{N} \rightarrow \mathbb{N} such that bn=af(m)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 aa_\bullet, and bb_\bullet is a subnet of aa_\bullet, then it is also true eventually for bb_\bullet.

Lemma: If aa_\bullet converges to xx and bb_\bullet is a subnet of aa_\bullet, then bb_\bullet also converges to xx.

Proof: If aa_\bullet converges to xx, then for all open subset UU containing xx, aUa_\bullet \in U eventually. Thus, bUb_\bullet \in U eventually. Thus, bb_\bullet converges to xx.

\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:DXa_\bullet: D \rightarrow X is a point xXx \in X such that for all open sets UU such that xUx \in U, and all iDi \in D, there exists jij \geq i such that ajUa_j \in U.

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

Lemma (Cluster Point iff Convergent Subnet): Let aa_\bullet be a net in XX. xXx \in X is a cluster point of UU if and only if there exists a subnet of aa_\bullet that converges to xx.

Proof: Suppose that bb_\bullet is a subnet of aa_\bullet that converges to xx. Let aa_\bullet be a net in XX. If UU is an open set containing xx, then bUb_\bullet \in U eventually. Thus, bUb_\bullet \in U infinitely often. Thus, aUa_\bullet \in U infinitely often. Thus, xx is a cluster point of aa_\bullet.

On the other hand, suppose that xx is a cluster point of a:DXa_\bullet: D \rightarrow X. Define E={(U,j):xU,jD,ajU}E = \{(U, j) : x\in U, j \in D, a_j \in U\} where UU is an open neighborhood of xx. Give EE the ordering (U,j)(U,j)(U, j) \leq (U', j') if UUU' \subset U. It is straightforward to show that EE is a directed set. Define the net b:EXb_\bullet: E \rightarrow X by b(U,j)=ajb_{(U, j)} = a_j.

It is again straightforward to show that bb_\bullet converges to xx. Also, bb_\bullet is a subnet of aa_\bullet with the function f:EDf: E \rightarrow D defined by f((U,j))=jf((U, j)) = j. Thus, a subnet of aa_\bullet converges to bb_\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 XX be a topological space. XX is compact if and only if every net in XX has a convergent subnet if and only if every net in XX has a cluster point.

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

Let XX be a compact topological space. Suppose for contradiction that aa_\bullet is a net in XX that has no cluster point. Then, for every xXx \in X, there exists an open neighborhood UU of xx such that a∉Ua_\bullet \not\in U eventually. Let C\mathcal{C} be the collection of all open sets UU such that aa_\bullet is eventually not in UU. We have just shown that C\mathcal{C} is an open cover of XX. Since XX is compact, C\mathcal{C} has a finite subcover (U1,U2Uk)(U_1, U_2 \dots U_k). aa_\bullet is eventually not in any of these open sets, so aa_\bullet is eventually not in XX. This is absurd. Thus, every net in XX must have a cluster point.

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

The net aa_\bullet has a cluster point xx. Let αI\alpha \in I such that xUαx \in U_\alpha. Eventually, aa_\bullet is not in UαU_\alpha. This is a contradiction. Thus, XX 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 (Xα)αI(X_\alpha)_{\alpha \in I} be a collection of compact topological spaces. Then, the product αIXα\prod_{\alpha \in I} X_\alpha is also compact.

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

To prove that aa is a cluster point of xx_\bullet, it suffices to prove that xπα1(W)x_\bullet \in \pi_\alpha^{-1}(W) infinitely often for each α\alpha and each neighborhood WW of aαa_\alpha. This is true because πα(x)W\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.