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compact spaces equivalently have converging subnet of every net

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

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Extra stuff, structure, properties

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topological homotopy theory

Analysis

Contents

Idea

For metric spaces with their metric topology it is true that sequentially compact metric spaces are equivalently compact metric spaces. The analogous statement fails for more general topological spaces: for them, being sequentially compact in general neither implies nor is implied by being compact (see the counter-examples here).

But the failure of this equivalence is not due to a deficit in the concept of convergence but in the concept of sequences and sub-sequences. If the latter are generalized to nets and sub-nets (beware that the definition of sub-nets is slightly non-obvious), then the analogue of the statement remains true in generality: A topological space is compact precisely if every net in it has a sub-net that convergence.

Statement

Proposition

(compact spaces are equivalently those for which every net has a converging subnet)

Assuming excluded middle and the axiom of choice, then:

A topological space (X,τ)(X,\tau) is compact precisely if every net in XX has a sub-net that converges.

We break this up into lemmas 1 and 2:

Lemma

(in a compact space, every net has a convergent subnet)

Let (X,τ)(X,\tau) be a compact topological space. Then every net in XX has a convergent subnet.

Proof

Let ν:AX\nu \colon A \to X be a net. We need to show that there is a subnet which converges.

For aAa \in A consider the topological closures Cl(S a)Cl(S_a) of the sets S aS_a of elements of the net beyond some fixed index:

S a{ν bX|ba}X. S_a \;\coloneqq\; \left\{ \nu_b \in X \;\vert\; b \geq a \right\} \subset X \,.

Observe that the set {S aX} aA\{S_a \subset X\}_{a \in A} and hence also the set {Cl(S a)X} aA\{Cl(S_a) \subset X\}_{a \in A} has the finite intersection property, by the fact that AA is a directed set. Therefore this prop. implies from the assumption of XX being compact that the intersection of all the Cl(S a)Cl(S_a) is non-empty, hence that there is an element

xaACl(S a). x \in \underset{a \in A}{\cap} Cl(S_a) \,.

In particular every neighbourhood U xU_x of xx intersects each of the Cl(S a)Cl(S_a), and hence also each of the S aS_a. By definition of the S aS_a, this means that for every aAa \in A there exists bab \geq a such that ν bU x\nu_b \in U_x, hence that xx is a cluster point of the net.

We will not produce a sub-net

B f A ν X \array{ B && \overset{f}{\longrightarrow} && A \\ & \searrow && \swarrow_{\nu} \\ && X }

that converges to this cluster point. To this end, we first need to build the domain directed set BB. Take it to be the sub-directed set of the Cartesian product directed set of AA with the directed neighbourhood set Nbhd X(x)Nbhd_X(x) of xx

BA ×Nbhd X(x) B \subset A_{\leq} \times Nbhd_X(x)_{\supset}

on those pairs such that the element of the net indexed by the first component is is contained in the second component:

B{(a,U x)|ν aU X}. B \;\coloneqq\; \left\{ (a,U_x) \,\vert \, \nu_a \in U_X \right\} \,.

It is clear BB is a preordered set. We need to check that it is indeed directed, in that every pair of elements (a 1,U 1)(a_1, U_1), (a 2,U 2)(a_2, U_2) has a common upper bound (a bd,U bd)(a_{bd}, U_{bd}). Now since AA itself is directed, there is an upper bound a 3a 1,a 2a_3 \geq a_1, a_2, and since xx is a cluster point of the net there is moreover an a bda 3a 1,a 3a_{bd} \geq a_3 \geq a_1, a_3 such that ν a bdU 1U 2\nu_{a_{bd}} \in U_1 \cap U_2. Hence with U bdU 1U 2U_{bd} \coloneqq U_1 \cap U_2 we have obtained the required pair.

Next take the function ff to be given by

B f A (a,U) AAA a. \array{ B &\overset{f}{\longrightarrow}& A \\ (a, U) &\overset{\phantom{AAA}}{\mapsto}& a } \,.

This is clearly order preserving, and it is cofinal since it is even a surjection. Hence we have defined a subnet νf\nu \circ f.

It now remains to see that νf\nu \circ f converges to xx, hence that for every open neighbourhood U xU_x of xx we may find (a,U)(a,U) such that for all (b,V)(b,V) with aba \leq b and UVU \supset V then ν(f(b,V))=ν(b)U x\nu(f(b,V)) = \nu(b) \in U_x. Now by the nature of xx there exists some aa with ν aU x\nu_a \in U_x, and hence if we take UU xU \coloneqq U_x then nature of BB implies that with (b,V)(a,U x)(b, V) \geq (a,U_x) then bVU xb \in V \subset U_x.

Lemma

Assuming excluded middle, then:

Let (X,τ)(X,\tau) be a topological space. If every net in XX has a subnet that converges, then (X,τ)(X,\tau) is a compact topological space.

Proof

By excluded middle we may equivalently prove the contrapositive: If (X,τ)(X,\tau) is not compact, then not every net in XX has a convergent subnet.

Hence assume that (X,τ)(X,\tau) is not compact. We need to produce a net without a convergent subnet.

Again by excluded middle, then by this prop. (X,τ)(X,\tau) not being compact means equivalently that there exists a set {C iX} iI\{C_i \subset X\}_{i \in I} of closed subsets satisfying the finite intersection property, but such that their intersection is empty: iIC i=\underset{i \in I}{\cap} C_i = \emptyset.

Consider then P fin(I)P_{fin}(I), the set of finite subsets of II. By the assumption that {C iX} iI\{C_i \subset X\}_{i \in I} satisfies the finite intersection property, we may choose for each JP fin(I)J \in P_{fin}(I) an element

x JiJIC i. x_J \in \underset{i \in J \subset I}{\cap} C_i \,.

Now P fin(X)P_{fin}(X) regarded as a preordered set under inclusion of subsets is clearly a directed set, with an upper bound of two finite subsets given by their union. Therefore we have defined a net

P fin(X) ν X J AAA x J. \array{ P_{fin}(X)_{\subset} &\overset{\nu}{\longrightarrow}& X \\ J &\overset{\phantom{AAA}}{\mapsto}& x_J } \,.

We will show that this net has no converging subnet.

Assume on the contrary that there were a subnet

B f P fin(X) ν X \array{ B && \overset{f}{\longrightarrow} && P_{fin}(X) \\ & \searrow && \swarrow_{\nu} \\ && X }

which converges to some xXx \in X.

By the assumption that iIC i=\underset{i \in I}{\cap} C_i = \emptyset, there would exist an i xIi_x \in I such that xC i xx \neq C_{i_x}, and because C iC_i is a closed subset, there would exist even an open neighbourhood U xU_x of xx such that U xC i x=U_x \cap C_{i_x} = \emptyset. This would imply that x JU xx_J \neq U_x for all J{i x}J \supset \{i_x\}.

Now since the function ff defining the subset is cofinal, there would exist b 1Bb_1 \in B such that {i x}f(b 1)\{i_x\} \subset f(b_1). Moreover, by the assumption that the subnet converges, there would also be b 2Bb_2 \in B such that ν b 2U x\nu_{b_2 \leq \bullet} \in U_x. Since BB is directed, there would then be an upper bound bb 1,b 2b \geq b_1, b_2 of these two elements. This hence satisfies both ν f(e)U x\nu_{f(e)} \in U_x as well as {i x}f(b 1)f(b)\{i_x\} \subset f(b_1) \subset f(b). But the latter of these two means that ν f(b)\nu_{f(b)} is not in U xU_x, which is a contradiction to the former. Thus we have a proof by contradiction.

Revised on April 23, 2017 14:27:38 by Urs Schreiber (92.218.150.85)