compact spaces equivalently have converging subnet of every net



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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.



(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:


(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.


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.


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.


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 (