# 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 $(X,\tau)$ is compact precisely if every net in $X$ 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 $(X,\tau)$ be a compact topological space. Then every net in $X$ has a convergent subnet.

###### Proof

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

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

$S_a \;\coloneqq\; \left\{ \nu_b \in X \;\vert\; b \geq a \right\} \subset X \,.$

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

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

In particular every neighbourhood $U_x$ of $x$ intersects each of the $Cl(S_a)$, and hence also each of the $S_a$. By definition of the $S_a$, this means that for every $a \in A$ there exists $b \geq a$ such that $\nu_b \in U_x$, hence that $x$ is a cluster point of the net.

We will not produce a sub-net

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

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

$B \subset A_{\leq} \times Nbhd_X(x)_{\supset}$

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

$B \;\coloneqq\; \left\{ (a,U_x) \,\vert \, \nu_a \in U_X \right\} \,.$

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

Next take the function $f$ to be given by

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

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

It now remains to see that $\nu \circ f$ converges to $x$, hence that for every open neighbourhood $U_x$ of $x$ we may find $(a,U)$ such that for all $(b,V)$ with $a \leq b$ and $U \supset V$ then $\nu(f(b,V)) = \nu(b) \in U_x$. Now by the nature of $x$ there exists some $a$ with $\nu_a \in U_x$, and hence if we take $U \coloneqq U_x$ then nature of $B$ implies that with $(b, V) \geq (a,U_x)$ then $b \in V \subset U_x$.

###### Lemma

Assuming excluded middle, then:

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

###### Proof

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

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

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

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

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

Now $P_{fin}(X)$ 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

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

We will show that this net has no converging subnet.

Assume on the contrary that there were a subnet

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

which converges to some $x \in X$.

By the assumption that $\underset{i \in I}{\cap} C_i = \emptyset$, there would exist an $i_x \in I$ such that $x \neq C_{i_x}$, and because $C_i$ is a closed subset, there would exist even an open neighbourhood $U_x$ of $x$ such that $U_x \cap C_{i_x} = \emptyset$. This would imply that $x_J \neq U_x$ for all $J \supset \{i_x\}$.

Now since the function $f$ defining the subset is cofinal, there would exist $b_1 \in B$ such that $\{i_x\} \subset f(b_1)$. Moreover, by the assumption that the subnet converges, there would also be $b_2 \in B$ such that $\nu_{b_2 \leq \bullet} \in U_x$. Since $B$ is directed, there would then be an upper bound $b \geq b_1, b_2$ of these two elements. This hence satisfies both $\nu_{f(e)} \in U_x$ as well as $\{i_x\} \subset f(b_1) \subset f(b)$. But the latter of these two means that $\nu_{f(b)}$ is not in $U_x$, 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)