nLab compact spaces equivalently have converging subnets

Idea

In classical mathematics, 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 classical mathematics: A topological space is compact precisely if every net in it has a sub-net that converges.

In constructive mathematics, on the other hand, this property implies the law of excluded middle, so is thus equivalent to the law of excluded middle.

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 and :

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 now 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 $x \notin C_{i_x}$, 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 $x_J \notin U_x$ 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.

Failure in constructive mathematics

In constructive mathematics, this statement is equivalent to excluded middle.

Proposition

The set of two elements $\{0, 1\}$ with its discrete topology is a compact space.

Proof

Let $U$ be an open cover of $\{0, 1\}$. Then there exists $i_0$ such that $0 \in U_{i_0}$ and $i_1$ such that $1 \in U_{i_1}$. Then $U_{i_0}, U_{i_1}$ form a finite subcover of $\{0, 1\}$: the universal property of the set of two elements implies that if $x \in \{0, 1\}$, then $x = 0$ or $x = 1$, which means that if $x = 0$, then $x \in U_{i_0}$, and if $x = 1$, then $x \in U_{i_1}$.

Proposition

Every inhabited subset of $\{0, 1\}$ is a directed poset.

Proof

Let $P$ be an inhabited subset of $\{0, 1\}$. Then $P$ is a directed poset: it is inhabited, it has a partial order inherited from the partial order defined on $\{0, 1\}$ by $0 \leq 1$, $0 \leq 0$, $1 \leq 1$, and there is an upper bound function $\beta:P \times P \to P$ defined for elements $a \in P$ and $b \in P$ as $\beta(a, b) = b$ if $a = 0$, $\beta(a, b) = a$ if $a = 1$, $\beta(a, b) = a$ if $b = 0$, and $\beta(a, b) = b$ if $b = 1$: this means that $a \leq \beta(a, b)$ and $b \leq \beta(a, b)$.

Lemma

Every subset inclusion $i:P \hookrightarrow \{0, 1\}$ of an inhabited subset $P \subseteq \{0, 1\}$ is a net in $\{0, 1\}$.

Proof

Since $P$ is a directed poset, it follows that any function $i:P \to \{0, 1\}$ is a net in $\{0, 1\}$. Thus the subset inclusion $i:P \hookrightarrow \{0, 1\}$ is a net.

Proposition

Given the set of two elements $\{0, 1\}$ with its discrete topology, if every net in $\{0, 1\}$ has a converging subnet, then the law of excluded middle is true.

Proof

Let $Q$ be a directed set (i.e. a directed preordered set). Assume that every net in $\{0, 1\}$ has a converging subnet, where $\{0, 1\}$ has the discrete topology. Because the subset inclusion $i \colon P \to \{0, 1\}$ for any subset $P \subseteq \{0, 1\}$ is a net, there is a cofinal function $f:Q \to P$ from $Q$ to $P$ such that $i \circ f$ is a net which converges to an element $x \in \{0, 1\}$, which by the universal property of $\{0, 1\}$ satisfies either $x = 0$ or $x = 1$.

Because $\{0, 1\}$ has the discrete topology, $\{1\}$ is an open of $\{0, 1\}$. If $x = 1$, then there is an element $q \in Q$ such that $f(q) \in \{1\}$, but $f(q) \in P$, so $1 \in P$.

If $x = 0$, then suppose that $1 \in P$. Because $f$ is cofinal, there is an element $q \in Q$ such that $f(q) \geq 1$, and by the definition of the partial order in $\{0, 1\}$, $f(q) = 1$, which means that for any other element $r \in Q$ such that $r \geq q$, $f(r) \geq f(q)$, $f(r) \geq 1$, and $f(r) = 1$, which contradicts the fact that $f$ converges to $0$. Thus, $1 \notin P$.

Thus, assuming that $\{0, 1\}$ comes with its discrete topology, if every net in $\{0, 1\}$ has a converging subnet, then for every inhabited subset $P$ of $\{0, 1\}$, either $1 \in P$ or $1 \notin P$. For the empty subset, $1 \notin \emptyset$, so for every subset $P$ of $\{0, 1\}$, either $1 \in P$ or $1 \notin P$.

Given any proposition $p$, let us define the subset of $\{0, 1\}$ as

$$P_p \;\coloneqq\; \Big\{ x \in \{0, 1\} \Big\vert (x = 0) \vee \big((x = 1) \wedge p\big) \Big\}$$

It follows that if $1 \in P_p$, then $p$, and if $1 \notin P_p$, then $\not p$. Thus, the law of excluded middle is true for all propositions $p$.

Related concepts

• sequentially compact metric spaces are equivalently compact metric spaces

• countably compact metric spaces are equivalently compact metric spaces

• sequentially compact metric spaces are totally bounded

• Lebesgue number lemma