# $X$ is compact iff every net in $X$ has a convergent subnet (using filters).

I'm trying to prove that a topological space $$X$$ is compact iff every net has a convergent subnet.

A topological space $$X$$ is compact iff every filter on $$X$$ has an adherent point and I'd like to use the connection between filters and nets to prove this statement.

So, I attempted like this:

Let $$X$$ be compact and let $$x:=(x_\alpha)_{\alpha\in I}$$ be a net in $$X$$. Then we can associate a filter $$\mathcal{F}_x$$ to this net by

$$\mathcal{F}_x:= \operatorname{stack}\{\{x_n:n \geq m\}: m \in I\}$$

Because $$X$$ is compact, it follows that there is $$y \in X$$ such that $$\mathcal{F}_x \dashv y$$. We then know that $$x = (x_\alpha)_{\alpha \in I} \dashv y$$ as well (by one of the properties of this associated filter). Consequently, $$x$$ has a convergent subnet converging to $$y$$.

Conversely, let $$\mathcal{F}$$ be any filter on $$X$$. We can associate a net with this filter by considering the directed set

$$I:= \{(x,F): x \in F, F \in \mathcal{F}\}$$

partially ordered via reverse inclusion, ignoring the first coordinate and the map

$$N_\mathcal{F}: I \to X: (x,F) \mapsto x$$

then gives the desired net.

By assumption, this net has a convergent subnet, which after an analaguous reasoning tells us that $$\mathcal{F}$$ has an adherent point as well, showing that $$X$$ is compact.

Is this correct?

• What is a stack here? Feb 14 '20 at 13:54
• Given $X$ a set and $\mathcal{A}$ a collection of subsets of $X$, we define $\operatorname{stack}(\mathcal{A}):=\{F\subseteq X\mid \exists A \in \mathcal{A}: A \subseteq F\}$. So basically all supersets of sets in $\mathcal{A}$. @Asaf Karagila
– user745578
Feb 14 '20 at 14:05

a. A cluster point of the net $$(x_a)_{a \in A}$$ in $$X$$ is a $$p$$ such that for every (open) neighbourhood $$O$$ of $$p$$ and every $$a \in A$$ there is some $$a' \ge a$$ such that $$x_{a'} \in O$$. (The net is frequently in every neighbourhood of $$p$$). This is probably what you denote by $$(x_a)_{a \in A} \dashv p$$.

b. It is well-known (e.g. Willard, chapter 11) that $$p$$ is a cluster point of a net iff there is a subnet of that net that converges to $$p$$. You seem to assume this fact as known.

c. To a net we associate its tail filter (as Willard also does in chapter 12) and $$p$$ is a cluster point (or adherence point) of the tail filter iff $$p$$ is a cluster point of the original net. This is an easy exercise in definitions.

d. Similarly we can define a net $$N_{\mathcal{F}}$$ from a filter $$\mathcal{F}$$ as you do (Willard chapter 12 construction again) and note that $$p$$ is a cluster point of that $$N_{\mathcal{F}}$$ iff $$p$$ is a cluster point of $$\mathcal{F}$$, again an easy exercise in definitions.

So assuming you know

1. $$X$$ is compact iff every filter on $$X$$ has a cluster point.

We can show the required

1. $$X$$ is compact iff every net has a convergent subnet.

using these correspondences and facts:

$$2$$, $$\Rightarrow$$: let $$(x_a)_{a \in A}$$ be a net in $$X$$ and $$X$$ compact. Its tail filter has a cluster point by "$$1$$, $$\Rightarrow$$" and that cluster point is also one for the net by c. Then b. tells us that $$(x_a)_{a \in A}$$ has a convergent subnet.

$$2$$, $$\Leftarrow$$: let $$\mathcal{F}$$ be a filter on $$X$$ (On $$X$$ we assume that every net has a convergent subnet), then $$N_{\mathcal{F}}$$ has a convergent subnet to some $$p$$. So by b. (reverse direction) $$p$$ is a cluster point of $$N_{\mathcal{F}}$$ and so by d. $$p$$ is a cluster point of $$\mathcal{F}$$. Then $$1$$,$$\Leftarrow$$ tells us that $$X$$ is compact (as the filter was arbitrary).

So your argument is in essence correct. I just made all the known facts more explicit. So if a-d are all known to you you can use the final proof; maybe you need more details filled in for d? You seem to skip over some details there.

• Thanks! I know about (a)-(b)-(c)-(d) so it's OK!
– user745578
Feb 15 '20 at 12:24
• @user745578 of course there are direct proofs for 2, not using filters, too. Feb 15 '20 at 12:25

A direct proof, not using the correspondences is also quite doable:

Suppose $$X$$ is compact, and $$(x_a)_{a \in A}$$ is any net. We only need to show that $$p$$ has a cluster point to get a convergent subnet. So suppose no point is a cluster point, and so we can pick for every $$x \in X$$ some open neighbourhood $$U_x$$ such that $$\exists a(x) \in A: \forall a \ge a(x): x_a \notin U_x\tag{1}$$

This defines an open cover of $$X$$ that has a finite subcover $$\{U_x: x \in F\}$$ for some finite subset $$F$$ of $$X$$. Now by directedness (applied finitely many times) we can find $$a_0 \in A$$ such that $$\forall x \in F: a_0 \ge a(x)$$. Now $$p=x_{a_0}$$ must lie in some $$U_x$$ for $$x \in F$$, but then $$a_0 \ge a(x)$$ directly contradicts $$(1)$$, as we have $$p \in U_x$$ and simultaneously $$p \notin U_x$$. This contradiction shows that the net does have a cluster point and we're done.

So suppose every net has a cluster point (or equivalently, a convergent subnet) and we'll show $$X$$ is compact: let $$\mathcal{U}$$ be an open cover of $$X$$ and suppose it has no finite subcover (going for a contradiction). Define a directed set by $$I = \{(\mathcal{U}', x): x \in X \setminus \bigcup \mathcal{U}', \mathcal{U}' \subseteq \mathcal{U} \text{ finite }\}$$ ordered by inclusion on the first component and a net $$n:I \to X$$ by $$n(\mathcal{U}', x)= x$$. This definition only works because the cover has no finite subcovers.

Then if $$x \in X$$, let $$U_x \in \mathcal{U}$$ so that $$x \in U_x$$, then pick any $$y \notin U_x$$ (otherwise $$U_x=X$$ and $$\mathcal{U}$$ would have had a finite subcover) and define $$i(x)=(\{U_x\},y)$$ and by definition if $$i \ge i_0$$, $$n(i) \notin U_x$$, so the pair $$U_x$$ and $$i_0$$ witness that $$x$$ is not a cluster point of $$n$$. So as $$x$$ was arbitrary, the net $$n$$ has no cluster points and we have our contradiction. So $$X$$ is compact.

• Thanks. This was easier than I thought, actually!
– user745578
Feb 16 '20 at 20:37
• @user745578 I just wanted to illustrate that theory on nets doesn't "need" filters; filters are useful in otehr situations (like studying compactifications). And there is a set of filters, not a set of nets on a space. Feb 17 '20 at 5:40