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

compactnessfiltersgeneral-topologynetssolution-verification

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

Here is what I already know about compactness/filters:

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?

Best Answer

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.

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