Munkres supplementary exercises chapter 3(nets) question 10

Question

Prove the following.

X is compact if and only if every net in X has a convergent subnet.

I am stuck in proving the (<=) side.

https://dbfin.com/topology/munkres/chapter-3/supplementary-exercises-nets/problem-10-solution/ I did find a solution here, but I think it has an error. The set $K$ of all finite subsets of $J$ should have the empty set as an element in order to have a well-defined partial relation(because two finite subsets of $J$ can be disjoint, but there still should be an element of $K$ that is a subset of both). And in the last part, from $ \beta \subset \{k\}$ it is possible that $\beta $ is the empty set, so it is impossible to conclude that $U \cap B_k \neq 0$.

Answer 1

I think they made a mistake in defining the order relation. You should order finite subset by inclusion instead of reverse inclusion. If you do that then the proof works fine.

Answer 2

Yes, if $\{A_i\mid i \in I\}$ is a family of closed sets with the FIP, let $I^\ast$ be the set of all non-empty finite subsets of $I$ under inclusion (so $I \le J \iff I \subseteq J$). This is a partial order and $I,J \in I^\ast$ have a common upperbound $I \cup J \in I^\ast$, so it's clearly a directed set.

The confusion probably comes from the fact that many natural nets are ordered by families of sets under reverse inclusion (like neighbourhood systems). So the tendency is to take reverse inclusion instead of the correct inclusion. Here larger subsets will bring us "closer" to the intersection so inclusion is also intuitively correct if you think about it.

The proof then writes itself (with AC): for each $I \in I^\ast$ pick (heavy use of AC) $x_I \in \bigcap_{i \in I} A_i$, using that the family has the FIP.

This defines a net $I^\ast \to X$ so this must have an accumulation point $p \in X$. The claim is that $p \in \bigcap_{i \in I} A_i$ which would finish the proof. To this end, fix $i \in I$ and let $U$ be any neighbourhood of $p$. Then $\{i\} \in I^\ast$ and so there is some $J \in I^\ast$ such that $\{i\} \le J$ (so $i \in J$) and $x_J \in U$. In particular, $$x_J \in\bigcap_{j \in J} A_j \subseteq A_i \text{ so } A_i \cap U \neq \emptyset$$

As $U$ was arbitrary, $p \in \overline{A_i} = A_i$. As $i \in I$ was arbitrary we are done.

As an alternative to this proof, let $\mathcal{U}$ be an open cover of $X$ and assume it has no finite subcover (or we'd be done). Take as partial order $I$ all finite subsets of $\mathcal{U}$ again under inclusion. The union of any two finite subfamilies is again one, so directedness is again clear. As a net pick $x_I \in X\setminus \bigcup I$, exploiting the fact that there is no finite subcover. If $p$ is an accumulation point of this net, it is in some $U_p \in \mathcal{U}$ (we started with a cover) and so for some $J \ge \{U_p\}$ we have $x_J \in U_p$ which gives an immediate contradiction: $x_J \notin \bigcup J$ while $x_J \in U_p \in J$. So no such open cover without a finite subcover can exist in a space where every net has an accumulation point, which shows the implication.