Net compactness and relative compactness

Question

I'm trying to understand the relation between the following conditions. I will assume that $X$ is a Hausdorff topological space and $A \subset X$.

1. $\overline{A}$ is compact;
2. Every net $\{x_{\lambda}\}_{\lambda \in \mathbb{L}} \subset A$ has a subnet converging to some point;

It is clear to me that $1 \Rightarrow 2$. I read that $2 \Rightarrow 1$ if $X$ is regular, but I am not able to find a proof. I would like to have a proof and, if possible, an explicit example in which the implication $2 \Rightarrow 1$ is false.

Answer 1

I find these things easier to think about in the language of filters. Using the usual correspondence between nets and filters, (2) is equivalent to saying that every filter on $X$ containing $A$ has an accumulation point in $X$.

So, suppose $\overline{A}$ is not compact, and we will find a filter containing $A$ with no accumulation point in $X$. Since $\overline{A}$ is not compact and is closed in $X$, there is a filter $F$ containing $\overline{A}$ with no accumulation point in $X$. Let $G$ be the filter generated by all open elements of $F$ together with $A$.

First, I claim $G$ is a proper filter. Indeed, suppose $U\in F$ is open. Then $U\cap \overline{A}\in F$ since $\overline{A}\in F$. Since $U$ is open and $A$ is dense in $\overline{A}$, this means $U\cap A$ is nonempty. Since every element of $G$ contains a set of the form $U\cap A$, this means $G$ is a proper filter.

Second, I claim $G$ has no accumulation point in $X$, and is thus our desired filter since $A\in G$. Indeed, let $x\in X$ be any point. Since $x$ is not an accumulation point of $F$, there is an open neighborhood $U$ of $x$ such that $X\setminus U\in F$. By regularity, there are disjoint open sets $V$ and $W$ such that $x\in V$ and $X\setminus U\subseteq W$. Then $W\in G$, and hence $X\setminus V\in G$, and hence $x$ is not an accumulation point of $G$.

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Here is another proof which is a bit more complicated at first glance but which nicely conceptualizes the role of regularity.

Recall that if $X$ is a set, then the set $\beta X$ of ultrafilters on $X$ has a natural compact Hausdorff topology, which has as a basis the sets $U_A=\{F\in\beta X:A\in F\}$ for each $A\subseteq X$. If $X$ is a Hausdorff space, we will write $C_X\subseteq\beta X$ for the set of ultrafilters that converge in $X$ and $L:C_X\to X$ for the map taking an ultrafilter to its limit. We then have the following remarkable characterization of regularity.

Theorem: Let $X$ be a Hausdorff space. Then $X$ is regular iff $L:C_X\to X$ is continuous.

Proof: Suppose $X$ is regular. Let $F\in C_X$, write $x=L(F)$, and suppose $U$ is a neighborhood of $x$. By regularity, let $A$ be a closed neighborhood of $x$ contained in $U$. Then $A\in F$ since $A$ is a neighborhood of $x$, and if $G\in C_X$ and $A\in G$ then $L(G)\in A$ since $A$ is closed. Thus $U_A\cap C_X$ is a neighborhood of $F$ in $C_X$ whose image under $L$ is contained in $U$, as desired.

Conversely, suppose $X$ is not regular; let $x\in X$ with a neighborhood $U$ which contains no closed neighborhood of $x$. For each neighborhood $V$ of $x$, its closure is not contained in $U$, so we can pick an ultrafilter $F_V$ which contains $V$ and converges to a point $L(F_V)\not\in U$. Consider these $(F_V)$ as a net in $\beta X$, indexed by the directed set of neighborhoods of $x$ ordered by reverse inclusion. By compactness of $\beta X$, this net has a subnet converging to an ultrafilter $F$. Since $V\in F_V$ for all $V$, this limit $F$ must contain every neighborhood of $x$; that is, $L(F)=x$. However, the net $(L(F_V))$ is entirely outside the neighborhood $U$ of $x$, so no subnet can converge to $x$. Thus $L$ fails to preserve the convergence of this subnet and is not continuous.

Using this theorem, proving $2\Rightarrow 1$ for regular spaces is quite natural. In terms of ultrafilters, (2) says that every ultrafilter containing $A$ has a limit in $X$. Now suppose this is true and let $(x_i)$ be a net in $\overline{A}$. For each $i$, we can pick an ultrafilter $F_i$ containing $A$ which converges to $x_i$. By compactness of $\beta X$, there is a subnet of $(F_i)$ that converges to some ultrafilter $F$, which will still contain $A$. By (2), $F$ converges to some $x\in\overline{A}$. Since $X$ is regular, the theorem says that the corresponding subnet of $(x_i)$ converges to $x$. Thus $(x_i)$ has a convergent subnet. Since $(x_i)$ was an arbitrary net in $\overline{A}$, this means $\overline{A}$ is compact.

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Finally, here is an example of how $2\Rightarrow 1$ can be false if $X$ is not regular. Let $X$ be the closed unit disk and let $A\subseteq X$ be the open unit disk, and say a subset $C\subseteq X$ is closed if it contains the closure of $C\cap A$ with respect to the usual topology. This defines a topology in $X$ (another way to describe it is you take the usual topology and then enlarge it by declaring that every subset of the unit circle $X\setminus A$ is closed, and take the topology that generates; so a closed set in $X$ is just a union of a closed set in the usual topology and an arbitrary subset of $X\setminus A$).

Now $\overline{A}=X$ is not compact, since $X\setminus A$ is closed in $X$ but is not compact since it is infinite and discrete. However, every net in $A$ has a limit in $X$. Indeed, every net in $A$ has a subnet converging to some point of $X$ with respect to the usual topology by compactness, and the same is true of the topology of $X$ since nets in $A$ which converge with respect to the usual topology still converge with respect to the topology of $X$.

Answer 2

IIRC the proof is along these lines:

If we have a net $x_i, i \in I$ that is defined on $\overline{A}$, we need to proof it has a convergent subnet (or cluster point) in $\overline{A}$. For each $i \in I$ we find some net $a_j, j \in N_i$ on $A$ that converges to $x_i$ (as $x_i \in \overline{A}$ this is possible). Then using a Kelly-like diagonal construction we combine these nets in a "super-net" on $A$ and then using the given we have some cluster point on $p \in \overline{A}$ of this super-net, and using closed neighbourhoods of $p$ we can find a subnet of the original net $(x_i)_{i \in I}$ that converges to $p$ too.