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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.

Bremen000
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  • Well I was also asking for an example in the non regular case. Your answer was quite helpful honestly. Can you write it back? Also the comment on the non regularity of the space: I was trying to understand it and everything disappeared! If you want also to provide a proof for he regular case, you are welcome! – Bremen000 May 18 '20 at 22:36
  • I have added a second proof to my answer which you may be interested in, and which I would consider the "right" way to think about regularity in this context. – Eric Wofsey May 19 '20 at 03:19

2 Answers2

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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 $G$.

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.

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 $X$.

Eric Wofsey
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  • Thank you very much for your answer. I still have to understand all the details. Is this topology regular? – Bremen000 May 18 '20 at 22:29
  • No; in fact it's a fairly well-known example of a Hausdorff space that is not regular. For instance, if you pick a point $p\in X\setminus A$ and a small open ball $B$ around $p$, then ${p}\cup(B\cap A)$ is open but does not contain any closed neighborhood of $p$ (any closed neighborhood would have to contain points near $p$ in $X\setminus A$). – Eric Wofsey May 18 '20 at 22:30
  • Thank you very much for having posted it again. I see, it is a very nice example! Do you know how to conclude the proof in the regular case i.e. how to prove that $p$ is a cluster point for the original net? – Bremen000 May 18 '20 at 22:47
  • I have added a proof in the regular case. – Eric Wofsey May 18 '20 at 22:53
  • Thank you very much for both proofs! I am not very familiar with filters, but I will try to understand everything. However, I'm still trying to use nets but maybe it's not possible in this particular case.. – Bremen000 May 19 '20 at 07:51
  • Thank you for sharing this theorem! The reason why we are assuming that $X$ is Hausdorff, is for the function $L$ to be well-defined. However, is it possible even if $X$ isn't Hausdorff, to define $L:C_X\to Y$, where $Y = 2^X$ is the hyperspace of non-empty subsets of $X$ with appropiate topology? Perhaps the Vietoris topology? So that the continuity of $L$ tells us something about the underlying space $X$. – Jakobian Apr 28 '23 at 16:39
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    @Jakobian I don't know of any result along those lines. One obstacle is that if $X$ is finite then $C_X$ is discrete so any map out of it is continuous, even though $X$ may not be regular. – Eric Wofsey Apr 28 '23 at 17:19
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    @Jakobian: Though, I guess one reformulation you can give for non-Hausdorff spaces is: $X$ is regular iff whenever $(F_i)$ is a net of ultrafilters converging to an ultrafilter $F$, if $F_i$ converges to $x_i$ and $F$ converges to $x$, then $(x_i)$ converges to $x$. (Note in particular that taking $F_i=F$ for all $i$, this implies that if an ultrafilter has two different limits in $X$ those limits must be topologically indistinguishable, so that the $T_0$ quotient of $X$ is Hausdorff, so this theorem is easily reduced to the Hausdorff case.) – Eric Wofsey Apr 28 '23 at 17:30
  • The reason this formulation doesn't easily translate into a continuity condition is that the $F_i$ and $F$ may not be distinct even if the $x_i$ and $x$ are so there is not a well-defined function whose continuity you are talking about. To get a continuity statement you have to replace the domain $C_X$ with a larger domain that distinguishes all the $F_i$ and $F$. (This is essentially what I did in my answer to your question, with the $F_i$ becoming the points in $S$ and $F$ becoming the point $\infty$.) – Eric Wofsey Apr 28 '23 at 17:36
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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.

Henno Brandsma
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