# Constructing all compact Hausdorff spaces from a generalised sequential-limit operator

Given a set $X$ and a partial function $\lim:X^{\mathbb{N}}\rightarrow X$, define for all $M\subset X$, $$\overline{M}=\{x\in X :\forall (x_n)_{\mathbb{N}}\in M^{\mathbb{N}}\,((x_n)_{\mathbb{N}},x)\in \lim \Rightarrow x\in M \}.$$ Now, suppose $\lim$ is such that the following are satisfied:

1. $\overline{\overline{L}\cup \overline{M}}=\overline{L}\cup \overline{M}$

2. $\overline{\displaystyle\bigcap_{i\in I}\overline{L_i}}=\displaystyle\bigcap_{i\in I}\overline{L_i}$

Is it possible to define a compact Hausdorff space $X$, using $\lim$ in the obvious way?

The same as above but for a relation $\operatorname{clust}:X^{\mathbb{N}}\rightarrow X$ and claiming a general definition for a compact space?

• I recommend wiki for this matter. Sep 4, 2014 at 7:49
• So you want the $\overline{M}$ to be a closure operation, and you get your topology from that, and you ask if the condition you state is equivalent to this topology being Hausdorff? Sep 4, 2014 at 7:51
• @Amitai Yuval: Thanks for the tips, but I couldn't find the proof there.
– Lehs
Sep 4, 2014 at 7:59
• @Tobias Kildetoft: I don't think it's possible to prove that it's a closure operation if (i) and (ii) is not included..?
– Lehs
Sep 4, 2014 at 8:01
• Something like this works for compact Hausdorff spaces, but it is necessary to work with arbitrarily large index sets and to consider limits of ultrafilters and not just sequences. The keyword here is "ultrafilter monad." Your proposal can't work as stated because in general topologies are not determined by convergence of sequences; one needs more general objects such as (ultra)filters or nets. Sep 4, 2014 at 8:17

The answer to the question in the post is "No" as can be seen by the following simple example. Let $X := \mathbb{N}$ and let $lim \subseteq (\mathbb{N}^\mathbb{N} \times \mathbb{N})$ be defined by
$((x_n)_{n\in \mathbb{N}}, x) \in lim :\Leftrightarrow (\exists N_0\in \mathbb{N})[k\geq N_0 \implies x_k = x].$
Then $\overline{M}=M$ for all $M\subseteq\mathbb{N}$, therefore the topology you get on $\mathbb{N}$ is the discrete topology which is not compact.