Reformulating the definition of compactness

Question

Consider the following alternative definition of a topological space.

Definition: A topological space is an ordered pair $(X, \lessdot)$ consisting of a set $X$ and a binary relation $\lessdot$ between the members of $X$ and the subsets of $X$. (We read $x \lessdot A$ as "$x$ touches $A$".) The touch relation $\lessdot$ must satisfy the following axioms:

• No element of $X$ touches the empty set.
• If $x \in A$, then $x \lessdot A$.
• If $x \lessdot (A \cup B)$, then $x \lessdot A$ or $x \lessdot B$.
• If $x \lessdot A$ and every element of $A$ touches $B$, then $x \lessdot B$.

This definition is equivalent to the usual definition via open sets and better captures the intuition that a topology on a set specifies which points are "infinitesimally close" to each other. It also makes a lot of definitions conceptually easier: for example, $f: X \to Y$ is continuous iff $x \lessdot A \implies f(x) \lessdot f(A)$.

Question: Can the definitions of compact spaces and proper maps be reformulated in terms of a touch relation in a simple way, similar in spirit to the definition of continuity above?

Of course, it's possible to define open sets in terms of a touch relation ($A \subseteq X$ is open iff no point of $A$ touches $X \setminus A$) and then restate the usual definition of compactness, but I have a suspicion that something simpler ought to be possible. The definition of continuity above is a lot shorter than "the inverse image under $f$ of every set whose points do not touch its complement is itself a set whose points do not touch its complement." Proper maps, in particular, have a nice intuitive characterization (faraway points are sent to faraway points) that seems compatible with the structure of a touch relation. (Note that $x \not\lessdot A \implies f(x) \not\lessdot f(A)$ doesn't work. Sigmoid functions $\mathbb{R} \to \mathbb{R}$ have this property but are not proper.)

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Aside: I believe this axiomatization of point-set topology first appeared in a 1977 article by David B. Gauld called "Nearness - A Better Approach To Topology". In this article, Gauld mentions that

"...one can formulate a definition of compactness involving nearness spaces, but it is rather unwieldy."

Unfortunately, Gauld does not state his definition, and I can't find any reference to it elsewhere. I caution readers that Gauld's terminology did not catch on, and "nearness space" typically refers to a different mathematical structure elsewhere in the literature.

Answer 1

Isn't $x \lessdot A$ not a fancy way to state $x \in \overline{A}$? So just define $\overline{A} = \{x: x \lessdot A\}$.

The axioms then become more familiar:

• $\overline{\emptyset} = \emptyset$
• $A \subseteq \overline{A}$
• $\overline{A \cup B} = \overline{A} \cup \overline{B}$
• $x \in \overline{A}$ and $A \subseteq \overline{B}$ then $x \in \overline{B}$

The last is a bit weird but the system of axioms is equivalent to the same system but with the 4th replaced by the standard

• $\overline{\overline{A}} = \overline{A}$ for all $A$.

So we just get Kuratowski's closure axioms which are well-known to characterise a topology.

I don't see a lot of independent merit in this particular axiomisation. The continuity characterisation is just the standard one in terms of closure

• $f:X \to Y$ is continuous iff for all $A \subseteq X$ we have $f[\overline{A}] \subseteq \overline{f[A]}$.

I don't know of any easy characterisation of compactness or properness of maps in terms of closures e.g. that one could try to translate...

Answer 2

I do not know, if the following is sufficiently nice for you, but compactness can be characterized as follows:

Definition. A nearness space $(X, \lessdot)$ is compact, iff for any familiy $\mathcal F$ of subsets of $X$ with the finite intersection property, there is some point of $X$ that touches all sets in $\mathcal F$.

We will show that this is equivalent to the usual open set characterisation: Suppose $(X, \lessdot)$ is topologically compact and let $\mathcal F$ a family with the finite intersection property. If $\{X \setminus \bar F: F \in \mathcal F\}$ were an open cover of $X$, is would have a finite subcover, contradicting the finite intersection property. Hence, $\bigcap_{F \in \mathcal F} \bar F \ne \emptyset$, any point in this intersection touches all sets in $\mathcal F$.

On the other side, let $X$ a compact nearness space and $\mathcal U$ an open cover. Suppose $\mathcal U$ doesn't have a finite subcover. Then $\mathcal F := \{X \setminus U: U \in \mathcal U\}$ has the finite intersection property and therefore there is $x \in X$ that touches all sets $X \setminus U$. As all sets $U$ is open, $x$ cannot be contained in any set $U$, Contradiction.