The axioms of a topological space are usually stated in the "open set" form: A topological space $X$ has a set of subsets $\tau$ whose members satisfy:

  1. $\emptyset$ and $X$ are in $\tau$.
  2. $\tau$ is closed under arbitrary unions.
  3. $\tau$ is closed under finite intersections.

We will call this the usual open set axiomatization of a topological space. The open set axiomatization is probably the most economical, but of course there are other ways as well, such as the Kuratowski closure axioms which axiomatizes the topology using the closure operator.

In his note here, Pete L. Clark gives several alternative characterizations of a topology, most of which have details given. It was mentioned, I think originally by Willard, that it is "possible, but unrewarding to characterize a topology completely by its [boundary]". I could find no further references on this, and I am wondering how it might be done (unrewarding as it may be).

So my question is this: How can a topological space by axiomatized using the boundary operation as the primitive notion? A complete answer should give a list of axioms required for the formulation, as well as a sketch of how they are equivalent to the usual open set formulation.

  • 4
    $\begingroup$ It is a fairly standard exercise that sets are closed iff they contain their boundary and open iff they are disjoint from their boundary. So in principle the reconstruction of axioms for topology in terms of a boundary operator will assure that the usual axioms (e.g. in terms of open sets) can be recovered. $\endgroup$ – hardmath Aug 8 '15 at 22:38
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    $\begingroup$ Read this pdf. $\endgroup$ – Tony Piccolo Aug 9 '15 at 6:24
  • 1
    $\begingroup$ @Tony: Well, that is an impressively on-the-nose answer! Perhaps I should cite this in my topology notes. $\endgroup$ – Pete L. Clark Jul 20 '18 at 1:07

As it known, $\partial A=c(A)\cap c(X\setminus A)$. Therefore, since $c(A)\supset A$, you can restore closure operator (and then topology) by the given boundary operator, when you let $c(A):=A\cup \partial A$.

For the boundary correctly determine the closure, following axioms have to hold:

  1. $\partial A=\partial(X\setminus A)$ for any $A$;$\,\,\,$ [for satisfy redefinition]

  2. $\partial \emptyset=\emptyset$;$\,\,\,$ [for satisfy KC1]

  3. $\partial(\partial A)\subset\partial A$ for any $A$;$\,\,\,$ [for satisfy KC3]

  4. $\partial(A\cup B)\subset \partial A\cup\partial B$ for any $A,B$;$\,\,\,$ [for satisfy KC4]

and KC2 will be satisfied automatically.

EDIT: for really satisfy CK4, besides axiom 4 we need also

  1. $A\subset B\Rightarrow \partial A\subset B\cup\partial B$.
  • $\begingroup$ Actually, this list of axioms is not full: statement $\partial(A\cup B)\subset\partial A\cup\partial B$ gives only inclusion $c(A\cup B)\subset c(A)\cup c(B)$; for making this inclusion an equality, we need also the axiom "5. $A\subset B\Rightarrow \partial A\subset B\cup\partial B$", as in pdf in Tony Piccolo's comment. $\endgroup$ – Andrey Ryabichev Aug 9 '15 at 10:50

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