We recall Kuratowski’s closure axioms for a set . These are a set of axioms which can be used to define a topology on a set. Here, one replaces the notion of open (or closed) sets with a closure operation. The closure operation has the advantage of being defined on all subsets of . For, one may think of the closure of , denoted , to be points that are “close to” .
First we recall what is meant by a topological space. We let be the power set of , that is the set of all subsets of .
Definition 1: Let be a set and . Then is a topology with open sets if
- (a) ,
- (b) if , then ,
- (c) if for all , then .
Furthermore, we say
are the closed sets of .
Note we could have defined (a), (b), (c) in terms of closed sets, utilizing DeMorgan’s law. We make use of this implicitly below. We now present the definition of Kuratowski closure.
Definition 2: Let be a set and . We say is a closure operation if
- (i) ,
- (ii) if , then ,
- (iii) if , then ,
- (iv) if , then .
As mentioned above, one can informally think of as the set of points in which are “close” to . We now explain that these two definitions are equivalent. To define a closure operation from Definition 1, we set
Note that is closed by (c). Given a closure operation, we may define the closed sets to be
Proposition 1: Let be a topological space as in Definition 1. Then the operation defined in (1) is a closure operation in the sense of Definition 2.
be the set of closed sets in . Let . By (1), we have and thus (ii) holds. As is open, and so (i) is satisfied. Note for any closed set, , we have by (ii) and that itself appears on the right hand side of (1). By (c), we have that
and so . This proves (iii) and so it remains to show (iv). Note that is a closed set containing , by (b). Thus
Now let that contains . Then and so
This proves the reverse inclusion of (3) and thus of Proposition 1.
We now show that a closure axiom can be used to define a topology.
Proposition 2: Let be a closure axiom as in Definition 2. Then the sets defined in (2) form the closed sets of a topology in the sense of Definition 1.
which establishes (b). Let for all some index set. Then we have
Indeed, the backwards inclusion follows from (ii). To see the forward direction, it is enough to show for any ,
The second equality is (iii). The first subset inequality follows from (iv) as if then
We apply this with and .
Recall if and are topological spaces, then
is said to be continuous if the pre-image of every open set in is open in , that is
for all open . As , (5) is equivalent to the analogous definition for closed sets. It turns out that we may equivalently define a map to be continuous if
This can be informally interpreted as points that are close to are mapped to points that are close to . We now prove this.
Proof: We start with the forward direction. Let . Then by (4), is closed. As , we have
Since the right hand side is closed, by (ii) and (iii), we have
Applying to both sides establishes (5).
Now we show the reverse implication. Let be a closed set. Then by (5),
As any element which is mapped to by must lie in , we have
By (ii), equality holds and thus is closed.