# Kuratowski Closure and Topological Spaces

This post was the result of conversations with Chris Gartland (from several years ago). In what follows we outline an alternate definition of a topological space due to Kuratowski.

We recall Kuratowski’s closure axioms for a set ${X}$. 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 ${X}$. For${A \subset X}$, one may think of the closure of ${A}$, denoted ${\overline{A}}$, to be points that are “close to” ${A}$.

First we recall what is meant by a topological space. We let ${\mathcal{P}(X)}$ be the power set of ${X}$, that is the set of all subsets of ${X}$.

${\ }$ Definition 1: Let ${X}$ be a set and ${T \subset \mathcal{P}(X)}$. Then ${(X,\mathcal{T})}$ is a topology with open sets ${\mathcal{T}}$ if

• (a) ${\emptyset , X \in \mathcal{T}}$,
• (b) if ${A, B \in \mathcal{T}}$, then ${A \cap B \in \mathcal{T}}$,
• (c) if ${A_i \in \mathcal{T}}$ for all ${i \in \mathcal{I}}$, then ${\cup_{i\in \mathcal{I}} A_i \in \mathcal{T}}$.

Furthermore, we say

$\displaystyle \mathcal{C} := \{X \setminus A : A \in \mathcal{T}\},$

are the closed sets of ${X}$. ${\spadesuit}$ ${\ }$

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 ${X}$ be a set and ${\overline{\cdot} : \mathcal{P}(X) \rightarrow \mathcal{P}(X)}$. We say ${\overline{\cdot}}$ is a closure operation if

• (i) ${\overline{\emptyset} = \emptyset}$,
• (ii) if ${A \subset X}$, then ${A \subset \overline{A}}$,
• (iii) if ${D = \overline{A}}$, then ${\overline{D} = D}$,
• (iv) if ${A,B \subset X}$, then ${\overline{A\cup B} = \overline{A} \cup \overline{B}}$. ${\spadesuit}$

${\ }$

As mentioned above, one can informally think of ${\overline{A}}$ as the set of points in ${X}$ which are “close” to ${A}$. We now explain that these two definitions are equivalent. To define a closure operation from Definition 1, we set

$\displaystyle \overline{A} := \bigcap_{A \subset C \ \text{closed}} C.\ \ \ \ \ (1)$

Note that ${\overline{A}}$ is closed by (c). Given a closure operation, we may define the closed sets to be

$\displaystyle \mathcal{C} : = \{\overline{A} : A \subset X\}.\ \ \ \ \ (2)$

${\ }$ Proposition 1: Let ${(X, \mathcal{T})}$ be a topological space as in Definition 1. Then the operation defined in (1) is a closure operation in the sense of Definition 2. ${\spadesuit}$ ${\ }$

Proof: Let

$\displaystyle \mathcal{C} = \{X \setminus A : A \in \mathcal{T}\},$

be the set of closed sets in ${X}$. Let ${A , B \subset X}$. By (1), we have ${A \subset \overline{A}}$ and thus (ii) holds. As ${X}$ is open, ${\emptyset \in \mathcal{C}}$ and so (i) is satisfied. Note for any closed set, ${C \in \mathcal{C}}$, we have ${\overline{C} = C}$ by (ii) and that ${C}$ itself appears on the right hand side of (1). By (c), we have that

$\displaystyle D = \bigcap_{A \subset C \ \text{closed}} C \in \mathcal{C},$

and so ${\overline{D} = D}$. This proves (iii) and so it remains to show (iv). Note that ${\overline{A} \cup \overline{B}}$ is a closed set containing ${A \cup B}$, by (b). Thus

$\displaystyle \overline{A \cup B} \subset \overline{A} \cup \overline{B}.\ \ \ \ \ (3)$

Now let ${C \in \mathcal{C}}$ that contains ${A \cup B}$. Then ${A, B \subset C}$ and so

$\displaystyle A , B \subset \ \bigcap_{(A\cup B) \subset C \ \text{closed}} C.$

This proves the reverse inclusion of (3) and thus of Proposition 1. ${\spadesuit}$

We now show that a closure axiom can be used to define a topology.

${\ }$ Proposition 2: Let ${\overline{\cdot} : \mathcal{P}(X) \rightarrow \mathcal{P}(X)}$ 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. ${\spadesuit}$ ${\ }$

Proof: We implicitly make use of DeMorgan’s law to transition bewtween closed and open sets. By ${(i)}$, we have ${\emptyset \in \mathcal{C}}$ and by (ii), we have ${X \in \mathcal{C}}$. Thus ${(a)}$ is established. Let ${A, B \in \mathcal{C}}$. By (iii), (iv) and (2), we have

$\displaystyle A \cup B = \overline{A \cup B} \in \mathcal{C},$

which establishes (b). Let ${C_i \in \mathcal{C}}$ for all ${i \in \mathcal{I}}$ some index set. Then we have

$\displaystyle \overline{\cap_{i \in \mathcal{I}} C_i}= \cap_{i \in \mathcal{I}} C_i.$

Indeed, the backwards inclusion follows from (ii). To see the forward direction, it is enough to show for any ${i \in \mathcal{I}}$,

$\displaystyle \overline{\cap_{i \in \mathcal{I}} C_i} \subset \overline{C_i} = C_i.$

The second equality is (iii). The first subset inequality follows from (iv) as if ${A \subset B}$ then

$\displaystyle \overline{A} \subset \overline{A} \cup \overline{B} = \overline{A \cup B} = \overline{B}.$

We apply this with ${A = \cup_{i\in \mathcal{I}} C_i}$ and ${B = C_i}$. ${\spadesuit}$

Recall if ${(X , \mathcal{T}_X)}$ and ${(Y , \mathcal{T}_Y)}$ are topological spaces, then

$\displaystyle f : X \rightarrow Y,$

is said to be continuous if the pre-image of every open set in ${Y}$ is open in ${X}$, that is

$\displaystyle f^{-1}(B) \in \mathcal{T}_X,\ \ \ \ \ (4)$

for all open ${Y}$. As ${f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)}$, (5) is equivalent to the analogous definition for closed sets. It turns out that we may equivalently define a map to be continuous if

$\displaystyle f(\overline{A}) \subset \overline{f(A)}.\ \ \ \ \ (5)$

This can be informally interpreted as points that are close to ${A}$ are mapped to points that are close to ${f(A)}$. We now prove this.

${\ }$ Proposition 3: Let ${f : X \rightarrow Y}$ be a function. Then ${f}$ is continuous in the sense of (4) if and only if it is continuous in the sense of (5). ${\spadesuit}$ ${\ }$

Proof: We start with the forward direction. Let ${A \subset X}$. Then by (4), ${f^{-1}(\overline{f(A)})}$ is closed. As ${f(A) \subset \overline{f(A)}}$, we have

$\displaystyle A \subset f^{-1}(\overline{f(A)}).$

Since the right hand side is closed, by (ii) and (iii), we have

$\displaystyle \overline{A} \subset f^{-1}(\overline{f(A)}).$

Applying ${f}$ to both sides establishes (5).

Now we show the reverse implication. Let ${B}$ be a closed set. Then by (5),

$\displaystyle f(\overline{f^{-1}(B)}) \subset \overline{f(f^{-1}(B))} =\overline{B} = B .$

As any element which is mapped to ${B}$ by ${f}$ must lie in ${f^{-1}(B)}$, we have

$\displaystyle \overline{f^{-1}(B)} \subset f^{-1}(B).$

By (ii), equality holds and thus ${f^{-1}(B)}$ is closed. ${\spadesuit}$