Preclosure operator

In topology, a preclosure operator or Čech closure operator is a map between subsets of a set, similar to a topological closure operator, except that it is not required to be idempotent. That is, a preclosure operator obeys only three of the four Kuratowski closure axioms.

A preclosure operator on a set ${\displaystyle X}$ is a map ${\displaystyle [\ \ ]_{p}}$

${\displaystyle [\ \ ]_{p}:{\mathcal {P}}(X)\to {\mathcal {P}}(X)}$

where ${\displaystyle {\mathcal {P}}(X)}$ is the power set of ${\displaystyle X.}$

The preclosure operator has to satisfy the following properties:

1. ${\displaystyle [\varnothing ]_{p}=\varnothing \!}$ (Preservation of nullary unions);
2. ${\displaystyle A\subseteq [A]_{p}}$ (Extensivity);
3. ${\displaystyle [A\cup B]_{p}=[A]_{p}\cup [B]_{p}}$ (Preservation of binary unions).

The last axiom implies the following:

4. ${\displaystyle A\subseteq B}$ implies ${\displaystyle [A]_{p}\subseteq [B]_{p}}$.

A set ${\displaystyle A}$ is closed (with respect to the preclosure) if ${\displaystyle [A]_{p}=A}$. A set ${\displaystyle U\subset X}$ is open (with respect to the preclosure) if its complement ${\displaystyle A=X\setminus U}$ is closed. The collection of all open sets generated by the preclosure operator is a topology;¹ however, the above topology does not capture the notion of convergence associated to the operator, one should consider a pretopology, instead.²

Given ${\displaystyle d}$ a premetric on ${\displaystyle X}$, then

${\displaystyle [A]_{p}=\{x\in X:d(x,A)=0\}}$

is a preclosure on ${\displaystyle X.}$

The sequential closure operator ${\displaystyle [\ \ ]_{\text{seq}}}$ is a preclosure operator. Given a topology ${\displaystyle {\mathcal {T}}}$ with respect to which the sequential closure operator is defined, the topological space ${\displaystyle (X,{\mathcal {T}})}$ is a sequential space if and only if the topology ${\displaystyle {\mathcal {T}}_{\text{seq}}}$ generated by ${\displaystyle [\ \ ]_{\text{seq}}}$ is equal to ${\displaystyle {\mathcal {T}},}$ that is, if ${\displaystyle {\mathcal {T}}_{\text{seq}}={\mathcal {T}}.}$