general topology on set of neighbourhoods

Question

Let $X$ be a set and Let $M\rightarrow \overline M$, be mapping of set of all neighbourhoods onto itself such that 1) $\overline \phi=\phi$, 2) for every $M\subset X$ we have $M\subset \overline M$, 3) for every $M\subset X$, we have $\overline{\overline M}=\overline M$, 4) for all $M\subset X$ and $N\subset X$ we have $\overline {M\cup N}=\overline M \cup \overline N$. Now show that there is unique topology on $X$ such that $\overline M$ is the closure of $M$ with respect to this topology for all $M\subset X$. (define topology by means of its closed set.) . this looks very simple but i am unable to put it in correct words. help required.

Answer 1

It is Kuratowski closure axioms. Let $\tau = \{C \mid \overline C = C\}$ then $X \in \tau$ by (2) and $\emptyset \in \tau$ by (1). By (4), $A \in \tau \land B \in \tau \implies A \cup B \in \tau$. Also, if $A \subset B, \overline B = \overline{A \cup B} = \overline A \cup \overline B$ so $\overline A \subset \overline B$. In other words, it preserves inclusion. For any subfamily $S$ of $\tau$, $$ \bigcap_{A \in S} \overline{A} = \bigcap_{A \in S} A \subset \overline{\bigcap_{A \in S} A} \subset \bigcap_{A \in S} \overline{A}$$ so $\kappa = \{X \setminus C \mid \overline C = C\}$ is a topology on $X$.(The second inclusion comes from the fact that it preserves inclusion.)

Now we claim that $\operatorname{cl}(A) = \overline A$. $\operatorname{cl}(A) \subset \overline A$ by (3) and $\overline A \subset \overline{\operatorname{cl}(A)} = \operatorname{cl}(A)$

Answer 2

Define a set $A$ to be closed iff $\overline{A}=A$.

We check the axioms for closed sets:

(i) $\emptyset$ is closed by 1). $X$ is closed as $X\subseteq \overline{X}(\subseteq X)$ by 2).

(ii) The union of two closed sets is closed by 4): $C,D$ closed means $\overline{C}=C$ and $\overline{D}=D$, then $\overline{C \cup D}= \overline{C} \cup \overline{D} = C \cup D$ and so $C \cup D$ is closed too. Finite unions follow by induction.

To see intersections, first prove

$A \subseteq B \implies \overline{A} \subseteq \overline{B}$ for all $A,B \subseteq X$.

$A \subseteq B$ iff $B = A \cup B$ which implies by 4) that $\overline{B} = \overline{A} \cup \overline{B}$ iff $\overline{A} \subseteq \overline{B}$. QED

Now if $C_i, i \in I$ are closed, define $C= \bigcap_{i \in I} C_i$. Because $C \subseteq C_i$ for all $i$ the previous fact implies $\overline{C} \subseteq \overline{C_i}$ for all $i$ which implies $\overline{C} \subseteq \bigcap_{i \in i} \overline{C_i} = \bigcap_{i \in i} C_i = C$. Also, $C \subseteq \overline{C}$ follows by (2). So together we have $\overline{C} = C$ and the intersection of any family of closed sets is closed.

This shows the closed sets that are so defined form the closed sets of a topology $\mathcal{T}$. (the complements of the closed sets).

Now we just have to show that $\operatorname{Cl}(A)$ the closure operation of this topology coincides with $\overline{A}$ for all subsets $A$ of $X$.

I'll use the definition $\operatorname{Cl}(A)$ as the smallest closed set that contains $A$, i.e. $$\operatorname{Cl}(A)=\bigcap\{ B : A \subseteq B, B \text{ closed in } \mathcal{T}\}$$

Now let $A \subseteq X$ be arbitrary. Axiom 3) tells us that $\overline{A}$ is closed in $\mathcal{T}$ and 2) tells us that $A \subseteq \overline{A}$. So $\overline{A}$ is one of the closed sets we are taking the intersection of in the definition of $\operatorname{Cl}(A)$, so $\operatorname{Cl}(A) \subseteq \overline{A}$.

On the other hand, if $B$ is any of the sets in the intersection defining $\operatorname{Cl}(A)$, so $B$ closed (i.e. $\overline{B}=B$) and $A \subseteq B$ then the aforementioned fact tells us that $\overline{A} \subseteq \overline{B}=B$. As $B$ was an arbitrary element of that intersection it follows that $\overline{A} \subseteq \operatorname{Cl}(A)$ and we indeed have the desired equality.