If $c:\mathcal P(X) \to \mathcal P(X)$ is a closure operator on $X$, then the set $\tau=\{U \in \mathcal P(X) : c(X \setminus U)=X \setminus U\}$ is a topology on $X$.

First let me write the properties of a closure operator

1) $c(\emptyset)=\emptyset$,

2) if $A \in \mathcal P(X)$, then $A \subset c(A)$,

3)if $A \in \mathcal P(X)$, then $c(c(A))=c(A)$,

4) if $A, B \in \mathcal P(X)$, then $c(A \cup B)=c(A) \cup c(B)$

Using these axioms, I could show $X \in \tau$ and finite intersection of elements in $\tau$ remains in $\tau$.

I couldn't show $\emptyset \in \tau: c(X \setminus \emptyset)=c(X)$, now I am not so sure which axioms could I use to prove $c(X)=X$.

And for arbitrary union of elements in $\tau$, I have to prove that $c(X \setminus \bigcup_{i \in I} U_i)=X \setminus \bigcup_{i \in I} U_i$. The inclusion $X \setminus \bigcup_{i \in I} U_i \subset c(X \setminus \bigcup_{i \in I} U_i)$ is satisfied by 2), how could I prove the other inclusion?

I would appreciate any help.


To prove $c(X) = X$, note that by 2, $X \subseteq c(X)$. $c : \mathcal{P}(X) \rightarrow \mathcal{P}(X)$. Thus $c(X)$ is a subset of $X$. The only subset of $X$ that contains $X$ is $X$ itself. Hence $c(X) = X$.

For arbitrary union, let $V_i = X \backslash U_i$. Note that $X - \bigcup_{i \in i} U_i = \bigcap V_i$. For each $j \in I$, $\bigcap_{i \in \omega} V_i \subseteq V_j$. Hence by 2, for all $j$, $c(\bigcap_{i \in \omega} V_i) \subseteq c(V_j) = V_j$. Hence $c(\bigcup_{i \in I}{V_i}) \subseteq \bigcap_{i \in I} V_i$. This is gives the direction you were looking for.

For the above proof, the following property of closure operator is required: if $A \subseteq B$, then $c(A) \subseteq c(B)$.

To prove this, note that $B = (B - A) \cup A$. Hence by 3, $c(B) = c(B - A) \cup c(A) \supseteq c(A)$.

  • $\begingroup$ Thanks for the answer, I got lost from: Hence by (2), for all $j$, $c(\bigcap_{i \in I} V_i) \subset c(V_j)$. (2) says that a set $A$ is always contained in its image $c(A)$, I don't see how this implies what you've affirmed. And then: "Hence c(⋃i∈IVi)⊆⋂i∈IVi." I don't see why that is true. $\endgroup$ – user100106 Aug 14 '14 at 20:42
  • $\begingroup$ @user100106 I forgot one additional property that follows from the closure property: If $A \subseteq B$, then $c(A) \subseteq c(B)$. I added a proof of this fact to the end of my answer. $\endgroup$ – William Aug 14 '14 at 22:30

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