I have interest in Kuratowski closure axioms for topology. I would like to know how to define Hausdorff space using only closure operator such that it is the same to definition of Hausdorff space of equivalent axiomatic framework of topological spaces using open sets.
In this post: Why is a topology made of open sets? there is an answer about Kuratowski closure axioms for topology and the fact that using them many equivalent definitions can be made.
There, user "Vectornaut" states that "(WARNING: I'm kinda rusty at this, so these definitions may not be correct.)" and that "The topological space $X$ is Hausdorff if for any two distinct points $w,x \in X$, there is a subset $A$ of $X$ such that $w$ doesn't touch $A$ and $x$ doesn't touch the complement of $A$." Translating this into closure operator axiomatic system (denoting closure operator by $cl$) we have that $X$ is Hausdorff if for any two distinct points $w,x \in X$ there is a subset $A$ of $X$ such that $w \notin cl(A)$ and $x \notin cl(X-A)$, where $-$ is set difference. Is this statement true? Can someone maybe hint me towards some reference about proving equivalence of this with usual definition?
I know I can prove the following that an equivalent definition is that $X$ is Hausdorff if for any two distinct points $w,x \in X$ there exists an open set $U$ such that $w \in U$ and $x \notin cl(U)$. I guess that equivalent to this is that we require that there exists an arbitrary set $V$ such that $w \in X-cl(V)$ and $x \notin cl(X – cl(V))$. This is the same as saying $w \notin cl(V)$. So I guess the first part about $w$ is exactly like in the definition I am not sure about, but then I would have to somehow show that $cl(X-V) = cl(X – cl(V))$. Is that possible?
I would be happy if you give any other equivalent definition for Hausdorfness that use only closure operator. Thanks!
Best Answer
Yes, the other definition is completely valid : call $X$ "closure Hausdorff" iff $$\forall x,y \in X: x \neq y \implies \left(\exists A \subseteq X: x \notin \operatorname{cl}(A) \land y \notin \operatorname{cl}(X-A)\right) \tag{cH}$$
If $X$ is cH, it is $T_2$: check that for $x \neq y$ the sets $X-\operatorname{cl}(A)$ is an open neighbourhood of $x$ and $X-\operatorname{cl}(X-A)$ one for $y$ that are disjoint, as
$$(X - \operatorname{cl}(A)) \cap (X-\operatorname{cl}(X-A)) = X-(\operatorname{cl}(A) \cup \operatorname{cl}(X-A)) = X-\operatorname{cl}(A \cup (X-A)))= X-\operatorname{cl}(X)=\emptyset$$
by de Morgan and laws for closures.
On the other hand, if $X$ is $T_2$ and $x \neq y$, pick disjoint open neighbourhoods $U_x$ of $x$ and $U_y$ of $y$ and note that $A=U_y$ is as required for cH: $U_x$ witnesses that $x \notin \operatorname{cl}(A)$ and as $X-A \subseteq U_x$ and $y \notin \operatorname{cl}(U_x)$ is witnessed by $U_y$ we clearly have $y \notin \operatorname{cl}(X-A)$, as required.