Axioms of Continuous Separation
For a $T_1$-space $X$, let $Cld(X)$ denote all its nonempty closed subsets and $T_4(X)=\{(F_1,F_2)\in Cld(X)^2:F_1\cap F_2=\emptyset\}$. In terms of closed sets, $X$ is $T_4$ iff for each $(F_1,F_2)\in T_4(X)$, there is a pair of closed sets $ϕ_1(F_1,F_2)$ and $ϕ_2(F_1,F_2)$ such that their union is $X$ and $F_j\cap ϕ_j(F_1,F_2)=\emptyset$ for $j=1,2$. Thus, in this paper, for a topology $τ$ on $Cld(X)$, we introduce the definition: A $T_1$-space $X$ is called $CT_4$ with respect to $τ$ if the above maps $ϕ_j:T_4(X)\to Cld(X)$ are continuous on $τ$. Similarly, for $i=1,2,3$, we can define a $T_1$-space to be $CT_i$ for $τ$. We only consider the Vietoris topology on $Cld(X)$ and show that every $CT_4$-space is countably compact, and every $CT_3$-space is a Fréchet-Urysohn space, every separable subspace of a $CT_3$-space is metrizable. We give relevant examples. Any finite-dimensional cubes, the infinite-dimensional cube, any finite-dimensional spheres, and all 0-dimensional compact metrizable spaces are $CT_4$. All infinite discrete spaces, all finite-dimensional Euclidean spaces and all countable limit ordinal spaces are $CT_3$ but not $CT_4$. Also, each metrizable space with a unique non-isolated point is $CT_3$, and is $CT_4$ if it is compact. Moreover, the infinite sum of $CT_3$ spaces is $CT_3$ but not $CT_4$. Every countable space with a unique non-isolated point is $CT_2$, and it is $CT_3$ if and only if it is metrizable. All subfields of real numbers and their complement spaces are $CT_2$ but not $CT_4$; and their $CT_3$ status remains unclear. All uncountable ordinal spaces are not $CT_2$. The one-point compactification of any uncountable discrete space is not $CT_2$. $CT_1$ and $T_1$ are equivalent, hence all spaces above are $CT_1$. Open Problems: is there a non-metrizable $CT_3$ or $CT_4$ space? Is any compact $CT_2$ space $CT_3$ or $CT_4$?