The starting point of this question is a slight reformulation of the [$T\_0$ separation axiom](https://en.wikipedia.org/wiki/Kolmogorov_space): A topological space $(X,\tau)$ is $T\_0$ if for all $x\neq y\in X$ there is a set $U\in \tau$ such that $$\{x,y\}\cap U \neq \emptyset \text{ and } \{x,y\}\not\subseteq U.$$


Given a cardinal $\kappa \geq 2$, we say that a space $(X,\tau)$ is $T^{\kappa}\_0$ if for all subsets $S\subseteq X$ with $|S|=\kappa$ there is a set $U\in \tau$ such that $U$ "splits" $S$, or more formally $$S\cap U \neq \emptyset \text{ and } S\not\subseteq U.$$ Obviously, if $\lambda\geq \kappa\geq 2$ and if $(X,\tau)$ is $T^\kappa\_0$, then $X$ is also $T^\lambda\_0$. We say, the space $(X,\tau)$ is *minimally* $T^\kappa\_0$ if it is $T^\kappa\_0$, but for all cardinals $\alpha<\kappa$ with $\alpha\geq 2$, the space $(X,\tau)$ is not $T^\alpha\_0$.


**Question.** Given cardinals $\lambda\geq\kappa\geq 2$, is there a topological space $(X,\tau)$ such that $|X|=\lambda$ and $(X,\tau)$ is minimally $T^\kappa\_0$?



Generalizing the $T_0$-axiom
Fleshing out [Will Brian](https://mathoverflow.net/users/70618/will-brian)'s [suggestion](https://mathoverflow.net/questions/312566/generalizing-the-t-0-axiom/312591#comment779594_312566) (and saving him the trouble of writing it up):


If $\kappa$ is finite you topologize $\lambda$ using the base $[\kappa-1,\lambda)$ (which is a perverted way of listing all the initial of $\lambda$ that contain $\kappa-1$. The initial segment $\kappa-1$ ensures that there is a set of size $\kappa-1$ that is not split; it and the other initial segments help to split all sets split all sets of size $\kappa$ or more.


If $\kappa=\omega$ you have to be a bit more careful: take $\{2^n:n\in\omega\}\cup[\omega,\lambda)$. The $2^n$ are needed to split all infinite subsets of $\omega$ and they are spread out enough to ensure unsplit subsets of arbitrary large finite cardinality. The rest ensures every infinite set is split.


If $\kappa$ is an infinite successor cardinal, say $\kappa=\mu^+$ then Will Brian's base works: $\{\mu\}$ together with all singletons above $\mu$. Continuing the perverse streak: $[\mu,\lambda)$ works too. In either case the set $\mu$ (and its subsets) is unsplit, everything with points above $\mu$ is split.


If $\kappa$ is a limit cardinal then one can let $A$ be the set of cardinals below $\kappa$ and use $A\cup[\kappa,\lambda)$ as a base. For every cardinal $\mu<\kappa$ the interval $[\mu,\mu^+)$ is unsplit; all subsets of $\kappa$ of cardinality $\kappa$ are split by the members of $A$.the rest ensure splitting of anything with points above $\kappa$. 


