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The starting point of this question is a slight reformulation of the $T_0$ separation axiom: 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 $T^\kappa_0$?

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  • $\begingroup$ Can't you just take the topology on $\lambda$ with basis $\{\kappa\} \cup \{\{\alpha\} : \kappa \leq \alpha < \lambda\}$? $\endgroup$
    – Will Brian
    Commented Oct 11, 2018 at 12:53
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    $\begingroup$ @WillBrian In this way it looks like $\kappa$ itself is not split, so this is not $T^\kappa_0$. What am I missing? $\endgroup$
    – KP Hart
    Commented Oct 11, 2018 at 13:30
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    $\begingroup$ Right -- thanks @KPHart. The space I defined is minimally $T^{\kappa+}_0$. It's not too hard to modify the example to find something minimally $T^\kappa_0$ as well. (I'll type out the details and post them later this morning.) $\endgroup$
    – Will Brian
    Commented Oct 11, 2018 at 13:51
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    $\begingroup$ If $\kappa$ is a limit take a cofinal set, $A$, of cardinals and have $A\cup\{\{\alpha\}:\kappa\le\alpha<\lambda\}$ as a base? $\endgroup$
    – KP Hart
    Commented Oct 11, 2018 at 14:01
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    $\begingroup$ I'd just like to point out that the following definition of $T_0$ is more sensible: a space is $T_0$ if any two points that have the same neighborhoods are equal. $\endgroup$
    – Andrej Bauer
    Commented Oct 23, 2018 at 21:43

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Fleshing out Will Brian's suggestion (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 $\kappa$.

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