# Generalizing the $T_0$-axiom

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 $$T^\kappa_0$$?

• Can't you just take the topology on $\lambda$ with basis $\{\kappa\} \cup \{\{\alpha\} : \kappa \leq \alpha < \lambda\}$? – Will Brian Oct 11 '18 at 12:53
• @WillBrian In this way it looks like $\kappa$ itself is not split, so this is not $T^\kappa_0$. What am I missing? – KP Hart Oct 11 '18 at 13:30
• 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.) – Will Brian Oct 11 '18 at 13:51
• If $\kappa$ is a limit take a cofinal set, $A$, of cardinals and have $A\cup\{\{\alpha\}:\kappa\le\alpha<\lambda\}$ as a base? – KP Hart Oct 11 '18 at 14:01
• 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. – Andrej Bauer Oct 23 '18 at 21:43

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 $$\kappa$$.