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If $X\neq \varnothing$ is a set we say that ${\frak P} \subseteq {\cal P}(X)$ is a partition of $X$ if

  1. $\bigcup{\frak P} = X$, and
  2. $P\neq Q \in {\frak P} \implies P\cap Q = \varnothing$.

Let $H = (V,E)$ be a hypergraph with $V \neq \varnothing$ and $\bigcup E = V$. A partition ${\frak P}$ of $V$ is said to be splitting if for all $e\in E$ and $P \in {\frak P}$ we have $|e \cap P| = 1$, and we call such a hypergraph admitting a splitting partition partite.

It is easy to see that every edge in a partite hypergraph $H$ has the same cardinality $\kappa$, and that $\kappa$ is also the cardinality of every splitting partition of $H$.

Let $\tau$ be the Euclidean topology on $\mathbb{R}$. It is easy to see that $(\mathbb{R}, \tau\setminus\{\varnothing\})$ cannot have a splitting partition since $\mathbb{R} \in \tau$.

Question. Is there a subbase ${\cal S}\subseteq (\tau\setminus\{\varnothing\})$ such that $(\mathbb{R}, {\cal S})$ is partite?

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    $\begingroup$ The restriction $X\neq\emptyset$ is non-standard, nevertheless it is usual to assume that components of a partition are nonempty. For instance for $X=\{1,2,3\}$ you currently allow both $\{\{1\},\{2,3\}\}$ and $\{\emptyset, \{1\},\{2,3\}\}$ as distinct partitions. Probably you want to remove $X\neq\emptyset$, and add $P\in\mathfrak{P}$ $\Rightarrow$ $P\neq\emptyset$ in the axioms. $\endgroup$
    – YCor
    Commented May 7, 2020 at 20:36
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    $\begingroup$ A near miss: $\mathcal S$ is all open intervals of length $1$, and $\mathfrak{P}$ is all shifts of $\mathbb Z$. $\endgroup$
    – Will Brian
    Commented May 8, 2020 at 0:19
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    $\begingroup$ May I ask what the motivation for your question is? and why do you ask for a subbase and not for a base? $\endgroup$
    – Ramiro de la Vega
    Commented May 8, 2020 at 1:15
  • $\begingroup$ A base would contain (an infinite chain of) sets nested by inclusion, but a partite hypergraph has no pair of nested edges. $\endgroup$
    – Jan Kyncl
    Commented May 8, 2020 at 1:25
  • $\begingroup$ If it can help, it is not too difficult to see that any $P$ in a possible $\mathfrak P$ would be closed discrete, and there would be for all $[-M,M]$ an $\varepsilon>0$ (depending only on $M$) such that the distance between two points of $P\cap [-M,M]$ would be at least $\varepsilon$ (the distance between two consecutive points of $P$ is locally bounded below, uniformly in $P$). $\endgroup$
    – Pierre PC
    Commented May 8, 2020 at 1:55

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