Let $X$ be a topological space. Call a partition $\pi$ of $X$ compatible with the topology, or just compatible, if the closure of each block is a union of blocks (in other words, closures of blocks are saturated with respect to $\pi).$

For example, a compatible finite partition $\pi$ of $\mathbb{R}$ is used here to represent the Kuratowski closure-complement problem in $\mathbb{R}$ as a problem in the finite quotient space $\mathbb{R}/\pi.$

Question: Does every finite partition $\pi$ of an arbitrary topological space $X$ have a compatible finite refinement?

A while back I conjectured here that every connected finite space $X$ is homeomorphic to $\mathbb{R}/\pi$ for some compatible partition $\pi$ of $\mathbb{R}.$ I verified in an answer that this holds for all connected $X$ such that $|X|\leq5.$


1 Answer 1


No. For instance, let $X=\mathbb{N}$ with the topology such that $C$ is closed iff $x\in C$ implies $y\in C$ for all $y\geq x$. Consider the partition $\pi$ consisting of the even numbers and the odd numbers. Given any finite refinement $\rho$ of $\pi$, let $A$ be the block of $\rho$ with the largest least element; let $n$ be the least element of $A$. Then $n+1\in\overline{A}$, but $n+1\not\in A$ since $n+1$ and $n$ have different parity. Let $B\in\rho$ be such that $n+1\in B$. By our choice of $A$, $n+1$ cannot be the least element of its block $B$, so there is some $m\in B$ such that $m<n$. But then $m\not\in\overline{A}$, so $B$ intersects $\overline{A}$ but is not contained in it. Thus $\rho$ is not compatible with the topology.

  • $\begingroup$ Obvious as this is, it might be good to write "$n+1$ cannot be the least element of its block $B$" so the reader doesn't wonder what $B$ is. $\endgroup$ May 13, 2019 at 18:15

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