Does every finite partition of a topological space have a finite refinement such that the closure of every block is a union of blocks?

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 $$|X|\leq5.$$

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 $$m. But then $$m\not\in\overline{A}$$, so $$B$$ intersects $$\overline{A}$$ but is not contained in it. Thus $$\rho$$ is not compatible with the topology.
• 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. May 13, 2019 at 18:15