Let $X$ be a locally compact Hausdorff space and suppose that $X$ can be written as the disjoint union of countably many nonempty closed subsets. Is at least one of the subsets clopen?

$\begingroup$ @YCor, I think the OP means infinitely many closed subsets. $\endgroup$– Ramiro de la VegaApr 12, 2018 at 14:50

1$\begingroup$ @RamirodelaVega thanks (this addressed an erased comment about the segment). Actually, Sierpinski proved in 1918 that a connected compact space never has a partition into countably infinitely many closed subsets. $\endgroup$– YCorApr 12, 2018 at 15:16

$\begingroup$ @YCor, do you have a reference for the Sierpiński result (maybe even in a textbook, rather than the original paper)? $\endgroup$– LSpiceApr 12, 2018 at 16:29

1$\begingroup$ @LSpice W. Sierpinski, Un théorème sur les continus, Tôhoku Math. J. 13 (1918), 300303. jstage.jst.go.jp/article/tmj1911/13/0/13_0_300/_article (He's quoted in matwbn.icm.edu.pl/ksiazki/fm/fm142/fm14216.pdf with this result. Actually the 1918 result is stated for a connected compact subset of a Euclidean space) $\endgroup$– YCorApr 12, 2018 at 16:31
1 Answer
Let $X$ be a countable compact Hausdorff space consisting of a unique accumulation point $0$ with discrete complement $X'$. Let $j$ be a bijection $\mathbf{N}\to X'\times\mathbf{N}$.
For $n\in\mathbf{N}$, write $F_n=\{(0,n),j(n)\}$. This is a 2element subset of $X\times\mathbf{N}$. Then $(F_n)_{n\in\mathbf{N}}$ is a countable closed partition of the Hausdorff locally compact space $X\times\mathbf{N}$, and none of the $F_n$ is clopen.
(If we 1point compactify, and add $F_\infty=\{\infty\}$, we even have a compact example, homeomorphic to $X^2$.)

2$\begingroup$ Equivalently, this is the ordinal $\omega^2$ with the order topology. It contains countably many limit ordinals and countably many successors, so you can partition it into sets $F_n$ containing one of each. The compactification is $\omega^2+1$. $\endgroup$ Apr 12, 2018 at 16:46

$\begingroup$ Yep, homeomorphic copies of this space and partition appear in zillions of ways. $\endgroup$– YCorApr 12, 2018 at 16:48

$\begingroup$ The quotient space would seem to have every point a limit point and to be nowhere first countable. Is this the same as the ArhangelskiiFranklin example (which came to mind after I had posed the question)? $\endgroup$ Apr 12, 2018 at 17:26

$\begingroup$ Ah, I didn't think of the quotient, but indeed it can't be Hausdorff because it's (quasi)compact, countable, with no isolated point. I don't know the ArhangelskiiFranklin example. I didn't even know that there could be countable spaces that are not firstcountable; I haven't thought about this one. $\endgroup$– YCorApr 12, 2018 at 18:15

$\begingroup$ This is the ArhnangelskiiFranklin paper projecteuclid.org/euclid.mmj/1029000034 $\endgroup$ Apr 12, 2018 at 21:42