# closed decomposition of locally compact Hausdorff space

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

• @YCor, I think the OP means infinitely many closed subsets. Apr 12, 2018 at 14:50
• @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.
– YCor
Apr 12, 2018 at 15:16
• @YCor, do you have a reference for the Sierpiński result (maybe even in a textbook, rather than the original paper)? Apr 12, 2018 at 16:29
• @LSpice W. Sierpinski, Un théorème sur les continus, Tôhoku Math. J. 13 (1918), 300--303. 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)
– YCor
Apr 12, 2018 at 16:31

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 2-element 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 1-point compactify, and add $F_\infty=\{\infty\}$, we even have a compact example, homeomorphic to $X^2$.)

• 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 $\omega^2+1$. Apr 12, 2018 at 16:46
• Yep, homeomorphic copies of this space and partition appear in zillions of ways.
– YCor
Apr 12, 2018 at 16:48
• The quotient space would seem to have every point a limit point and to be nowhere first countable. Is this the same as the Arhangelskii-Franklin example (which came to mind after I had posed the question)? Apr 12, 2018 at 17:26
• 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 Arhangelskii-Franklin example. I didn't even know that there could be countable spaces that are not first-countable; I haven't thought about this one.
– YCor
Apr 12, 2018 at 18:15
• This is the Arhnangelskii-Franklin paper projecteuclid.org/euclid.mmj/1029000034 Apr 12, 2018 at 21:42