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Can any totally disconnected locally compact Hausdorff space be written as a disjoint union of subsets that are both compact and open?

If this is true, does anyone know of a good reference?

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  • $\begingroup$ If you have a group, then yes. $\endgroup$ – Marc Palm Dec 30 '13 at 11:09
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The set $\omega_1$ of countable ordinals with the order topology is a totally disconnected locally compact Hausdorff space which can not be written as a disjoint union of subsets that are both compact and open. This follows from the fact that the space is sequentially compact but not compact.

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I claim that every paracompact locally compact totally disconnected space can be written as a disjoint union of compact open sets.

Every paracompact locally compact space can be partitioned into a collection of $\sigma$-compact open sets (for a proof of this result see my answer to another question here).

Take note that every locally compact totally disconnected space is zero-dimensional (i.e. Every locally compact totally disconnected space has a basis of clopen sets. See this answer for a proof).

Suppose that $X$ is a locally compact paracompact totally disconnected space. Then $X$ can be written as a disjoint union $\bigcup_{i\in I}X_{i}$ of $\sigma$-compact open sets. Since each $X_{i}$ is $\sigma$-compact, the sets $X_{i}$ can be written as countable unions $\bigcup_{n}C_{i,n}$ of compact clopen sets. If $D_{i,n}=C_{i,n}\setminus\bigcup_{m<n}C_{i,m}$, then $D_{i,m}$ is a partition of $X_{i}$ into compact clopen sets. Therefore $\{D_{i,n}|i\in I,n\in\mathbb{N}\}$ is a partition of $X$ into compact clopen sets.

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