3
$\begingroup$

Let $\Omega$ be a locally compact Hausdorff set. Is the sigma algebra generated by compact sets is the same as the sigma algebra generated by open sets?

$\endgroup$
9
$\begingroup$

No. For example, take any non-countable infinite set with the discrete topology. Then the $\sigma$-algebra generated by open sets is all of $\cal P (\Omega)$, but the compact subsets are just the finite subsets, so the $\sigma\text{-algebra}$ generated by compact sets is the collections of countable and countable-complement subsets.

$\endgroup$
  • $\begingroup$ Yes, absolutely right. $\endgroup$ – Ali Bagheri Jul 16 '16 at 13:50
2
$\begingroup$

Yes if, say, $\Omega$ is $\sigma$-compact. No in general.

For example let $\Omega = \omega_1$, the set of countable ordinals. If $K$ is compact then $K$ is bounded above (otherwise the intervals $[0,\alpha)$ would be an open conver with no finite subcover). So any compact set is countable. The class of sets $E$ such that either $E$ or $E^c$ is countable is a $\sigma$-algebra, which hence contains the $\sigma$-algebra generated by the compact sets.

But $V=\{\alpha+2:\alpha\in\omega_1\}$ is an uncountable open set such that $V^c$ is also uncountable (the successor of any limit ordinal is in $V^c$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.