0
$\begingroup$

In P Halmos measure theory book it written

A Sigma ring S generated by C, the class of compact subsets of X where X is locally compact Hausdorff Space. Then S has open set and sets of S is called Borel Sets.

I am very confused. How a sigma ring generated by the class of compact set has open set.

P Halmos Measure theory book:

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Looking at the linked excerpt, it appears you may be misinterpreting Halmos here. He is not claiming that $\mathbf S$ contains every open set. He is specifically defining $\mathbf U$ to be the class of open sets contained in $\mathbf S$, so clearly he anticipates not every open set necessarily belongs to $\mathbf S$.

Now, certainly there are some open sets in $\mathbf{S}$, and in fact you can conclude that every relatively compact open set is in $\mathbf{S}$, since such a set's boundary and closure lie in $\mathbf{S}$.

If the claim were that all open sets are in $\mathbf S$, then this would indeed be false without additional countability assumptions. However, that does not appear to be what Halmos is claiming.


Remark

To see that in general, we can't expect $\mathbf S$ to contain all open sets, let $\omega_1$ be the first uncountable ordinal (aka, the set of countable ordinals), equipped with the order topology. Then $\omega_1$ is locally compact Hausdorff, (for each $\alpha\in \omega_1$, $[0,\alpha]$ is a compact neighborhood), yet compact subsets of $\omega_1$ must be bounded, since if $A\subseteq \omega_1$ is unbounded, then the open cover $\{[0,\alpha]\mid \alpha\in A\}$ has no finite subcover.

Therefore each compact set is countable, and so every member of $\mathbf S$ is countable. In particular, $\mathbf S$ cannot contain any unbounded open set, such as $\omega_1$ itself, for example.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .