# Sigma ring generated by the class of compact subsets of X

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:

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.
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.