# $\sigma$-algebra generated by open sets coincides with $\sigma$-ring generated by open sets.

Under the topic of Metric spaces in my measure theory book I came across this definition that says: "Denote by $B$ the $\sigma$-ring generated by the class of all the open sets of X. The sets of $B$ are called Borel-sets."

"Note that $B$ coincides with the $\sigma$-algebra generated by the open sets of X."

I saw a nice example about a sigma-algebra being generated, but this note I don't really get, also maybe because I'm not sure how to generate a sigma-ring.

Thanks for any hint or tip!

Let $B$ denote the $\sigma$-ring generated by the open sets and $C$ the $\sigma$-algebra generated by them. Obviously $B \subseteq C$, it suffices to show that $X \in B$, as a $\sigma$-ring containing the base set is a $\sigma$-algebra. But $X$ is open.
• thanks, that made it more clear. One follow up: what do you mean by $\sigma$-ring containing the "base set" is a $\sigma$- algebra. what is meant by base set. Commented Oct 7, 2013 at 17:04
• A $\sigma$-algebra on a set $X$ is ... I didn't find a better name for $X$ Commented Oct 7, 2013 at 17:10