Every Open Set is a Borel Set I know this should be simple but I'm having a lot of trouble wrapping my head around it. I also want to prove that every closed set is a borel set but I believe I'll have to use the original proof in my title to prove it.

Reading your comments it seems like you might have a misunderstanding as to what algebra generated by a set means. So let's start from the here:

Let $\Omega$ be an $\sigma$-algebra of a set X then it is defined to be a
collection of subsets of X with the following propreties:

*

*$X,\emptyset\in \Omega$

*If $A   \in \Omega$ then $X-A\in \Omega$

*If $A_i \in \Omega$ then $\bigcup\limits_{i=1}^{\infty} A_i, \bigcap\limits_{i=1}^{\infty} A_i \in \Omega$

If we are talking about the $\sigma$-algebra generated by a collection $\tau$ (any collection of subsets of X) then we are talking about
$$\Omega_\tau = \bigcap_{\Omega\in C(\tau)} \Omega$$
where $$C(\tau)= \{\Omega \text{ }|\text{ } \tau \subseteq \Omega \text{ and } \Omega \text{ is a } \sigma \text{-algebra}\}$$ is the collection of all $\sigma$-algebra of X that contain $\tau$.
Two things to note about this collection.

*

*It is a $\sigma$-algebra

*That the way it is defined implies that
$$\tau \subseteq \Omega\subseteq \Omega_\tau$$
Thus, almost by definition, every $\sigma$-algebra generated by a collection $\tau$ contains all sets in $\tau$.

Notes:

*

*Borel algebras are just a special case of this generated $\sigma$-algebras where the collection $\tau$ is made to be the topology of X.


*Another way of thinking about the $\sigma$-algebra generated by $\tau$ is to think of it as the smallest $\sigma$-algebra containing $\tau$
Example:
$$X = \{1,2,3,4,5\}$$
$$\tau = \{ \{1\} \}$$
$$\Omega_\tau = \{ X,\emptyset, \{1\}, \{2,3,4,5\}\}$$
