Why is the sigma algebra generated by the set of all closed subsets a subset of the Borel sigma algebra on $\mathbb{R}$?

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I have that, by definiton, the Borel $\sigma$-algebra (denoted B($\mathbb{R}$)) is generated by the set of open subsets of $\mathbb{R}$. Now I want to prove that the set can also be generated by the sets:

a) the set of closed subsets of $\mathbb{R}$,

b) {$(-\infty,b]: b \in \mathbb{R}$},

c) {$(a,b]: a,b \in \mathbb{R}$}.

Let $A, B$ and $C$ be the respective $\sigma$-algebras of a), b) and c).

I am at the stage that I can prove that the Borel algebra can be generated by the set a), what I can't seem to prove is that $A \subseteq B(\mathbb{R})$.

Any help here would be appreciated.

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Let $A$ be the sigma algebra generated by the closed sets. If $C$ is closed in $\mathbb{R}$, then $C^c$ is open so that $C^c\in B(\mathbb{R})$ and thus $C\in B(\mathbb{R})$. Thus $B(\mathbb{R})$ contains the closed sets. But $A$ is the smallest sigma algebra containing the closed sets. Hence $$ A\subset B(\mathbb{R}). $$