How can I prove that $\mathbb R$ contains no more then $\mathfrak c$ $F_\sigma$ sets

Question

How can I prove that $\mathbb R$ contains no more then $\mathfrak c$ $F_\sigma$ sets? (or equivalently, that $\mathbb R$ contains no more then $\mathfrak c$ $G_\sigma$ sets?

The more general argument that I am trying to prove is that $cof(\mathcal M) \leq \mathfrak c$ (where $\mathcal M$ is the set of all meager sets in $\mathbb R$ and $cof(\mathcal M)$ is $min\{|\mathcal A||\mathcal A \in \mathcal M \forall B \in \mathcal M,\exists C \in \mathcal A(B \subseteq C)\}$)

Thank you!

Answer 1

Here's the general outline:

1. There are $\frak c$ open sets.
2. Each $G_\delta$ is an intersection of a sequence of open sets, there are $\mathfrak c^{\aleph_0}=\frak c$ of those.
3. Every open set is $G_\delta$, so there are at least $\frak c$ of those.
4. Every $F_\sigma$ is a complement of $G_\delta$. (Optional)