Is the random graph ideal an $F_\sigma$-ideal?

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Let $\{X_n: n \in \omega\}$ be an independent family of subsets of $\omega$ such that $n\in X_m$ iff $m\in X_n$, for all $n,m \in \omega$. The random graph can be defined by $\mathcal{R}=(\mathbb{N}, E)$ where $E:=\{\{n,m\} : n \in X_m\}$.

The random graph ideal $I_{\mathcal{R}}$ is the ideal generated by cliques and free sets in the random graph, i.e., $A\in I_{\mathcal{R}}$ if and only if there exists $A_1, \dots, A_n$ sets homogeneous for $c:[\omega]^2 \rightarrow \{0,1\}$ defined by $c(\{n,m\})=0$ iff $\{n,m\} \in E$ such that $A \subseteq \bigcup_{i=1}^n A_i$.

My question is: Is $I_{\mathcal{R}}$ an $F_\sigma$ subset of $2^\omega$?

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I managed to proof something stronger: If $c:[\omega]^2 \rightarrow \{0,1\}$ is a coloring then the ideal $I_{hom(c)}$, the ideal generated by $hom(c)$ the $c$-homogeneous sets, is a $F_{\sigma}$ subset of $2^\omega$.

For $n \in \omega$ we can defined $F_0=hom(c)$ and

$$F_{n+1}=\left\{B \subseteq \omega: B \subseteq A_1 \cup A_2 \,\, where\,\, A_i \in F_n \right\}=\bar\cup(F_n \times F_n)$$

where $\bar \cup:2^{\omega} \times 2^\omega \rightarrow 2^\omega$ defined by $(A, B) \mapsto A \cup B$. It is clear that $$I_{hom(c)}=\bigcup_{n \in \omega} F_n$$ and $F_n$ is closed because $hom(c)$ is a closed subset of $2^\omega$ and $\bar{\cup}$ is a closed function.