Assume the context of a first undergraduate course in real analysis, i.e. all closed sets are closed sets of the real line.
Let $\{F_n: n \in \mathbb{N}\}$ be a countable collection of $F_\sigma$ sets.
Then for each $F_j$ , there exists a countable collection of closed sets C$_{j,k}$ s.t. $\bigcup\limits_{k=1}^{\infty} C_{j,k} = F_j$.
Then $\bigcup\limits_{n=1}^{\infty} ( \bigcup\limits_{k=1}^{\infty} C_{n,k} )$ = $\bigcup\limits_{n=1}^{\infty} F_{n}$
Now suppose x $\in$ ($\bigcup\limits_{n=1}^{\infty} ( \bigcup\limits_{k=1}^{\infty} C_{n,k} )$) $^c$, then x $\notin$ $\bigcup\limits_{n=1}^{\infty} ( \bigcup\limits_{k=1}^{\infty} C_{n,k} )$
which implies $\forall$ n $\in$ N, x $\notin$ {F$_n$} = $\bigcup\limits_{k=1}^{\infty} C_{n,k}$,
implies $\forall$ k $\in$ N, x $\notin$ C$_{k,n}$.
We conclude that x is not a closed set.
Therefore, $\forall$ x $\in$ $\bigcup\limits_{n=1}^{\infty} F_{n}$, x is closed.
Thus, $\bigcup\limits_{n=1}^{\infty} F_{n}$is an F$_\sigma$ set.
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There's probably a lot wrong with this proof. Any obvious ways to improve/fix it?
Best Answer
Note that a countable union of a countable union is a countable union.
Now adhering to your notation, you just need to prove that $\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty C_{n,k}=\bigcup_{(n,k)\in\mathbb{N\times N}}C_{n,k}$.