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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?

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    $\begingroup$ You could more easily observe that each $D_n=\cup_{i=1}^n \cup_{j=1}^nC_{i,j}$ is closed, and that $\cup_{n=1}^{\infty}D_n=\cup_{n=1}^{\infty}F_n.$ $\endgroup$ Apr 28, 2017 at 4:39
  • $\begingroup$ "$x$ is closed": This statement makes no sense, as points are not sets. All the $C_{j,k}$ are closed and that's all you know. It's not clear what you're trying to prove when you start $x$ in the complement of the union of unions.... By the definions $\cup_n F_n = \cup_n \cup_k C_{n,k}$ is true. What you have to show is that the latter can also be written as a countable union of closed sets. @Janitha357 has the simplest way to se this IMHO. $\endgroup$ Apr 29, 2017 at 5:30

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Note that a countable union of a countable union is a countable union.

Now adhering to your notation, you just need to prove that $\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty C_{n,k}=\bigcup_{(n,k)\in\mathbb{N\times N}}C_{n,k}$.

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  • $\begingroup$ And the only real set theory needed besides this (almost trival) equality is that $\mathbb{N} \times \mathbb{N}$ is countable as well. @Daniel $\endgroup$ Apr 29, 2017 at 5:21
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It's very simple. The countable union of a countable union is a countable union. So then a countable union of a countable union of closed sets is a countable union of closed sets.

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    $\begingroup$ I'm not sure I can use that given the theorems in my text (not that I doubt you). What I can use is that the countable union of countable sets is countable, but I'm not sure it makes sense to treat the countable union of closed sets as a countable set. $\endgroup$
    – user413923
    Apr 27, 2017 at 23:41
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    $\begingroup$ Hey @Daniel. I'm confused. Have you dealt with countability yet? If so, you'll have learned that the countable union of countable sets is countable right? Break it up into pieces: countable union of $F_\sigma$ sets is the countable union of the countable union of closed sets. When you see countable union of countable, you should just think countable. $\endgroup$ Apr 28, 2017 at 1:17

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