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Friends

I was just curious while reading about $G_{\delta}$ and $F_{\sigma}$ sets,

Where $G_{\delta}$ set is defined as countable intersection of Open sets and

$F_{\sigma}$ set is defined as countable union of closed sets

Just by seeing the definition i concluded $(F_{\sigma})^{c} = G_{\delta}$, where $A^{c}$ means complement of $A$.

Is this correct or i am lacking something ?

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    If by $(F_{\sigma})^{c} = G_{\delta}$, you mean the complement of an $F_\sigma$ set is a $G_\delta$ set, then yes you are correct, and the proof is trivial. – 2017-02-20
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    If $A^c$ means the complement of $A$, then $(F_\sigma)^c$ should mean the complement of $F_\sigma$. So you have to first tell us what this set $F_\sigma$ is, and in what larger set are we taking its complement. (Or perhaps you didn't write what you meant.) – 2017-02-20
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    Saying that larger set is any set $X$ would not be a correct way , so let it be defined in $\mathbb{R}$ @arjafi – 2017-02-20
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    And i think this is correct , is it -- "$F_{\sigma}$ set is defined as collection of countable union of closed sets" – 2017-02-20
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    So are you asking "If a set is not $F_\sigma$, then it is $G_\delta$?" i.e. taking complements in the power set of $\mathbb R$? If that's the case, the answer is no, as any singleton is both $F_\sigma$ and $G_\delta$. – 2017-02-20
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    Now i am confused , as i saw the statement stated in the question in my notebook but i never thought about larger set ,what do you think about the larger set @Aweygan – 2017-02-20
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    I think the burden of clarification is on you. Either cite the book, or pick a context yourself. – 2017-02-20

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Not exactly. If you identify $F_{\sigma},G_{\delta}$ with sets family it's wrong. What is true is that $$ A\in F_{\sigma}\iff A^c\in G_{\delta}$$