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I've thought about it, and I'm not sure such sets exist. I mean, obviously the finite sets satisfy both definitions, buy I'm searching a non-trivial example.

The definitions are: A set $M \subseteq \mathbb{R}$ its a meager set if exists a sequence $\{F_n\}$ of nowhere dense (i.e., with empty interior) with $M = \bigcup F_n$; and a set $X \subseteq \mathbb{R}$ its a $G_\delta$ set if exists a sequence $\{A_n\}$ of open sets with $M = \bigcap A_n$.

If any have an idea or some reference, I would appreciate the information. Thanks!

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  • $\begingroup$ Right you are... $\endgroup$ Sep 10 '20 at 16:15
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    $\begingroup$ FYI, a $G_{\delta}$ meager set is nowhere dense. The converse doesn't hold, but not because of size considerations (because every nowhere dense set is contained in a closed nowhere dense set, and a closed nowhere dense set is certainly a $G_{\delta}$ meager set), but because of "descriptive set complexity" considerations (there exist nowhere dense sets that are not $G_{\delta},$ indeed not even Borel). $\endgroup$ Sep 10 '20 at 18:21
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A more interesting example is the Cantor set, it is a nowhere dense (hence meager) subset of the reals with the same cardinality as the whole of $\Bbb R$, and it also a $G_\delta$ set since every closed set is $G_\delta$ in a metric space. By considering a fat Cantor set instead of the standard middle-thirds construction you can even get examples with positive measure.

Note that in general, by the Baire category theorem, it is much easier to produce comeager $G_\delta$ sets (hence meager $F_\sigma$ sets)

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