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Can anyone give an example of a Meagre subset of $\mathbb{R}$ (i.e., of the first Baire Category), which isn't an $F_\sigma$?

Simple cardinality arguments show that such a thing exists, but I can't really think of one. Maybe it's impossible to explicitly construct one? A proof (not using cardinalities) would be welcome too (Axiom of Choice, perhaps?).

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up vote 4 down vote accepted

Take the Cantor set $C$. It is closed and nowhere dense in $\mathbb R$, so $C$ and all its subsets are meager in $\mathbb R$. Now take a countable set $T$ dense in $C$, and consider $A = C \setminus T$. Then $A$ is meager in $\mathbb R$, but is not an $F_\sigma$.

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Why is not an $F_\sigma$? –  FPP Apr 3 '12 at 23:15
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Baire catetory theorem in the space $C$. –  GEdgar Apr 3 '12 at 23:53
    
For a "real life example", I believe the set of normal numbers (or all numbers simply normal to base $10$, or all numbers satisfying most any reasonable normality condition) is also an example. See Lars Olsen's 2004 paper Extremely non-normal numbers for some very strong results that imply meagerness for "most any reasonable normality condition". As for not being $F_{\sigma}$, I don't know off-hand if there's an easy way to show this, but the stronger result of not being $G_{\delta \sigma}$ is proved in Ki/Linton's 1994 paper Normal numbers and subsets of $\mathbb N$ with given densities. –  Dave L. Renfro Apr 4 '12 at 15:21
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