Does there exist a first category set that is not $F_\sigma$?

Question

Suppose $A$ is of first category in $X$,then $A=\cup_n A_n$ where each $A_n$ is nowhere dense.Now my question is can we write $A=\cup_n \bar A_n$ or can a category I set be expressed as a countable union of closed sets with empty interior?

Addendum I think what I though was wrong because consider the cantor set $C$,which is of first category,so $C-\{x\}$,for some $x\in C$ is also first category.But $C-\{x\}=\cup _n (C-\{x\})$,although $C-\{x\}\neq \cup_n C$,note that $Closure(C-\{x\})=C$. Does there exist a first category set that is not $F_\sigma$?We can find $(0,1)$ as an example $F_\sigma$ set which is not of first category.But I am having problem in finding example of a first category set that is not $F_\sigma$.

Answer 1

Let $E$ be any first category subset of ${\mathbb R}$ such that the cardinality of $E$ is $c = 2^{\aleph_0}.$ Then each subset of $E$ is a first category subset of ${\mathbb R}.$ Since there are $2^c$-many subsets of $E$ and only $c$-many $F_{\sigma}$ subsets of ${\mathbb R}$ (proof in a moment), it follows that most (in the sense of cardinality) subsets of $E$ are first category subsets of $\mathbb R$ that are not $F_{\sigma}$ subsets of ${\mathbb R}.$

Proof there are exactly $c$-many $F_{\sigma}$ subsets of ${\mathbb R}$: [$\geq c$ many] There are at least $c$-many such subsets, since for each positive real number $r,$ the closed interval $[0,r]$ is an $F_{\sigma}$ subset of ${\mathbb R}.$ Note that the mapping defined by $r \mapsto [0,r]$ is an injection from the positive real numbers (has cardinality $c)$ to the set of $F_{\sigma}$ subsets of ${\mathbb R}.$ [$\leq c$ many] There are $c$-many sequences of closed subsets of $\mathbb R$ since there are $c$-many closed subsets (because there are $c$-many open subsets, which I'll just assume as known since I don't want to digress too much here) and, given any set of $c$-many elements, there are $c$-many sequences of elements in that set (follows from some simple cardinal arithmetic and definition of "sequence" as a function from the natural numbers to the set --- $c^{\aleph_0} = \left(2^{\aleph_0}\right)^{\aleph_0} = 2^{ {\aleph_0} \cdot {\aleph_0}} = 2^{\aleph_0} = c).$ Thus, since the mapping defined by taking each $F_{\sigma}$ subset to a sequence of closed subsets whose elements have union equal to that $F_{\sigma}$ subset is an injection, it follows that the cardinality of the $F_{\sigma}$ subsets is less than or equal to the cardinality of the set of sequences of closed subsets, or from what was just shown, the cardinality of the $F_{\sigma}$ subsets is less than or equal to $c.$

It turns out that since there are only $c$-many Borel subsets of ${\mathbb R}$ (proofs of this are much more difficult than for the $F_{\sigma}$ case), it follows that most $F_{\sigma}$ subsets of $E$ are not Borel subsets of ${\mathbb R}.$

In particular, since the Cantor middle thirds set $C$ is a first category subset of ${\mathbb R}$ that has cardinality $c$ (indeed, $C$ is a nowhere dense subset of ${\mathbb R}$ that has cardinality $c),$ we know that most subsets of $C$ are first category subsets of $\mathbb R$ that are not $F_{\sigma}$ (or even Borel) subsets of ${\mathbb R}.$ An explicit (term used informally here, not in some precise constructive sense) example of a non-$F_{\sigma}$ subset of $C$ is the set of irrational numbers in $C.$

The above (excluding the last paragraph) continues to hold if $\mathbb R$ is replaced by any fixed Polish space, and more generally still (e.g. a trivial generalization are metric spaces that contain a Polish space as a closed subset; and I'm not sure whether we need separability). However, it is easy to see that the result is not true for any metric space, since there are metric spaces that have no non-$F_{\sigma}$ subsets. For example, every subset of a discrete metric space is open, and hence every subset of a discrete metric space is $F_{\sigma}$ (recall that in metric spaces, every open set is an $F_{\sigma}$ set). I suspect there have been papers written that deal with defining relatively large classes of metric spaces, or even that deal with defining relatively large classes of topological spaces, for which the type of examples you want exist, and perhaps even characterizations (seemingly not involving the notions themselves) for the existence of such an example, but off-hand I don't know any references.

Related Stack Exchange question, found after I wrote most of the above:

Does there exist a nowhere dense set in $\mathbb R$ which is not $F_\sigma$

Answer 2

Since Cantor set is uncountable,so it has a condensation point.Now consider a condensation point $x_0$.Consider some $\epsilon$-nbd of $x_0$ ,$N(x_0,\epsilon)$.Now consider the set $A=N(x_0,\epsilon)\cap C-\{x_0\}(C$ being cantor set$)$.This set should be uncountable.Now suppose this set is $F_\sigma$,then $A=\cup_n F_n$ where $F_n$ is closed,$\forall n\in\mathbb N$.Now $x_0$ is a condensation point of $A$,hence it must be a condensation point of some $F_n$,(If not,then the nbd contains countably many points of each $F_n$ and countable union of countable sets is countable,so $(\implies\impliedby))$.$F_n$ is closed,so it must contain $x_0$,then $A$ contains $x_0$ but $A$ does not contain $x_0$ by the construction of $A$.So,$A$ is not $F_\sigma$ but is of first category as $C$ is of first category.

Addendum There is an error in the solution as pointed out in comment,I will try to find out a correct solution.