Let $A⊆\mathbb{R}$ be the set $A=\{x\in(0,1)$: the decimal expansion of x contains infinitely many 7's}.
$A^c⊆\mathbb{R}$ be the set $A=\{x\in(0,1)$: the decimal expansion of x contains finitely many 7's}.
$A$ is borel set , is it correct that $A^c$ is borel set as well ?
I denote $A_n := \{x\in(0,1)\mid \text{the $n$th digit of the decimal expression of $x$ is 7}\}.$
$A_n := \{x\in(0,1)\mid \text{the $n$th digit of the decimal expression of $x$ is 7}\}.$
$A'_n:= \{x\in(0,1)\mid \text{the $n$th digit of the decimal expression of $x$ is 7 where the 7's digits is finite }\}.$
I consider $A'_n\subset A_n$.
Since $A_n$ is Borel as it is a finite union of half-intervals and $A'_n\subset A_n$ implies $A'_n$ is Borel.
Is it correct ?
In general ,Any subset of Borel set is Borel ? any complement of borel set is Borel ?
Appreciaite any help.