showing that a set is not $G_\delta$ Let $A = \{x\in [0,1]\setminus\Bbb Q\mid \exists n\in \Bbb N $ such that $x_n=7\}$
where $x_n$ is the n'th digit of $x$ in its decimal representation.
prove or disprove that A is $G_\delta$ as a subspace of $[0,1]$.
I couldn't solve this problem, and I would be thankful if someone can give me a hint.
Thank you very much.
 A: Some possible hints:


*

*Observe that $A$ is $G_δ$ in $[0, 1]$ if and only if it is $G_δ$ in $[0, 1] \setminus ℚ$.

*How about $A_1 = \{x ∈ [0, 1] \setminus ℚ \mid x_1 = 7\}$? Can you say something about its complexity in $[0, 1]$ or in $[0, 1] \setminus ℚ$?

*Do you know the fact that irrationals are homeomorphic to $ℕ^ℕ$?


Ad $ℝ \setminus ℚ \cong ℕ^ℕ$ (or $ℤ^ℕ$): The idea is to partition the irrationals into $ℤ$-many consecutive relatively clopen intervals. Each of them is again homeomorphic to the irrationals, so you inductively repeat this process and build a tree of shape $ℤ^{<ω}$. If you are careful, you can do this in a such way that diameter of the chosen sets along every branch go to $0$, so by completeness its intersection is a singleton. This gives a homeomorphism between the original irrationals and the branches of $ℤ^{<ω}$, which are $ℤ^ω$ or $ℤ^ℕ$.
This is actually the same idea as when showing that the standard Cantor set is homeomorphic to $2^ω$. You can also see https://en.wikipedia.org/wiki/Baire_space_%28set_theory%29.
