One point set in $[0,1]^{A}$ is not $G_\delta$ when A is not countable I need to prove that one point set in  $[0,1]^{A}$ is not $G_\delta$  when A is not countable
I tried something like this: assume that $\{x\}$ is $G_\delta$ for some x $\in$ $[0,1]^{A}$
then  $\{x\} = \bigcap_{i=1}^{\infty } U_i$
I know that each basis element of $[0,1]^{A}$  is of the form $\prod_{\alpha \in A} V_\alpha$ when $V_\alpha = [0,1]$ for all but finite $\alpha \in A$
and I want some how to say the same about $U_i$  because then I'll get a contradiction
 A: You started off fine: assume that $\{x\}=\bigcap_{n\in\Bbb N}U_n$, where each $U_n$ is open in the product. For each $n\in\Bbb N$ there must be a basic open set $V_n$ such that $x\in V_n\subseteq U_n$, where a basic open set is one that has the form $\prod_{\alpha\in A}W_\alpha$, where each $W_\alpha$ is open in $[0,1]$, and $\{\alpha\in A:W_\alpha\ne[0,1]\}$ is finite.
For each $n\in\Bbb N$ let $A_n=\{\alpha\in A:W_\alpha\ne[0,1]\}$, and let $C=\bigcup_{n\in\Bbb N}A_n$; each $A_n$ is finite, so $C$ is countable. $A$, however, is uncountable, so there is an $\alpha\in A\setminus C$. Let $y$ be a point of $[0,1]^A$ that agrees with $x$ on every coordinate except $\alpha$. Clearly $y\ne x$, but you should be able to show that $y\in\bigcap_{n\in\Bbb N}V_n\subseteq\bigcap_{n\in\Bbb N}U_n$ to get your contradiction.
A: Hint: If you have countably many basic open sets $V_n = \prod_{\alpha \in A} V_\alpha^n$, ($n \in \mathbb{N}$) and $A$ is uncountable, what can you say about the set $$\{ \alpha \in A : ( \exists n \in \mathbb{N} ) ( V_\alpha^n \neq [0,1] ) \}?$$
