How to show that $\mathbb{Q}$ is not $G_\delta$ I read a section of a book and it made mention of the set of rationals not being a $G_\delta$. However, it gave no proof. I read on wikipedia about using contradiction, but it made use of the Baire category theorem, which is unfamiliar to me.
I was wondering if anyone could offer me a different proof; perhaps using the fact that the complement of $G_\delta$ is $F_\sigma$.
Thanks.
 A: Another silly argument which I think also works:
Suppose $\mathbb Q$ is a $G_\delta$, so that there exists a sequence $(A_n)_{n\geq1}$ of open subsets of $\mathbb R$ such that $\mathbb Q=\bigcap_{n\geq1} A_n$. 
Recall that each $A_n$ is a disjoint union of open intervals. For each $n\geq1$ let $\mathcal I_n$ be the set of the intervals making up $A_n$.
Without loss of generality, we can assume that $\mathcal I_1$ contains at least two elements — call them $I(0)$ and $I(1)$ — each of length at most $2^{-1}$
We can also assume that $\mathcal I_2$ it contains two elements contained in $I(0)$ each of length $2^{-1}$ — call them $I(00)$ and $I(01)$ — and two elements contained in $I(1)$, also of length at most $2^{-1}$ — call them $I(10)$ and $I(11)$.
Of course, we can continue in this way indefinitely... We thus obtain intervals $I(w)$, one for each finite word $w$ written in the letters $0$ and $1$, such that the length of $I(w)$ is at most $2^{-\mathrm{length}(w)}$, and such that whenever $w'$ is a prefix of $w$, then $I(w')\supseteq I(w)$.
Now pick any infinite sequence $w$ of zeroes and ones, and for each $n\geq1$ let $w_n$ be the prefix of $w$ of length $n$, and pick a point $x_n$ in the interval $I(w_n)$. It is easy to check that the limit $y_w=\lim_{n\to\infty}x_n$ exists and belongs to $\mathbb Q$, and that $y_w\neq y_{w'}$ if $w$ and $w'$ are distinct infinite sequences of zeroes and ones. This is absurd, as $\mathbb Q$ is countable yet there are uncountably many infinite sequences of zeroes and ones.
$$♦♦♦$$
Another way to implement the above idea.
Suppose $\mathbb Q=\cap_{n\geq1}A_n$ with $A_n$ open in $\mathbb R$. 
We massage the open sets a bit first. 


*

*For each $b\geq1$ let $B_n=\bigcap_{i=1}^nA_n$, so that $(B_n)_{n\geq1}$ is a decreasing sequence of open sets whose intersection is also $\mathbb Q$. 

*Next, for each $n\geq1$ let $C_n=B_n\cap\Big(\mathbb R\setminus(\pi+\tfrac1{2^n}\mathbb Z)\Big)$, so that $(C_n)$ is also a decreasing sequence of open sets whose intersection is $\mathbb Q$, with the added nice property that for all $n\geq1$ each connected component of $C_n$ is of length at most $\tfrac1{2^n}$.
Let $\mathcal I_n$ be set of connected components of $C_n$. If $n\geq1$, then $C_n\supseteq C_{n+1}$ so there is a function $\phi_n:\mathcal I_{n+1}\to\mathcal I_n$ sending each element of $\mathcal I_{n+1}$ to the unique element of $\mathcal I_n$ which contains it. Since $\bigcap_{n\geq1}C_n=\mathbb Q$, it is easy to see that $\phi_n$ is surjective.
Let $$X=\varprojlim(\mathcal I_n,\phi_n)=\Big\{(i_n)_{n\geq1}\in\prod_{n\geq1}\mathcal I_n:\phi_n(i_{n+1})=i_n,\quad\forall n\geq1\Big\}$$ be the inverse limit of the sets $\mathcal I_n$ along the maps $\phi_n$. This is an uncountable set, and it is easy to construct an injective function $f:X\to\mathbb Q$. Indeed, if $\xi=(i_n)_{n\geq1}\in X$ pick, for each $n\geq1$, a point $x_n\in i_n$; then one can show that $f(\xi)=\lim_{n\to\infty}x_n$ exists, and that this defines an injective function.
A: I suspect that just about any proof that doesn’t directly use the Baire category theorem either uses a consequence of it or proves a special case of it. I’ve chosen the second course.
Suppose that $\mathbb{Q} = \bigcap\limits_{k\in\omega}V_k$, where each $V_k$ is open in the usual topology on $\mathbb{R}$. Clearly each $V_k$ is dense in $\mathbb{R}$. Let $\mathbb{Q}=\{q_k:k\in\omega\}$ be an enumeration of the rationals, and for each $k\in\omega$ let $W_k=V_k\setminus \{q_k\}$; clearly each $W_k$ is dense and open in $\mathbb{R}$, and $\bigcap\limits_{k\in\omega}W_k = \varnothing$.
Let $(a_0,b_0)$ be any non-empty open interval such that $[a_0,b_0]\subseteq W_0$. Given a non-empty open interval $(a_k,b_k)$, let $r_k=\frac14(b_k-a_k)$; clearly $a_k<a_k+r_k<b_k-r_k<b_k$. Since $W_{k+1}$ is dense and open, there is a non-empty open interval $(a_{k+1},b_{k+1})$ such that $$(a_{k+1},b_{k+1}) \subseteq [a_{k+1},b_{k+1}] \subseteq W_{k+1}\cap (a_k+r_k,b_k-r_k),$$ and the construction can continue.
For $k\in\omega$ let $J_k = [a_k,b_k] \subseteq W_k$. For each $k \in \omega$ we have $J_k \supseteq J_{k+1}$, so $\{J_k:k\in\omega\}$ is a decreasing nest of non-empty closed intervals. Let $J = \bigcap\limits_{k\in\omega}J_k$; $J\subseteq J_k \subseteq W_k$ for each $k\in\omega$, so $J \subseteq \bigcap\limits_{k\in\omega}W_k = \varnothing$. But the nested intervals theorem guarantees that $J \ne \varnothing$, so we have a contradiction. Thus, $\mathbb{Q}$ cannot be a $G_\delta$-set in $\mathbb{R}$.
A: You can use the fact that the complement of $\mathcal{G}_\delta$ is $\mathcal{F}_\sigma$. We have that the set of rationals $\mathbb{Q}$ is a $\mathcal{G}_\delta$ set if and only if the set of irrationals $\mathbb{R}\backslash\mathbb{Q}$ is an $\mathcal{F}_\sigma$ set. Suppose that is true, so $\mathbb{R}\backslash\mathbb{Q}=\bigcup\limits_{n=1}^\infty B_n$, where $B_n$ are closed. None of the $B_n$ contains any nondegenerate interval, since that would contain a rational number. Let the rational numbers be enumerated as $\{q_1,q_2,\ldots\}$ and let $C_n=B_n\cup\{q_n\}$; note that $C_n$ is closed, as a finite union of closed sets, and contains no nondegenerate interval. We have $$\mathbb{R}=\mathbb{Q}\cup(\mathbb{R}\backslash\mathbb{Q})=\{q_1,q_2,\ldots\}\cup \bigcup\limits_{n=1}^\infty B_n=\bigcup\limits_{n=1}^\infty C_n.$$ Hence $\mathbb{R}$ is the union of closed sets $C_n$, none of which contain a nondegenerate interval.
Let $I$ be a closed interval such that $I\cap C_1=\varnothing$. This exists, since $C_1$ is closed, so for all $x\notin C_1$, we can find an open interval containing $x$ lying outside $C_1$, and hence a closed interval within that, or else $C_1=\mathbb{R}$, which is impossible since $C_1$ contains no nondegenerate interval. Similarly, define a closed interval $I_2\subset I_1$ such that $I_2\cap C_2=\varnothing$, possible since $I_1\not\subset C_2$ and $C_2$ is closed and contains no nondegenerate interval. Apply this repeatedly to obtain a nested sequence of closed intervals $I_1\supset I_2\supset\ldots$, where $I_n\cap C_n=\varnothing$.
Consider the set $I=\bigcap\limits_{n=1}^\infty I_n$. We know that $I$ is nonempty, but for $x\in I$, we have $x\notin C_n$ for all $n\in\mathbb{N}$, contradicting the assumption that $\mathbb{R}=\bigcup\limits_{n=1}^\infty C_n$. Hence, by contradiction, $\mathbb{R}\backslash\mathbb{Q}$ is not $\mathcal{F}_\sigma$, so that $\mathbb{Q}$ is not $\mathcal{G}_\delta$.
A: A strange argument follows:
Suppose $\mathbb Q$ is a $G_\delta$. Mazurkiewicz's theorem (you can read about it here ) tells us that there exists then a metric $d$ on $\mathbb Q$, equivalent to the original one,  such that $\mathbb Q$ is complete with respect to $d$.
Now, a complete metric space which is countable as a set has an isolated point (to prove this one needs Baire's theorem) so we conclude that $\mathbb Q$, in its usual topology, has an isolated point. This is of course absurd.
A: We  prove the following by direct elementary methods without "Baire" or its immediate corollaries: 
(1). A non-empty completely metrizable space with no isolated points is uncountable.( We construct  a subset  that is a bijective image of the set of functions from $\mathbb N$ to $\{0,1\}.$ In fact we construct a subset that is   homeomorphic to the Cantor set.) 
(2).(a). If X is Hausdorff and $Y$ is the intersection of a countable family of completely metrizable subspaces of $X$, then $Y$ is  completely metrizable.
(2).(b). If $X$ is completely metrizable  and $Y$ is an open subset of $X$ then Y is completely metrizable. 
(2).(c). Corollary to (2)(a) and (2)(b): A $G_{\delta}$ subset of $\mathbb R$ is completely metrizable.
So if $S$ is a  $G_{\delta}$ subset of $\mathbb R$ and $S$ is dense in $\mathbb R,$ then by (2)(c), $S$ is completely metrizable.And $S$  has no isolated points (because it is dense in $\mathbb R$). So by (1), $S$ is uncountable.
The results (1) and (2) are of interest, but the usefulness of the Baire category theorem is illustrated by using it to directly prove that $\mathbb Q $ is not $G_{\delta}$ in $\mathbb R.$
A: Let $\mathbb Q$ be $G_\delta$ set
$\implies \mathbb Q = \bigcap_{i \in \mathbb N} U_i$ where $U_i$ is open set
$\implies \mathbb Q \subseteq U_i \forall i  $
Since $\mathbb Q$ is dense $\implies U_i$ is also dense
As $\mathbb Q$ is countable then $\mathbb Q  = \{q_k | k \in \mathbb N \}$
Observe that each $ U_i \backslash \{q_i\}$ is open and dense
By Baire Category theorem ( A countable intersection of open dense set in $\mathbb R$ is again dense)
$\implies \bigcap U_i \backslash \{q_i\} $ is open and dense but observe that $\bigcap U_i \backslash \{q_i\} = \varnothing \;\;\;\;\;\;\Rightarrow\Leftarrow$
