Is the product of uncountably many copies of $\mathbb{R}$ with the discrete topology a Baire space? If $X$ is the product of uncountable many copies of the space $(\mathbb{R}, \text{disc})$, where $\mathrm{disc}$ denotes the discrete topology, then can we conclude that $X$ is a Baire space?
 A: Let $X$ be any set and $\tau$ the discrete topology on $X$. The topology $\tau$ is generated by the discrete metric
$$d(x,y)=\begin{cases}
0,\text{if }x=y\\
1,\text{otherwise}\;,
\end{cases}$$
which is complete. J.C. Oxtoby, Cartesian products of Baire spaces, Fund. Math., $\mathbf{49} (1961)$, $157$-$166$, defines a notion of pseudo-completeness for topological spaces, observes that every complete metric space is pseudo-complete, proves that every pseudo-complete space is Baire, and further proves that arbitrary products of pseudo-complete spaces are pseudo-complete. Thus, arbitrary products of discrete spaces are Baire.
It’s actually not hard to prove directly that your $X$ is Baire. For $n\in\Bbb N$ let $G_n$ be a dense open set in $X$. For each finite $F\subseteq\Bbb R$ and function $\varphi:F\to\Bbb R$ let 
$$B(F,\varphi)=\{x\in X:x\upharpoonright F=\varphi\}\;;$$
the sets $B(F,\varphi)$ form a base $\mathscr{B}$ for the product topology on $X$.
Let $B(F,\varphi)\in\mathscr{B}$ be arbitrary. $B(F,\varphi)\cap G_0\ne\varnothing$, so there are a finite $F_0\subseteq\Bbb R$ and a function $\varphi_0:F_0\to\Bbb R$ such that $B(F_0,\varphi_0)\subseteq B(F,\varphi)\cap G_0$. Clearly $\varphi_0\upharpoonright(F_0\cap F)=\varphi\upharpoonright(F_0\cap F)$, so we may assume that $F_0\supseteq F$ and $\varphi=\varphi_0\upharpoonright F$. Given a finite $F_n\subseteq\Bbb R$ and a function $\varphi_n:F_n\to\Bbb R$ such that $F_k\subseteq F_n$ and $\varphi_k=\varphi_n\upharpoonright F_k$ for $k\le n$, note that $B(F_n,\varphi_n)\cap G_{n+1}\ne\varnothing$, so there are a finite $F_{n+1}\subseteq\Bbb R$ and a function $\varphi_{n+1}:F_{n+1}\to\Bbb R$ such that $F_n\subseteq F_{n+1}$ and $\varphi_k=\varphi_{n+1}\upharpoonright F_k$ for each $k\le n+1$. Thus, we can recursively construct $F_n$ and $\varphi_n$ for each $n\in\Bbb N$.
Now let $C=\bigcup_{n\in\Bbb N}F_n$, a countable subset of $\Bbb R$, and let $\varphi=\bigcup_{n\in\Bbb N}\varphi_n:C\to\Bbb R$. Let $x\in X$ be defined by
$$x(r)=\begin{cases}
\varphi(r),&\text{if }r\in C\\
0,&\text{otherwise}\;.
\end{cases}$$
Clearly $x\in B(F,\varphi)\cap\bigcap_{n\in\Bbb N}G_n$, so $\bigcap_{n\in\Bbb N}G_n$ is dense in $X$, and $X$ is Baire.
