Let $X= \Bbb R^{\Bbb R}$ and suppose that $X$ has the product topology. Let $A$ be the set of functions $f: \Bbb R \to \Bbb R$ for which $f(x) \in \Bbb Q$ for all $x \in \Bbb R$ and $f(x)=0$ for all but finitely many $x$. Let $B$ be the set of $f: \Bbb R \to \Bbb R$ for which $f(x)=0$ when $x \in \Bbb Q$. Is $A$ or $B$ dense in $X$?
Since for $f_n \in A$ we have that $f_n \to 0$ as there is only finitely many $x$ for which $f(x)\ne 0$ and the zero function is in $X$. I think that $A$ is dense in $X$.
For $B$ since in any sequence $g_n \in B$ the elements $g_i(x)$ only evaluate to $0$ countably many times. I don't think this sequence converges to any $g \in A$.
Is my hunch correct here or am I making some kind of obvious mistake?
$A$ is dense as is easy to see (it intersects every basic open set that is non-empty). $B$ is not as it's closed and not the whole space.
That $B$ is closed follows from $B = \bigcap_{q \in \Bbb Q} \pi_q^{-1}[\{0\}]$