Is this "epsilon expansion" of a closed set in $\mathbb{Q}^\mathbb{N}$ closed? Consider $\mathbb{Q}$ as a discrete topological space, and form the infinite (Tychonoff) product space $\mathbb{Q}^{\mathbb{N}}$. Let $F\subseteq\mathbb{Q}^{\mathbb{N}}$ be closed.
Question: Given a sequence $(\epsilon_n)$ of positive numbers, is the set $F_{(\epsilon_n)}=\{(x_n)\in\mathbb{Q}^{\mathbb{N}}:\exists(y_n)\in F(\forall n|x_n-y_n|\leq\epsilon_n)\}$ also closed in $\mathbb{Q}^{\mathbb{N}}$?
Edit: If not, is it Borel? (It is clearly analytic.)
 A: It need not be closed.
Let $X=\Bbb Q^{\Bbb N}$. For $m\ge 2$ define $y^{(m)}=\left\langle y_n^{(m)}:n\in\Bbb N\right\rangle\in X$ by
$$y_n^{(m)}=\begin{cases}
1,&\text{if }n\le m\\
m,&\text{if }n>m\;,
\end{cases}$$
so that
$$\begin{align*}
y^{(2)}&=\langle 1,1,1,2,2,2,2,\ldots\rangle\\
y^{(3)}&=\langle 1,1,1,1,3,3,3,\ldots\rangle\\
y^{(4)}&=\langle 1,1,1,1,1,4,4,\ldots\rangle\;,
\end{align*}$$
and so on. (My $\Bbb N$ contains $0$.) For $m\ge 2$ and $n\in\Bbb N$ let
$$z_n^{(m)}=\begin{cases}
0,&\text{if }n\le m\\
m,&\text{if }n>m\;,
\end{cases}$$
and for $m\ge 2$ let $z^{(m)}=\left\langle z_n^{(m)}:n\in\Bbb N\right\rangle$.
Let $F=\left\{y^{(m)}:m\ge 2\right\}$; $F$ is a closed discrete set in $X$. Let $\epsilon$, be the constant $1$ sequence in $X$, so that $\epsilon_n=1$ for each $n\in\Bbb N$; clearly $z^{(m)}\in F_\epsilon$ for each $m\ge 2$. Let $\mathbf{0}$ be the constant zero sequence in $X$, and let $U$ be any open nbhd of $\mathbf{0}$. By the definition of the product topology there is an $m\in\Bbb N$ such that the basic open nbhd
$$B_m(\mathbf{0})=\{x\in X:x_n=0\text{ for all }n\le m\}\;,$$
of $\mathbf{0}$ is a subset of $U$. Clearly $z^{(k)}\in B_m(\mathbf{0})$ whenever $k\ge m$, so every open nbhd of $\mathbf{0}$ meets $F_\epsilon$, and $\mathbf{0}\in\operatorname{cl}F_\epsilon$. But $\mathbf{0}\notin F_\epsilon$, so $F_\epsilon$ is not closed.
