Example where $A$ and $B$ closed, $A+B$ not an $F_{\sigma}$ This question arises from the following:
How can we prove that in $\mathbb{R}^n$ if $A$ and $B$ are both closed subsets, then $A+B$ is an $F_\sigma$ set?
If $A$ and $B$ are closed subsets of metrizable topological vector space (or more generally of a Hausdorff topological vector space) is $A+B$ an $F_{\sigma}$ set?  I believe it need not be but I haven't been able to construct a  counterexample. Thanks for your time. 
 A: $\mathbb{R}^\mathbb{R}$ with product topology is a Hausdorff topological vector space. Let $A = \{f_v : v \in \mathbb{R}\setminus\mathbb{Q}\}$, where, for $x \in \mathbb{R}$, $f_x \in \mathbb{R}^\mathbb{R}$ is defined as $f_x(0) = x$, $f_x(x) = 1$, and $f_x(y) = 0$ if $y \not \in \{0,x\}$. Let $B = \{g_v : v \in \mathbb{R}\setminus\mathbb{Q}\}$, where, for $x \in \mathbb{R}$, $g_x \in \mathbb{R}^\mathbb{R}$ is defined as $g_x(0) = 0, g_x(x) = -1$, and $g_x(y) = 0$ if $y \not \in \{0,x\}$. Then $A$ and $B$ are closed (this is obvious), and $A+B = \{h_v : v \in \mathbb{R}\setminus\mathbb{Q}\}\cup\{f_v+g_{v'} : v\not = v', v,v' \in \mathbb{R}\setminus \mathbb{Q}\}$, where, for $x \in \mathbb{R}$, $h_x \in \mathbb{R}^\mathbb{R}$ is defined as $h_x(0) = x$ and $h_x(y) = 0$ if $y \not = x$. Suppose that $A+B$ is an $F_\sigma$ set; say $A+B = \cup_n F_n$ with each $F_n \in \mathbb{R}^\mathbb{R}$ closed. Then, since $E := \{l \in \mathbb{R}^\mathbb{R} : l(x) = 0 \text{ for all } x \not = 0\}$ is closed, $\{h_v : v \in \mathbb{R}\setminus \mathbb{Q}\} = (A+B)\cap E = (\cup_n F_n)\cap E = \cup_n (F_n \cap E)$ is an $F_\sigma$ set. This is equivalent to saying that $\mathbb{R}\setminus\mathbb{Q}$ is an $F_\sigma$ set in $\mathbb{R}$, which is false.

If you want metrizable, just note that everything above is happening in the (metrizable) subset of functions from $\mathbb{R}$ to $\mathbb{R}$ with finite support. Below are the details.
Let $V = \{f \in \mathbb{R}^\mathbb{R} : \#\{x : f(x) \not = 0\} < \infty\}$. Then $V$ is a vector space. Define, for $f,g \in V$, $d(f,g) = \sum_{x \in \mathbb{R}} |f(x)-g(x)|$, which is well-defined, since it is a finite sum. Then, $d(f,f) = 0$, $d(f,g) = 0$ implies $f=g$, $d(f,g) \ge 0$ for each $f,g$, $0 \le d(f,g) < \infty$ for each $f,g$, and $d(f,h) \le d(f,g)+d(g,h)$ for each $f,g,h \in V$. Hence, $d$ is a metric on $V$. Of course, addition and multiplication by a scalar are continuous with respect to this metric, so $(V,d)$ is a metrizable, topological vector space. Define $A$ and $B$ as above (note they do lie inside $V$). Note that $A$ and $B$ are closed with respect to $d$, as well as $E$ (as defined above). So, the (by contradiction) argument above implies that $\{h_v : v \in \mathbb{R}/\mathbb{Q}\}$ is an $F_\sigma$ set, which, since $d$ restricted to $E$ is the $L^1$ norm on $\mathbb{R}$, still implies that $\mathbb{R}/\mathbb{Q}$ is an $F_\sigma$ set in $\mathbb{R}$, with the standard topology, which, once again, is false.
