$A$ and $B$ precompact , closed and disjoint subset of $E$. Let $E:=C_{00}$. For all $ n \in \mathbb{N}$,
$$ x_n = \left\{ 1, \frac{1}{2}, \frac{1}{3}, \cdots \frac{1}{n},0,0,0,\cdots \right\} $$
then
$$ A = \{ x_{2n -1} : n \in \mathbb{N} \} \quad \textrm{and} \quad  B = \{x_{2n} : n \in \mathbb{N} \} $$
then the following holds,
$ (a)$ $A$ and $B$ precompact , closed and disjoint subset of $E$.
$ (b)$ For all V neighborhood of zero , $ (A + V) \cap (B + V) \neq \emptyset $.
Firstly i check that A and B is disjoints subset of E:  Assume that $ A \cap B \neq \emptyset $ then there exists $ x \in A \cap B $ then $ x = x_{2n} $ and $x = x_{2n-1} $ so we have a contradiction with $ \frac{1}{2n} = 0 $. But i stack others. Can you give any hint? Thanks for all.
 A: Let's try to show that $A \subseteq c_{00}$ is closed. Recall that
$$c_{00} := \{(x_n)_{n \geq 0} \ | \ \text{ there exists an M } \geq 0 \text{ so that } (x_n)_{n \geq 0} = (x_1, \ldots, x_M, 0, \ldots)\}.$$
We're equipping this with the topology given by the norm
$$ \|(x_n) \|_\infty := \sup\{ |x_n| \ | \ n \geq 0\}. $$
Let
$$A := \left\{ \left(1, \frac{1}{2}, \ldots, \frac{1}{2n}, 0 \ldots \right) \ | \ n \geq 1 \right\}.$$
Let $(y_n) \subseteq A$ and suppose $y_n \rightarrow y \in c_{00}$ in the supremum norm. Since $y \in c_{00}$, there is some $M$ so that $y = (x_1, \ldots, x_M, 0, \ldots).$ Observe that $y_n$ must eventually be constant in $n$, otherwise it cannot converge to a point in $c_{00}$ (think about what our choices for $y_n$ are). Since it is eventually constant, we must get that $y \in A$, so $A$ is closed. The same trick works for $B$.
To see precompact, we need to discuss in terms of what space. Generally we consider
$$ c := \{(x_n)_{n \geq 0} \ | \ \lim_{n \rightarrow \infty} x_n \text{ exists}\}.$$
In this case, the closure of $A$ is going to be
$$\overline{A} = A \cup \left\{ \left( \frac{1}{n} \right)_{n \geq 1} \right\}.$$
To show it is compact, take any sequence $(y_n) \subseteq A$ and show that you can find a convergent subsequence (alternatively you can do it with open covers).
For the last one, it suffices to work with balls. Let
$$B_n := \left\{(x_n) \in c \ | \ \|x\|_\infty < \frac{1}{n}\right\}.$$
Notice that
$$ \left(1, \frac{1}{2}, \ldots, \frac{1}{2n}, 0, \ldots \right) + \left(0, 0, \ldots, 0, \frac{1}{2n+1}, \ldots \right) \in (A + B_{2n+1}) \cap (B + B_{2n+1}).$$
We can do this for every $n$, so it works for any (open) neighborhood of the origin.
