Does the following statement hold?


Let $p$ be a rational prime. For each $x\in\prod\_{\ell}{\mathbb Z}\_{\ell}$, there exist a natural number $M\_x$ such that $x$ lies in the closure of $\{e\_1p^{n\_1}+e\_2p^{n\_2}+\cdots+e\_{M\_x}p^{n\_{M\_x}}: \text{$n\_i\ge 0$ and $e\_i\in\{-1,0,1\}$ for each $i$}\}$ in $\prod\_{\ell} {\mathbb Z}\_{\ell}$. 


If it does not hold, what if we only consider those $x\in\prod\_{\ell}{\mathbb Z}\_{\ell}$ whose component at $p$ is zero?



Convergence in a product of $p$-adic groups
The answer is negative and we can tell this just by looking at $\mathbb Z\_p$.


For each $M$, the set of sums of $\leq M$ terms of the form $\pm p^n$ for arbitrary nonnegative $n$ is closed in $\mathbb Z\_p$. Because convolutions of closed sets are closed, it is sufficient to prove this for $\{\pm p^n\} \cup \{0\}$, where it is obvious - $\{0\}$ is the only limit point of $\{\pm p^n\}$.


So the union over all $M$ of the closure of this set is simply the union over all $M$ of this set and hence consists only (and exactly) of integers.




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The answer to the second question is also negative. Numbers of the form $e\_i p^{n\_i}$ can take at most $2k+1$ possible values modulo $p^k-1$, as the value only depends on $e\_i$ and $n$ mod $k$. Hence sums of $M$ terms of that type can only take at most $(2k+1)^M$ values modulo $p^{k}-1$. The set elemeents in $\prod\_{\ell\neq p} \mathbb Z\_\ell$ that lie in a given residue class is closed, so the closure of the set of sums of $M$ terms of that type has measure at most $\frac{(2k+1)^M}{p^{k}-1}$. Taking $k$ to $\infty$, it has measure $0$ in $\prod\_{\ell\neq p} \mathbb Z\_\ell$.


Hence the union over all $M$ also has measure $0$ in $\prod\_{\ell\neq p} \mathbb Z\_\ell$.


