Additive function $f: \mathbb{Z}^\infty \rightarrow \mathbb{Z}$ is zero everywhere.

Question

Let $f: \mathbb{Z}^\infty \rightarrow \mathbb{Z}$ be an additive function ($f(x+y)=f(x)+f(y)$ for every $x,y \in \mathbb{Z}^\infty$). In addition for every $x=(0,\dots, 0,1,0, \dots)$ we have $f(x)=0$. Prove that for every $x\in \mathbb{Z}^\infty$ we have $f(x)=0$.

Known (can be proved that): $$f(1,a,a^2,a^3,\dots) = 0 , \quad \forall 1 < a \in \mathbb{Z} .$$

Answer 1

A rather extensive HINT:

• Show that the hypothesis that $f(e_n)=0$ for each $n\in\Bbb N$ implies that if $x,y\in\Bbb Z^\infty$ differ in at most finitely many coordinates, then $f(x)=f(y)$.

Now let $\sigma_a=\langle a^k:k\in\Bbb N\rangle$, where $1<a\in\Bbb Z^+$. For any $n\in\Bbb N$ let

$$\sigma_a^{(n)}=\langle 0,\ldots,0,a^n,a^{n+1},\ldots\rangle=\sigma_a-\sum_{k<n}a_ke_k\;.$$

• Show that for each $n\in\Bbb N$ there is a sequence $\tau\in\Bbb Z^\infty$ such that $a^n\tau=\sigma_a^{(n)}$. Use this to conclude that $f(\sigma_a)=0$.

• Show that for any $\tau=\langle t_k:k\in\Bbb N\rangle\in\Bbb Z^\infty$ there are $\lambda=\langle\ell_k:k\in\Bbb N\rangle$ and $\mu=\langle m_k:k\in\Bbb N\rangle$ in $\Bbb Z^\infty$ such that $$\begin{align*}\tau&=\left\langle 2^k\ell_k+3^km_k:k\in\Bbb N\right\rangle\\&=\left\langle2^k\ell_k:k\in\Bbb N\right\rangle+\left\langle 3^km_k:k\in\Bbb N\right\rangle\;.\tag{1}\end{align*}$$ This uses a rather basic number-theoretic fact about relatively prime integers.

• Show that $f$ sends each of the sequences in $(1)$ to $0$; what you did for the second bullet point should help you here.