2 Minor clarifications in proof of Lemma 2: subsums of powers of b are to be taken with the same signs as the t_i

This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.

Lemma 1: Suppose that $0 < r < 1$. Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}_{\ge 0} \rbrace$. Fix $k \ge 1$. Let $S_k$ be the set of sums of the form $s_1+\cdots+s_k$ such that $s_i \in S$ and $|s_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum_{i \in I} s_i = 0$. Then $0$ is not in the closure of $S_k$.

Proof: Use induction on $k$. The base case is trivial: $S_1=\lbrace -1,1\rbrace$. Now suppose $k \ge 2$. If a sequence $(x_i)$ in $S_k$ converges to $0$, then the smallest summand in the sum giving $x_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S_k$, which are all nonzero. Discarding the finitely many $x_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x_i$ yields a sequence $(y_i)$ in $S_{k-1}$ tending to $0$, contradicting the inductive hypothesis.

Now fix $b>1$ and $k$. Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}_{\ge 0} \rbrace$. Let $T_k$ be the set of sums of the form $t_1+\cdots+t_k$ with $t_i \in T$.

Lemma 2: Each $t=t_1+\cdots+t_k \in T_k$ equals $u_1+\cdots+u_\ell+\delta$ for some $\ell \le k$ and some $u_i \in T$ with $u_i = O(t)$ and $\delta = O(1)$.

Proof: Examine the powers of $b$ used in the $t_i$. If any nonempty subsum (with signs) of these powers equals $0$, the corresponding $t_i$ sum to $O(1)$. If $b^n$ is the largest power that remains after removing all such subsums, divide all the remaining $t_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t_i$, which are $O(b^n)$, are $O(t)$.

Corollary: The number of elements of $T_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.

Corollary: $T_k \ne \mathbb{Z}$.

1

This is an answer to your "actual question" (2), building on some of the ideas in Douglas Zare's answer.

Lemma 1: Suppose that $0 < r < 1$. Let $S=\lbrace \epsilon r^i : \epsilon = \pm 1 \text{ and } i \in \mathbb{Z}_{\ge 0} \rbrace$. Fix $k \ge 1$. Let $S_k$ be the set of sums of the form $s_1+\cdots+s_k$ such that $s_i \in S$ and $|s_1|=1$ and there is no nonempty subset $I \subset \lbrace 1,\ldots,k \rbrace$ with $\sum_{i \in I} s_i = 0$. Then $0$ is not in the closure of $S_k$.

Proof: Use induction on $k$. The base case is trivial: $S_1=\lbrace -1,1\rbrace$. Now suppose $k \ge 2$. If a sequence $(x_i)$ in $S_k$ converges to $0$, then the smallest summand in the sum giving $x_i$ must tend to $0$, since a lower bound on the absolute values of the summands rules out all but finitely many elements of $S_k$, which are all nonzero. Discarding the finitely many $x_i$ for which the smallest summand is $\pm 1$ and removing the smallest summand from each remaining $x_i$ yields a sequence $(y_i)$ in $S_{k-1}$ tending to $0$, contradicting the inductive hypothesis.

Now fix $b>1$ and $k$. Let $T=\lbrace \epsilon \lfloor b^n + 1/2 \rfloor : \epsilon = \pm 1 \text{ and } n \in \mathbb{Z}_{\ge 0} \rbrace$. Let $T_k$ be the set of sums of the form $t_1+\cdots+t_k$ with $t_i \in T$.

Lemma 2: Each $t=t_1+\cdots+t_k \in T_k$ equals $u_1+\cdots+u_\ell+\delta$ for some $\ell \le k$ and some $u_i \in T$ with $u_i = O(t)$ and $\delta = O(1)$.

Proof: Examine the powers of $b$ used in the $t_i$. If any nonempty subsum of these powers equals $0$, the corresponding $t_i$ sum to $O(1)$. If $b^n$ is the largest power that remains, divide the remaining $t_i$ by $b^n$, and apply Lemma 1 with $r=1/b$ to see that $|t|/b^n$ is bounded away from $0$, so all these remaining $t_i$, which are $O(b^n)$, are $O(t)$.

Corollary: The number of elements of $T_k$ of absolute value less than $B$ is $O((\log B)^k)$ as $B \to \infty$.

Corollary: $T_k \ne \mathbb{Z}$.