Does this condition characterise intervals, among subsets of the real line?

Question

For a real number, $c\in \left]0,1\right[$, consider the following property $\mathbf(\mathbf P_c\mathbf)$ of subsets $A$ of $\mathbb R$:

• $\mathbf(\mathbf P_c\mathbf)$ For every bounded set $B\subset \mathbb R$ $$ B\subset A+cB\implies (1-c)B\subset A. $$ This may be restated equivalently as:
• $\mathbf(\mathbf Q_c\mathbf)$ For every bounded sequence $(a_j)_{j\ge0} \subset A$ one has $(1-c)\sum_{j=0}^\infty a_j c^j\in A$.

Indeed, if $A\subset \mathbb R$ satisfies $\mathbf(\mathbf P_c\mathbf)$, then take $B:=\bigl\{\sum_{j=0}^\infty a_jc^j: (a_j)_{j\ge0}\subset A, \text{ bounded sequence}\bigr\}$, so $B\subset A+cB$, and by $\mathbf(\mathbf P_c\mathbf)$ we have $(1-c)\sum_{j=0}^\infty a_j c^j\in A$. Conversely, if $A$ satisfies $\mathbf(\mathbf Q_c\mathbf)$, and $B$ is a bounded subset, say $B\subset [-M,M]$, such that $B\subset A+cB$, then, for every $b\in B$ we may form inductively a sequence $(b_j)_{j\ge0}\subset B$ and a sequence $(a_j)_{j\ge0}\subset A\cap [-2m,2m]$ so that $b_0=b$ and $b_n=a_n+cb_{n+1}$ for all $n\in\mathbb N$. So $b=\sum_{j=0}^na_jc^j+b_{n+1}c^{n+1},$ and since $b_{n+1}c^{n+1}=o(1)$, we have $(1-c)b=(1-c)\sum_{j=0}^\infty a_j c^j\in A$ by $\mathbf(\mathbf Q_c\mathbf)$, whence $(1-c)B\subset A$. In particular, we see that every real interval $A$ satisfies $\mathbf(\mathbf Q_c\mathbf)$.

The converse is also true, I think, which is no surprise: Every $A\subset \mathbb R$ verifying $\mathbf(\mathbf P_c\mathbf)$ (or $\mathbf(\mathbf Q_c\mathbf)$) is an interval. Yet I could not find a quick argument, and I'd like to see one, which is what I'm asking here. Both short existence proofs and constructive ones are welcome!

Note that if $A$ satisfies $\mathbf(\mathbf P_c\mathbf)$, so does every affine image of it, so if $A$ has more that one point, we can assume $\min A=0$ and $\max A=1 $. If $c=1/2$, property $\mathbf(\mathbf Q_c\mathbf)$ then says all binary expansions are in $A$, thus we can conclude that $A=[0,1]$, and the same observation is sufficient to treat the case of $1/2\le c\le 1$. For $0<c<1/2$, these points would only produce a Cantor set, but we can plug them again as sequences in $\mathbf(\mathbf Q_c\mathbf)$, so actually we have $(1-c)^2\sum_{j=0}^\infty\sum_{j=0}^\infty a_{ij} c^{i+j} \in A$ for all bounded double sequences $(a_{ij})_{(ij)\in\mathbb N^2}\subset A$, and more generally, iterating,

$$(1-c)^m\sum_{k=0}^\infty \left(\sum_{\substack{j\in\mathbb N^m\\ |j|=m}} a_{j_1,\dots,j_m}\right) c^m\in A $$ for every $m\in\mathbb N$ and every bounded family $(a_j)_{j\in \mathbb N^m}\subset A$. In particular, also

$$(1-c)^m\sum_{k=0}^\infty \binom{m+k-1}k a_k c^k\in A $$ for every bounded sequence $(a_k)_{k\in \mathbb N}$ and integers $0\le n_k\le \binom{m+k-1}k$. These seem sufficient to conclude, but is it necessary to follow this path?

Answer 1

(not to leave this unanswered) As observed in comments, the answer is quick as we know that $(1-c)^m\sum_{k\ge0} n_k c^k\in A$ for all $m\in\mathbb N$ and all choices of integer coefficients $0\le n_k\le \binom{m+k-1}k$; in fact a single $m$ such that $(m+1)c\ge1$ is sufficient. To simplify the description, we may see any such series $(1-c)^m\sum_{k\ge0} n_k c^k$ equivalently as a sum $\sum_{j\in J}w_j$ of a sub-family of a sequence $(w_j)_{j\in\mathbb N}$, where the $w_j$'s are just the $(1-c)^mc^k$, each repeated $\binom{m+k-1}k$ times, in decreasing order. The condition $(m+1)c\ge1$ ensures that $w_j \le \sum_{i>j} w_i$ for all $j\ge0$.

A general basic principle (classic stuff I think) to refer to:

Let $(w_j)_{j\in\mathbb N}$ be a decreasing sequence of positive weights summing to $\sum_{j=0}^\infty w_j=1$. Denote $r_j:=\sum_{i>j} w_i$, and $S:=\bigl\{\sum_{j\in J}w_j: J\subset \mathbb N\bigr\}$, the set of all sums of sub-families of the weights. Then $S=[0,1]$ if and only if $w_j\le r_j$ for all $j\ge0$.

Sketch: For every $x\in[0,1]$ we can form a binary sequence $(\epsilon_j)_{j\in\mathbb N}\in \{0,1\}^\mathbb N$ defining inductively $\epsilon_j=1$ iff $\bigl(\sum_{i=0}^{n-1}\epsilon_iw_i\bigr)+w_j\le x$. By construction, $\epsilon_j=1$ eventually, only if $e_j=1$ for all $j\ge0$, and this implies that $x= \sum_{i=0}^{\infty}\epsilon_iw_i$. On the other hand, if for some $j$ one has $r_j<w_j$, then $S\cap\mathopen]r_j,w_j\mathclose]=\emptyset $.