Does there exist a Bernstein set such that a finite sum still Bernstein?

Question

Recall that $B\subset\Bbb R$ is a Bernstein set if $B\cap P\neq\emptyset \neq P\setminus B$ for every perfect set $P\subset\Bbb R.$ It can be constructed by an easy transfinite induction. Moreover, there is some construction for Bernstein set to satisfy specific conditions like there is a Bernstein set $B\subset\Bbb R$ such that $B+B$ is a Bernstein set, see the this answer https://math.stackexchange.com/a/1534151/707884. I think we can also show that there is a Bernstein set $B$ such that $B+B+\cdots+B$ (n-times) is a Bernstein set. Also, $r\cdot B$ is a Bernstein set for every $r\in\Bbb R\setminus\{0\}.$ Here is my question

Question 1. Can we construct a Bernstein set $B\subset\Bbb R$ such that if $$a_1 B+a_2B+\cdots+a_nB\neq 0,$$ then $a_1 B+a_2B+\cdots+a_nB$ is a Bernstein set for every $a_1,a_2,\cdots,a_n\in\Bbb R$?

if Question 1 is impoissble. I will make a weak assumption

Question 2 Can we construct a set $D\subset\Bbb R$ such that $D$ intersect each perfect set and if $$a_1 D+a_2D+\cdots+a_nD\neq 0,$$ then $a_1 D+a_2D+\cdots+a_nD\neq\Bbb R$ for every $a_1,a_2,\cdots,a_n\in\Bbb R$?

It seems that the question wants to ensure that any linear combination of $B$ is still Bernstein set. I know there is an example of a Bernstein set $B\subset\Bbb R$ such that $B$ is linearly independent. But I do not think this would be sufficient.

Any help would be appreciated greatly.

Answer 1

Yes, we can build a Bernstein set such that any non-trivial linear combination is a Bernstein set. Let A be the set of all the pairs $(t,C)$ where $ t = (t_i) $ is a $n$-tuple of non-zero reals and $C$ is a perfect subset. Then, we want a subset $B$ such that for any pair $ (t,C) \in A$, $B \cap C \neq \emptyset$ and $ C - (t_1 B + t_2 B + \ldots + t_n B) \neq \emptyset$. Because $ \mathrm{Card}(A) = \mathfrak{c} $, we can simply do this by transfinite induction, similarly to the usual construction of a Bernstein set.

Edit: Let's do a more complete proof. We put on $A = (p_\alpha)_{\alpha < \mathfrak{c}}$ an order isomorphic to $ \mathfrak{c} $. We will define two subsets $B$ and $ D $ by transfinite induction. Let $ p_0 = (t_0, C_0) $ be the first pair in $ A $, we choose $ b_0 \in C_0$, and $ d_0 \in C_0 $ different from $t_{0,1} b_0 + \ldots + t_{0,n_0} b_0$. Now, suppose that we have defined the sequences $ (b_\beta)_{\beta < \alpha} $ and $ (d_\beta)_{\beta < \alpha} $ for some ordinal $ \alpha < \mathfrak{c} $, such that for any $ \gamma < \alpha $, with the pair $p_\gamma = (t_\gamma,C_\gamma)$ and the set $B_\alpha = \{ b_\beta \mid \beta < \alpha \}$, we have $ d_\gamma \notin t_{\gamma,1} B_\alpha + \ldots + t_{\gamma,n_\gamma} B_\alpha$. Then, we want to find $ b_\alpha \in C_\alpha$ such that for any $\gamma < \alpha$, we have $d_\gamma \notin t_{\gamma,1} (B_\alpha \cup \{b_\alpha\}) + \ldots + t_{\gamma,n_\gamma} (B_\alpha \cup \{b_\alpha\})$. Another way of saying this is that for any subset $S \subseteq \{1,\ldots,n_\gamma\}$ and its complementary $T = \{1,\ldots,n_\gamma\} - S$, $$ d\gamma \notin \sum_{i \in S} t_{\gamma,i} B_\alpha + (\sum_{i \in T} t_{\gamma,i}) b_\alpha$$ For any reals $r,s$ the equation $ rx + s = d_\gamma$ has only one solution if $r \neq 0$, but if $r = 0$ and $s = d_\gamma$, any $x$ is a solution. So here, if we fix a subset $S$, we have two cases:

1. if $(\sum_{i \in T} t_{\gamma,i}) \neq 0$, there are only $\mathrm{Card}(B_\alpha) < \mathfrak{c}$ elements that we cannot choose, so no problem.
2. if $(\sum_{i \in T} t_{\gamma,i}) = 0$, we can choose any $b_\alpha$ that we want, because for any $ b \in B_\alpha $ we have by hypothesis that $$ d_\gamma \notin \sum_{i \in S} t_{\gamma,i} B_\alpha + (\sum_{i \in T} t_{\gamma,i}) b = \sum_{i \in S} t_{\gamma,i} B_\alpha $$

Because the set $\{1,\ldots,n_\gamma\}$ has only finitely many subsets, and because there are less than $\mathfrak{c}$ ordinals smaller than $\alpha$, there are only $\mathrm{Card}(B_\alpha)$ elements that we cannot choose in $C_\alpha$, therefore we can find a $b_\alpha$ which verifies the condition, and we define $B_{\alpha + 1} = B_\alpha \cup \{b_\alpha\}$. Then, we choose $d_\alpha \in C_\alpha - (t_{\alpha,1} B_{\alpha + 1} + \ldots + t_{\alpha,n_\alpha} B_{\alpha + 1})$. You define $B = \bigcup B_\beta$. Then, for any pair $p_\beta = (t_\beta,C_\beta)$, you have that $d_\beta \in C_\beta - (t_{\beta,1} B + \ldots t_{\beta,n_\beta} B)$, and therefore for any non-trivial tuple $t$, $t_1 B + \ldots + t_n B$ is a Bernstein set.