I have a question related to set theory, of which I am a beginner. Please add/change tags if you have better references. Other than describing my main question, I'm also highlighting with the symbol [?] the parts I'm really not sure about.
Set up of the problem
Fix $r\in \mathbb{N}$ and $d\equiv r+\binom{r}{2}$ and consider the region (plane [?]) $$ \begin{aligned} \mathcal{B}\equiv \{(b_1,b_2,..., b_d)\in \mathbb{R}^{d}: \text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \\ &b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\\ &...,\\ & b_d=b_{r-1}-b_r\} \end{aligned} $$ For example, when $r=2$ ($d=3$) we have the surface $$ \begin{aligned} \mathcal{B}\equiv \{(b_1,b_2,b_3)\in \mathbb{R}^{3}: \text{ } & b_3=b_1-b_2\} \end{aligned} $$ When $r=3$ ($d=6$) we have $$ \begin{aligned} \mathcal{B}\equiv \{(b_1,..., b_6)\in \mathbb{R}^{6}: \text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3\} \end{aligned} $$
My goal: I want to write down the region [?] that is complement to $\mathcal{B}$ in $\mathbb{R}^d$ as a union of "boxes".
This is how I thought to proceed for $r=2$
Define these two boxes given $(b_1, b_2)\in \mathbb{R}^2$ $$ B(b_1, b_2)\equiv \{(x,y,z) \text{ s.t. } x\leq b_1, -y\leq -b_2, z> b_1-b_2\} $$ $$ Q(b_1, b_2)\equiv \{(x,y,z) \text{ s.t. } x> b_1, -y> -b_2, z\leq b_1-b_2\} $$ Then, $$ \cup_{(b_1, b_2)\in \mathbb{R}^2} B(b_1, b_2) $$ should be the region (plane [?]) above $\mathcal{B}$ and $$ \cup_{(b_1, b_2)\in \mathbb{R}^2} Q(b_1, b_2) $$ should be the region (plane [?]) below $\mathcal{B}$.
Hence, $$ \Big\{\cup_{(b_1, b_2)\in \mathbb{R}^2} B(b_1, b_2)\Big\} \cup \Big\{\cup_{(b_1, b_2)\in \mathbb{R}^2} Q(b_1, b_2)\Big \} $$ is the region [?] that is complement to $\mathcal{B}$ in $\mathbb{R}^3$.
Is this correct? How do I generalise this to any $r$?