Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial is bounded on S, then its lowest non-constant degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if S' is Zariski dense in $\mathbb{P}^{n-1}$, then B=Z (classically, sets with similar properties were called "generic"). On the other hand, S' could be defined over $\overline{\mathbb{Q}}$ even when S is not (e.g. $y^2=\pi x$).

Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and B is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible S not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.

Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial with zero constant term is bounded on S, then its highest degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if S' is Zariski dense in $\mathbb{P}^{n-1}$ over Z, then B=Z (classically, sets with similar properties were called "generic"). On the other hand, S' could be defined over $\overline{\mathbb{Q}}$ even when S is not (e.g. $y^2=\pi x$).

Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and B is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible S not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.

Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial is bounded on S, then its lowest degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if S' is Zariski dense in $\mathbb{P}^{n-1}$, then B=Z (classically, sets with similar properties were called "generic"). On the other hand, S' could be defined over $\overline{\mathbb{Q}}$ even when S is not (e.g. $y^2=\pi x$).

Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and B is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible S not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.

Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial is bounded on S, then its lowest non-constant degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if S' is Zariski dense in $\mathbb{P}^{n-1}$, then B=Z (classically, sets with similar properties were called "generic"). On the other hand, S' could be defined over $\overline{\mathbb{Q}}$ even when S is not (e.g. $y^2=\pi x$).

Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and B is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible S not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.