Comparison of capacity of sets in $\mathbb{R}^n$ This is mainly in reference to this MSE post. Let $B_r \subset \mathbb{R}^n$ denote the ball of radius $r$ centered at the origin. Consider any set $F \subset B_1$. For all sets $\Omega \subset \mathbb{R}^n$ such that $F \subset \Omega$, define $2$-capacity by 
$$cap_2 (F, \Omega) = \inf_{\phi|_F = 1, \phi \in C^\infty_0(\Omega)}\int_\Omega |\nabla \phi|^2 dV$$
It is clear that $cap_2(F, B_2) \geq cap_2(F, B_3)$. The answer there by soup proves that we have a constant $C$ such that $cap_2(F, B_2) \leq Ccap_2(F, B_3)$. This is done by considering a diffeomorphism $F$ between $B_2$ and $B_3$ 
such that both $F$ and $F^{-1}$ have bounded derivatives. 
My question, is what happens if we substitute $B_3$ for the whole space $\mathbb{R}^n$? Could we still say that $cap_2(F, B_2) \leq Ccap_2(F, \mathbb{R}^n)$?
 A: Yes, this still holds, and I would prove it in a different (maybe equivalent) way. The key is the existence of $\psi \in C_0^\infty(B_3)$ with $\psi \equiv 1$ on $B_1$.
Then, for all $\varepsilon > 0$, there is $\phi_\varepsilon \in C_0^\infty(B_2)$ with $\phi_\varepsilon \equiv 1$ on $F$
and $$\int |\nabla \phi_\varepsilon|^2 \, \mathrm{d}V \le cap_2(F,B_2) + \varepsilon.$$
Hence, $\psi \cdot \phi_\varepsilon \in C_0^\infty(B_3)$ and $\psi \cdot \phi_\varepsilon \equiv 1$ on $F$.
Moreover, there is a constant $C > 0$ (independent of $\varepsilon$) with
$$\int|\nabla(\psi \cdot \phi_\varepsilon)|^2 \, \mathrm{d}V
\le C \, \int|\nabla\phi_\varepsilon|^2 \, \mathrm{d}V.$$
Thus,
$$cap_2(F,B_3) \le \int|\nabla(\psi \cdot \phi_\varepsilon)|^2 \, \mathrm{d}V \le C \, (cap_2(F,B_2) + \varepsilon).$$
The limit $\varepsilon \to 0$ finishes the inequality.
By the way: I think your definition of capacity is slightly wrong. Indeed, the set $\mathbb{Q}^n \cap B_1$ is countable and, thus, should have capacity zero (for $n > 1$). But with your definition, the capacity is $+\infty$.
May I ask, where do you found this definition?
