&\leq C+C\sum_{n=0}^{N-1}\tau\|\psi_h^{n+1}\|_{W^{1,q}}^2 {\nonumber}\\
&\leq C+C_\epsilon\sum_{n=0}^{N-1}\tau\|\psi_h^{n+1}\|_{H^{1}}^2
+\epsilon\sum_{n=0}^{N-1}\tau\|\psi_h^{n+1}\|_{W^{1,q+\delta}}^2  ,
\quad\forall\,\epsilon\in(0,1) .\end{aligned}$$ The last two inequalities imply $$\begin{aligned}
\sum_{n=0}^{N-1}\tau\|\psi_h^{n+1}\|_{W^{1,q+\delta}}^2\leq 
C+C\sum_{n=0}^{N-1}\tau\|\psi_h^{n+1}\|_{H^{1}}^2
+C\|\psi_h^0\|_{H^1}^2
\leq C. \end{aligned}$$

The compactness and convergence of the finite element solution are proved based on the uniform estimates established. On one hand, in both and we need some constant $\delta>0$ (which depends on the given curved polyhedron) to prove the convergence of the finite element solution. On the other hand, both and are sharp: for any $\delta>0$ there exists a polyhedron such that and do not hold.

Proof of Theorem \[MainTHM\] {#SecProof}
============================

By substituting $\chi_h=\phi_h^{n+1}$ and ${\bf a}_h={\bf A}_h^{n+1}$ into the equations $$\begin{aligned}
& (\phi_h ^{n+1}, \chi_h) 
- ({\bf A}_h^{n+1},\nabla \chi_h ) = 0 \, , 
&&\forall\, \chi_h\in {\mathbb V}_{h}^{k+1}, \\[10pt]
&\frac{1}{\tau} ({\bf A}_{h}^{n+1},{\bf a}_{h}) 
+(\nabla\phi_{h}^{n+1} \, , {\bf a}_{h} )
+ (\nabla\times{\bf A}_h ^{n+1} \, , \nabla\times {\bf a}_{h}) =0 ,
  &&\forall\, {\bf a}_h\in  {\mathbb N}_{h}^{k} ,\end{aligned}$$ we see that the two equations above have only zero solution. Hence, for any given $(\psi_h^{n},{\bf A}_h^{n})\in {\mathbb S}_{h}^{r} \times {\mathbb N}_{h}^{k}$, the linear system - has a unique solution $(\phi_h^{n+1},{\bf A}_h^{n+1})\in 
{\mathbb V}_{h}^{k+1} \times {\mathbb N}_{h}^{k}$.

Under the condition $\tau<\eta$, it is easy to see that for any given ${\bf A}_h^{n+1}\in{\mathbb 