I know a similar question says that $H_0^1(\Omega)=H_0(div;\Omega)\cap H_0(curl;\Omega)$, which was shown in Lemma~ 2.5 of the book "Finite Elements Methods for Navier-Stokes Equations" by Girault and Raviart. And $H_0^1(\Omega)$ is obviously compactly embedded in $L^2(\Omega)$. In the case of weakened boundary conditions, I know that there is such a conclusion ("$\hookrightarrow\hookrightarrow $" means compactly embedding)
$$ H_0(div;\Omega)\cap H(curl;\Omega) \hookrightarrow\hookrightarrow L^2(\Omega) $$ and $$ H(div;\Omega)\cap H_0(curl;\Omega) \hookrightarrow\hookrightarrow L^2(\Omega), $$ which has been shown in Corollary 3.49 of the book "Finite Element Methods for Maxwell's Equations" by Peter Monk .
But I want to know, when the boundary conditions are completely removed, the space is still compact embedded in $L^2(\Omega)$ ? Associated graphic norm $\|\cdot\|^2_{V}:=\|\nabla\times\cdot\|^2+\|\nabla\cdot \cdot\|^2+\|\cdot\|^2$ on $V:=H(div; \Omega ) \cap H(curl; \Omega )$, can we get the following results?
$$V \hookrightarrow\hookrightarrow L^2(\Omega).$$