From the article [Weyl's law](http://www-personal.umich.edu/~stevmatt/weyl_law.pdf) of Matt Stevenson page $3$, there's a lemma where I'd like some help to understand the proof. 



> 
> **Lemma.** Let $\mu\_n$ be the Dirichlet eigenvalues on $\Omega$ and $\nu\_n$ be the Neumann eigenvalues on $\Omega$, then $\nu\_n \leq
>  \mu\_n$.
> 
> 
> *Pf :* Remark that $H\_0^1 (\Omega) \subset H^1 (\Omega)$, so $\Phi(H\_0^1 (\Omega)) \subset \Phi(H^1 (\Omega)).$ Then, version $2$
>  of the Minimax Principle says that $\nu\_n$ is given by minimizing the
>  same expression as for $\mu\_n$ over a larger space, and hence $\nu\_n
>  \leq \mu\_n$.
> 
> 
> 


**Question :** Could anyone be able to explain to me with clear words, rigorously, why the Minimax Principle says that $\nu\_n$ is given by minimizing the same expression (i.e. $ \inf\_{X \in \Phi\_n(V)} \sup\_{u \in X} \frac{\|\nabla u(\Omega)\|^2}{\|u(\Omega)\|^2}$) as for $\mu\_n$ over a larger space ?



Why the Minimax Principle says that $\nu_n$ is given by minimizing the same expression as for $\mu_n$ over a larger space
To save my fingers some typing, denote by $\mathcal{R}$ the Rayleigh quotient, $\mathcal{R}(u) = \|\nabla u\|^2/\|u\|^2$. Let 
$$R(H) = \{\sup\_{u\in X^n}\mathcal{R}(u)\ |\ X\in\Phi(H)\}$$
where $H\in \{H^1\_0,H^1\}$. Notice that $R$ is the image of a function $\Phi(H)\to\mathbb{R}$ taking $X\mapsto\sup\_{u\in X}\mathcal{R}(u)$. As $\Phi(H^1\_0)\subset\Phi(H^1)$, we have that $R(H^1\_0)\subseteq R(H^1)$. That $\mu\_n\geq\nu\_n$ follows from taking infima.


