Courant-Fischer-Weyl Theorem says:
If $A$ is $n$ by $n$ Hamiltonian matrix with eigenvalues $\lambda_1\ge\cdots\ge\lambda_n$, then $$\lambda_k=\max_S\left\{\min_x\left\{x^*Ax\big\vert\ |x|=1,x\in S\right\}\big\vert S\le \mathbb{C}^n,\dim{S}=k\right\}.$$
I have a question for this proof. The set $\left\{x^*Ax\big\vert\ |x|=1,x\in S\right\}$ has a minimum, since $|x|=1,x\in S$ gives compactness and $x^*Ax$ is continous. However, I don't know why mximum is taken over $S$ instead of supremum. Is it guaranteed that the set $\left\{\min_x\left\{x^*Ax\big\vert\ |x|=1,x\in S\right\}\big\vert S\le \mathbb{C}^n,\dim{S}=k\right\}$ contains its supremum?
I found a document which explains maximum for the proof, but it is too short to accept.
Is there any interpretation for this?
One can indeed take the maximum. Fix $k$. Consider $O_k$, the set of all $k$-unit vectors $(v_1, \cdots +v_k)$ so that $\langle v_i, v_j\rangle = 0$ when $i\neq j$. For each $(v_1, \cdots, v_k)\in O_k$, one has the $k$-plane $S= \operatorname{span}\{ v_1, \cdots, v_k\}$. Let $$U = \{ \vec t:=(t_1, \cdots, t_k) \in \mathbb C^k : |t_1|^2 + \cdots +|t_k|^2=1\}.$$
Thus one can consider a map \begin{align} F : O_k \times S &\to \mathbb R, \\ F((v_1, \cdots, v_k), \vec t) &= x^*Ax, \end{align} where $x = t_1 v_1 + \cdots t_k v_k$. Clearly $F$ is continuous. Then for each $S = \operatorname{span}\{v_1, \cdots, v_k\}$,
$$\min_x\{x^*Ax\big\vert\ |x|=1,x\in S\big\vert\} = \min_{\vec t\in U} F((v_1,\cdots, v_k), \vec t).$$
Since $U$ is a closed and bounded subset in $\mathbb C^k$,
$$ G : O_k \to \mathbb R, \ \ G(v_1, \cdots, v_k) = \min_{\vec t\in U} F((v_1,\cdots, v_k), \vec t)$$ is a continuous function. Since $O_k$ is also a closed and bounded subset in $\mathbb C^{nk}$, $G$ attains a maximum. Let $(v_1, \cdots, v_k)$ be a maximum of $G$, then $S = \operatorname{span} \{v_1, \cdots, v_k\}$ attains the maximum for
$$\left\{\min_x\left\{x^*Ax\big\vert\ |x|=1,x\in S\right\}\big\vert S\le \mathbb{C}^n,\dim{S}=k\right\}.$$