Hilbert vs. De Morgan Problem
Given a Hilbert space $\mathcal{H}$.

Then it holds:
  $$\overline{\left\langle\bigcap_{\lambda\in\Lambda}A_\lambda^\perp\right\rangle}=\left(\bigcup_{\lambda\in\Lambda}A_\lambda\right)^\perp\quad\overline{\left\langle\bigcup_{\lambda\in\Lambda}A_\lambda^\perp\right\rangle}=\left(\bigcap_{\lambda\in\Lambda}A_\lambda\right)^\perp$$
  How can I prove this?

Span, closure, orthogonal complement: $\langle A\rangle$, $\overline{A}$, $A^\perp$
Application
This applies in: Beppo-Levi
 A: Meanwhile I got it.. :)
Equality
On the one hand:
$$\varphi\in\big(\bigcup_{\lambda\in\Lambda}A_\lambda\big)^\perp\iff\varphi\perp\bigcup_{\lambda\in\Lambda}A_\lambda\iff\varphi\perp A_\lambda\quad(\forall\lambda\in\Lambda)$$
$$\varphi\perp A_\lambda\quad(\forall\lambda\in\Lambda)\iff\varphi\in A_\lambda^\perp\quad(\forall\lambda\in\Lambda)\iff\varphi\in\bigcap_{\lambda\in\Lambda}A_\lambda^\perp$$
So one obtains:
$$\overline{\left\langle\bigcap_{\lambda\in\Lambda}A_\lambda^\perp\right\rangle}=\overline{\bigcap_{\lambda\in\Lambda}A_\lambda^\perp}=\bigcap_{\lambda\in\Lambda}A_\lambda^\perp=\left(\bigcup_{\lambda\in\Lambda}A_\lambda\right)^\perp$$
Concluding equality.
Inclusion
On the other hand:
$$\varphi\in\bigcup_{\lambda\in\Lambda}A_\lambda^\perp\iff\varphi\in A_{\lambda_0}^\perp\quad(\exists\lambda_0\in\Lambda)\iff\varphi\perp A_{\lambda_0}\quad(\exists\lambda_0\in\Lambda)$$
$$\varphi\perp A_{\lambda_0}\quad(\exists\lambda_0\in\Lambda)\implies\varphi\perp\bigcap_{\lambda\in\Lambda}A_\lambda\iff\varphi\in\big(\bigcap_{\lambda\in\Lambda}A_\lambda\big)^\perp$$
So one obtains:
$$\bigcup_{\lambda\in\Lambda}A_\lambda^\perp\subseteq\left(\bigcap_{\lambda\in\Lambda}A_\lambda\right)^\perp\implies\overline{\left\langle\bigcup_{\lambda\in\Lambda}A_\lambda^\perp\right\rangle}\subseteq\left(\bigcap_{\lambda\in\Lambda}A_\lambda\right)^\perp$$
Concluding inclusion.
Strictness
Regard dense spaces:*
$$A_\pm:=\mathcal{D}_\pm<\overline{\mathcal{D}_\pm}=\mathcal{H}:\quad \mathcal{D}_+\cap\mathcal{D}_-=\{0\}$$
Then one obtains:
$$\overline{\langle A_+^\perp\cup A_-^\perp\rangle}=\overline{\langle\{0\}\cup\{0\}\rangle}=\{0\}\subsetneq\mathcal{H}=\{0\}^\perp=\left(A_+\cap A_-\right)^\perp$$
Concluding strictness.
*See thread: Dense Spaces
