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Let $H$ a Hilbert space, and $\mathcal C$ an arbitrary set of closed subspaces of $H$. Is it true that $$\left( \bigcap_{Z\in \mathcal C}Z\right)^\perp = \overline{\sum_{Z\in \mathcal C} Z^\perp}$$ where $\sum$ means finite sums of elements in the terms. The inclusion $\supset$ is clear, so I really am interested in $\subset$. I have found a similar question asked here: Orthogonal complements of intersections of closed subspaces

But my question is more general than this. I know it is true in finite dimensions. Also, in the thread above Jochen Wengenroth claims it is true for finite $\mathcal C$, but I do not understand the argument. What about countable $\mathcal C$? Uncountable $\mathcal C$? Does the situation become easier if we assume $\mathcal C$ totally ordered by inclusion?

I don't know how to approach this so really any help is appreciated.

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    $\begingroup$ Welcome to MathOverflow! Dosn't the inclusion you want follow from the fact that the closure of a vector subspace $V$ can be otained by taking the orthogonal complement twice? $\endgroup$
    – Jochen Glueck
    Commented Jun 13, 2023 at 8:07
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    $\begingroup$ To add some detail to Jochen's comment: in general $\overline{L(A)}=A^{\perp\perp}$, so the RHS of your identity is $(\bigcup Z^{\perp})^{\perp\perp}=(\bigcap Z^{\perp\perp})^{\perp}=LHS$. Here we use $(\bigcup A_j)^{\perp}=\bigcap A_j^{\perp}$, which is clear from the definition of the orthogonal complement. $\endgroup$
    – Christian Remling
    Commented Jun 13, 2023 at 16:36
  • $\begingroup$ That makes sense, thanks! $\endgroup$
    – Nathaël
    Commented Jun 14, 2023 at 1:13

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