If $A$ and $B$ are closed subspaces of a Hilbert Space, why is $A\cap B=(A^\perp+B^\perp)^\perp$ not true, while $(A\cap B)^\perp=A^\perp+B^\perp$ is? Does this not violate $(A^\perp)^\perp=A$ for all closed subspaces of Hilbert spaces?
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It would obviously violates it. Why wouldn't it be true? – Berci Sep 29 '18 at 00:31
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It isn't, its mentioned in the lecture and we are tasked to prove it. – Sep 29 '18 at 01:24
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@Berci The second equation is wrong. Please see my answer. – Kavi Rama Murthy Sep 29 '18 at 23:52
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Hmm.. It surprised me, I thought sum of closed subspaces must be closed.. – Berci Sep 30 '18 at 09:49
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The mistake you are doing in deriving the second identity from the first is in assuming that $A^{\perp}+B^{\perp}$ is closed. Sum of two closed subspaces in a Hilbert space need not be closed. Let $M$ and $N$ be two closed subspaces such that $M+N$ is not closed. Let $A=M^{\perp}$ and $B=N^{\perp}$. This gives a counter-example to the second identity. Let me know if you need help in verifying this. [ LHS is the closure of RHS in this case].

Kavi Rama Murthy
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For an example of two closed subspaces whose sum is not closed see https://math.stackexchange.com/questions/135471/the-direct-sum-of-two-closed-subspace-is-closed-hilbert-space – Kavi Rama Murthy Sep 29 '18 at 12:25