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I have proven this property: Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space V. Then $W^\perp_1 +W^\perp_2=(W_1\cap W_2)^\perp$

So I was wondering is this result also works on vector spaces with infinite dimension. I was trying to come up with a counter-example but I got stuck quickly.

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What does your argument look like? Isn't it obvious that $(W_1 \cap W_2)^\perp \subseteq W_1^\perp + W_2^\perp$? Then the other direction needs to be shown. So let $a+b\in W_1^\perp + W_2^\perp$... – nayrb Oct 22 '13 at 23:31
@nayrb No, $W_1^\perp + W_2^\perp \subseteq (W_1\cap W_2)^\perp$ is the obvious one. The inclusion $(W_1 \cap W_2)^\perp \subseteq W_1^\perp + W_2^\perp$ isn't obvious. – Daniel Fischer Oct 22 '13 at 23:34
@Cath, what structure have you to define $W^\perp$, and how do you define it? Do you know anything about topological concepts yet? – Daniel Fischer Oct 22 '13 at 23:37
To define $W^\perp$ I use the internal product and $W^\perp$ is the subspace of all the vectors that are ortogonal with all the vectors of $W$. – Cath Oct 22 '13 at 23:56
and b) every subspace has the same "orthogonal complement" as its closure, $\overline{W}^\perp = W^\perp$. (A subspace $W$ being closed means it contains all limits of convergent sequences whose terms belong to $W$.) Furthermore, if the space is complete, one has $W^{\perp\perp} = \overline{W}$, the orthogonal complement of the orthogonal complement of a subspace is the closure of the subspace. Now, the fact of the matter is that we do not always have $W_1^\perp + W_2^\perp = (W_1\cap W_2)^\perp$. For example, one can have two subspaces with trivial intersection, which are both dense in the – Daniel Fischer Oct 23 '13 at 0:12

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