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I proved that $(A^\bot)^\bot \supset \overline{\operatorname{span}A}$. To prove the reverse inclusion, let $x \in (A^\bot)^\bot$, then $\langle x, y\rangle = 0$ for all $y \in A^\bot$. How can I proceed? Can anyone suggest a proof without refering to the answer in double Orthogonal complement is equal to topological closure?

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ywx
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  • I think the answer given in http://math.stackexchange.com/questions/1043940/double-orthogonal-complement-is-equal-to-topological-closure is the "right" one. And in fact both answers below basically equal that. – user126154 Apr 22 '16 at 10:30
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    So, why do you want an answer different from that? what exactly are you searching for? – user126154 Apr 22 '16 at 10:30
  • I notice all the proofs seem to assume the identity $(A^\bot)^\bot = A$ if $A$ is a closed subset, and I was wondering whether you have to use it in this problem. – ywx Apr 22 '16 at 12:15

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$A^\perp = \operatorname{span}(A)^\perp = \overline{\operatorname{span}(A)}^\perp = \overline{A}^\perp$ and $A^\perp$ is closed.

This is clear from the following observations: $x$ is orthogonal to $A$ iff it is orthogonal to the span of $A$, by linearity of inner products.

By continuity of inner products, if $x_n \rightarrow x$ and all $x_n$ obey $\langle x_n,b \rangle = 0$ for all $b \in B$, the same holds for $x$. So sets of the form $B^\perp$ are always closed. If $x \in B^\perp$ and $p \in \overline{B}$ then some sequence $(x_n)$ from $B$ converges to $p$, then $\langle x_n,x \rangle = 0$ for all $n$ so also (continuity!) $\langle x,p \rangle = 0$, so $x \in \overline{B}^\perp$. The reverse inclusion is obvious, so $\overline{B}^\perp = B^\perp$ for all $B$. The last two imply that $\perp$ and closure (as operations on sets) commute.

Use this to finish off the proof.

Henno Brandsma
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