Show $A \subset B \Leftrightarrow B^{\perp} \subset A^{\perp}$ with $A,B$ subsets of a Hilbert space.

Question

Show $A \subset B \Leftrightarrow B^{\perp} \subset A^{\perp}$ with $A,B$ subsets of a Hilbert space.

The forward direction is easy: Assume $A \subset B$. For any $a \in A$ and $x \in B^\perp$ since $a \in B$ then ${\langle x,a \rangle} = 0$, since this is for any $a \in A$ then $x \in A^\perp$, therefore $B^\perp \subset A^\perp$.

For the converse direction: Assume $B^\perp \subset A^\perp$. For any $a \in A$ and $x \in B^\perp$, since $x \in A^\perp$ we have ${\langle x,a \rangle} = 0$. Since this is for any $x \in B^\perp$, then $a \in B^{\perp \perp}$. But how do we show $a \in B?$

We know that $B \subset B^{\perp \perp}$ but I don't see how that helps.

Answer 1

Consider $\mathbb R^2$ with the usual inner product. Let $A,B$ be one-element sets $$ A = \{(1,0)\},\qquad B = \{(2,0)\} . $$ Compute $A^\perp = B^\perp$, but of course $A \not\subseteq B$.

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We can deduce this converse if we assume $B$ is a closed linear subspace.

It is not enough (in the infinite-dimensional case) to assume $B$ is a linear subspace.