Examples of $(\mathcal K^\bot)^\bot=\mathcal K$ failing.

Question

For a closed linear subspace, $\mathcal K$, of a Hilbert space $\mathcal H$, it is well known that,

$$(\mathcal K^\bot)^\bot=\mathcal K$$

where "$\bot$" denotes the orthogonal complement of the set in question.

What are some examples of equality failing here when we drop the requirement on $\mathcal K$ to be a closed subspace of $\mathcal H$?

Answer 1

Let $\mathcal{K}$ be dense in $\mathcal{H}$. Then $\mathcal{K}^{\perp}=\left\{0\right\}$, so $$\mathcal{K}\neq \mathcal{H}=(\mathcal{K}^{\perp})^{\perp} $$

Answer 2

Any linear subspace which is not closed should work here. E.g., take any separable infinite-dimensional Hilbert space with an orthonormal (Schauder) basis, and let $\mathcal{K}$ be finite linear combinations of the basis elements. Since $\mathcal{K}$ is dense but not the entire space, $\mathcal{K}$ is not closed. We can check directly that this works: clearly $\mathcal{K}^\perp = 0$ (because any nonzero vector has at least one nonzero component, so the inner product with the corresponding basis element, which is in $\mathcal{K}$, is nonzero), so $(\mathcal{K}^\perp)^\perp = \mathcal{H}$