# Hilbert space with all subspaces closed

Does there exist an infinite-dimensional Hilbert space with all subspaces closed?

## 2 Answers

If $H$ is an infinite dimensional Hilbert space and $\{x_n:n\geq 1\}$ is a countably infinite set of linearly independent elements, then the abstract span of these elements, which is an infinite dimensional subspace, cannot be closed. If it were closed, it would be a Hilbert space itself, and infinite dimensional Hilbert spaces must have uncountably infinite dimension.

If $H$ is an infinite-dimensional Hilbert space, then $H$ contains $\ell^2(\mathbb{N})$ isometrically. And the property you mention fails for the latter. For instance $$F:=\mbox{span} \{e_k=(0,\ldots,0,1,0,\ldots)\;;\; k\in\mathbb{N}\}\subsetneq \overline{F}=\ell^2(\mathbb{N}).$$

Note: this is the lowtech argument for Hilbert spaces. But Keenan Kidwell's approach, which uses Baire, works for a general infinite dimensional Banach space. And since the converse is clearly true, we get: a Banach space has finite dimension if and only if every subspace is closed.