Is a subspace of a separable Hilbert space always a Borel set?

Question

This question seems more subtle than I expected..

For a separable Hilbert space $H$, which is possibly infinite-dimensional, we can think of the Borel $\sigma-$algebra generated by the inner product topology on it.

Then, is "any" vector subspace of $H$ an element of this Borel $\sigma-$algbera?

I suspect strongly that the answer is yes, but cannot prove rigorously. Could anyone please help me?

Answer 1

An infinite dimensional separable Hilbert space (or more generally Banach space) has a Hamel basis of cardinality $2^{\aleph_0}$. For each subset of such a Hamel basis, we obtain a subspace of $H$ so we have $2^{2^{\aleph_0}}$ subspaces.

However, the cardinality of the Borel $\sigma$-algebra on a separable metric space is at most $2^{\aleph_0}$ (see here) so it cannot be that every subspace of $H$ is borel measurable when $H$ is infinite dimensional.

In finite dimensions however, each subspace is closed and in particular Borel.

Answer 2

In addition to the answer given - every subspace is sandwitched between an $F_\sigma$ and its closure. If $V$ is a subspace, it is separable (every subset of a separable metric space is separable) - so starting from a dense subset of $V$ - one may form an increasing countable union of finite dimensional sub-spaces of $V$ which is an $F_\sigma$ and is dense in $V$.