Fermionic and Bosonic physical Hilbert spaces - are they actually Hilbert spaces?

Question

Consider two identical particles $A$ and $B$. The combined Hilbert space $\mathcal{H}_A\otimes \mathcal{H}_B \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left<#1\right|} $ is a valid Hilbert space. But what about the physical Hilbert spaces: $$\mathcal{H}_A \odot \mathcal{H}_B=\textrm{Span}\{ \ket{i}\otimes \ket{j}+\ket{j}\otimes\ket{i}\}\tag{Bosons}$$ $$\mathcal{H}_A \wedge \mathcal{H}_B=\textrm{Span}\{ \ket{i}\otimes \ket{j}-\ket{j}\otimes\ket{i}\}\tag{Fermions}$$ How do we go about showing that they are valid Hilbert spaces (if indeed they are)?

It is trivial to show they are inner product spaces but I wouldn't know where to start in show completeness.

Answer 1

The (anti-)symmetrized spaces are the quotient of the full tensor product space $H\otimes H$ (since the particles are identical I'm dropping the subscript on the spaces) by a closed subspace, and the quotient of a Hilbert space by a closed subspace is again a Hilbert space, cf. e.g. this math.SE question.

What's left is to justify the claim that we are indeed quotienting by a closed (or "complete") subspace. In the case of the anti-symmetrized space, we're quotienting by the ideal $I$ generated by elements of the form $v\otimes v$ for all $v\in H$. Elements of this ideal have the form $\sum_i (v_i\otimes v_i)$, and if a limit of this exists at all, it will be of the form $v\otimes v$ for $v = \lim_{i\to\infty} v_i$, which is clearly again an element of the ideal, so the ideal is closed. A very similar reasoning works for the ideal we have to divide out in the symmetric case.