I was thinking a bit about isometric embeddings into Hilbert spaces and got the following idea.

First, as we recall, many vector spaces over the reals are isomorphic to $\mathbb{R}^{\alpha}$ for some cardinal number $\alpha$. [EDIT: As Carl Mummert remarked below, not every vector space is of this form, as I had erroneously supposed; see comments.] So if we want to embed every single set-dimensional vector space in some kind of vector space, it would be nice to have something like $\mathbb{R}^\mathbf{ON} = \lbrace f:\mathbf{ON}\to\mathbb{R}|\textrm{ }\mathtt{True}\rbrace$ with operations defined as usual. Here $\mathbf{ON}$ is the class of all ordinal numbers. So my question is:


  Does the notion of class-dimensional vector space make sense in some appropriate variant of set theory?


Next, we could possibly modify this definition a bit, definining a class-dimensional Hilbert space: $\ell^2(\mathbf{ON})= \lbrace f:\mathbf{ON}\to\mathbb{R}|\textrm{ }\sum_{\alpha\in\mathbf{ON}}f(\alpha)^2<\infty\rbrace$, where the sum is as usual taken to be the supremum of the finite sums and the other operations are defined as usual. So if such a definition made sense, we would perhaps be able to isometrically embed every set-dimensional Hilbert space in such a space. Therefore I ask also for this case:


  Is it possible to consistently define such spaces? If such a definition is consistent, is there some literature in which a theory of such spaces is developed?


Thank you in advance.

[ADDED: As Asaf Karagila remarks below, every $\mathbb{R}$-linear space is of the form $\bigoplus_{\kappa}\mathbb{R}$ for some cardinal $\kappa$. So by analogy, I might also be interested in $\bigoplus_{\mathbf{ON}}\mathbb{R}$.]

In the usual topology of $\mathbb{R}$, $\mathbb{Q}$ is neither open nor closed.

The interior of $\mathbb{Q}$ is empty (any nonempty interval contains irrationals, so no nonempty open set can be contained in $\mathbb{Q}$). Since $\mathbb{Q}$ does not equal its interior, $\mathbb{Q}$ is not open.

The closure of $\mathbb{Q}$ is all of $\mathbb{R}$: every real number is the limit of a sequence of rationals, so every real number lies in the closure of $\mathbb{Q}$. Since $\mathbb{Q}$ does not equal its closure, it is not closed.

Naturally, since $\mathbb{Q}$ is not open, its complement is not closed; since $\mathbb{Q}$ is not closed, its complement is not open.

But this is in the usual topology. $\mathbb{R}$ can be endowed with lots of topologies, and it is certainly possible for $\mathbb{Q}$ to be open (or closed) in some of them. For example, in the discrete topology, where every subset of $\mathbb{R}$ is both open and closed, $\mathbb{Q}$ is both open and closed.
