Is there a well-defined notion of class-dimensional vector spaces? 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 $\bigoplus_{\mathbf{ON}}\mathbb{R}$.]
 A: The following is not so hard. Let $C$ be the class of all finite sets of pairs $(s, r)$ where $s$ is a set and $r$ is a real. Then $C$ is a sort of class-sized vector space over the entire universe of sets. To add two elements of $C$, or perform scalar multiplication, just treat elements of $C$ as formal $\mathbb{R}$-linear combinations of sets and add accordingly. The class $C$ itself is easily definable in ZFC, as are the classes that encode the addition and scalar multiplication operations.  Moreover, any basis of a vector space over $\mathbb{R}$ induces an embedding of that space into $C$.  I don't know why this space would be very interesting, however. 
A: It is well definable within ZFC the class of all finite functions from ordinals into $\mathbb R$. These would simply be finite sets of pairs $(\alpha,r)$ with addition of $(\alpha,r)+(\alpha,t)=(\alpha,r+t)$.
If we add two "vectors" then we take pointwise addition on common domain and union of the disjoint parts, so for example:
$$\{(\omega,5),(\omega_3+\omega+4,\pi)\}{\ \large+}\ \{(\omega,5)\}=\{(\omega,10),(\omega_3+\omega+4,\pi)\}$$
The definition of scalar multiplication is similar.
Indeed every real vector space is of the form $\bigoplus_\alpha\mathbb R$ for some ordinal $\alpha$ can be embedded naturally into this space. The problem with this sort of vector space is that one of the common construction - the dual space - becomes problematic. 
This is because $\mathbb R^{(\mathbf{Ord})}$ is spanned by $\{(\alpha,1)\}$ for $\alpha\in\mathbf{Ord}$. The dual space would be functions from all ordinals into $\mathbb R$. That would be a collection of classes, not of sets. We can define each "vector" but not the whole space.
Similarly for $\ell_2(\mathbf{Ord})$ as you defined it. You cannot have a collection of functions from all the ordinals into $\mathbb R$. Furthermore the sum of a proper class of indices is an abomination. If we have uncountably many elements we already require all but countably many to be zero - so this would just be a very very long string of mostly zeros. Even if we instead of finite functions take countable functions and require them to converge, we again run into the problem of dual space; endomorphisms of the space; etc.
I'd think that it is better to simply assume the existence of an inaccessible cardinal and consider the universe below it as your model - and functions from the cardinal itself as $\mathbb R^{(\mathbf{Ord})}$. From there it's a reasonable leap of faith to take Tarski-Grothendieck axiom of universes as well.
