Let $G$ be a locally compact group. I know that the collection of all unitary representations of $G$ is not a set, since there are unitary representations on inner product spaces with bases of any cardinality, and the collection of all cardinals is not a set. However, apparently if we only consider unitary representations on Hilbert spaces, the collection $\tilde{G}$ of these representations becomes a set, and this allows to put a topology on it for example (the Fell topology). This is mentioned in the book by Bekka-de la Harpe-Valette on property (T). Why is this the case? I was thinking that maybe the collection of all Hilbert spaces is itself a set, but don't know how to prove that either.
Edit: Actually Bekka, de la Harpe and Valette restrict to Hilbert spaces of dimension bounded by some cardinal number. The same argument shows that the bounded dimension is necessary. But why is it a set then? And why the further restriction of completeness: would it not work for inner product spaces of bounded dimension?
So let me gather all the comments into one answer.
(1) Since every inner product space $V$ can be completed to a Hilbert space $\overline{V}$ and $V$ can be treated as a subspace of $\overline{V}$ then $\dim(\overline{V})\geq\dim(V)$. In particular Hilbert spaces don't have bounded dimension and thus they don't form a set.
But the book additionally assumes that dimensions of those Hilbert spaces are bounded by some fixed cardinal.
(2) Having a bounded dimension is enough for Hilbert spaces to form a set. Indeed, for a cardinal $M$ the collection of all vector spaces with $\dim(V)<M$ is a set. That's because there is only one vector space (up to isomorphism) for each dimension (as a cardinal number). And since inner product is just a function $V\times V\to k$ then we conclude that Hilbert spaces are a subcollection of $(V, V\times V\to k)$ pairs which is a set if the collection of $V$'s is.
(3) We can replace "bounded dimension" condition by "irreducible representations". Indeed, if $V$ is an irreducible representation then $V$ is simple as a $k[G]$-module (here $k[G]$ denotes the group algebra). In particular there is a module epimorphism $k[G]\to V$. Indeed, you map $1$ to some non-zero vector. It extends to module homomorphism because $k[G]$ is a free $k[G]$-module. It is surjective because $V$ is simple. In particular $V$ is a quotient of $k[G]$. Thus the dimension of $V$ is at most $|G|$. Point (2) applies now or you can simply conclude that they form a set because all quotients of a given module is a set.
EDIT
(4) So there is another defintion: $V$ is an irreducible representation if it doesn't contain a proper, nontrivial closed subrepresentation. In that situation, just like in (3) we have a module homomorphism $k[G]\to V$ but now it doesn't have to be onto. However the image, name it $W$, has to be dense in $V$.
Now if $V$ is a normed space and $W\subseteq V$ is a dense subspace then every element of $V$ can be written as $\lim x_n$ for some sequence $(x_n)\subseteq W$. If we denote $l(W, V)$ the set of all sequences in $W$ convergent in $V$ then we obtain a vector space epimorphism $\lim:l(W,V)\to V$. Note that $l(W,V)$ is a subspace of $S(W)$ = all sequences in $W$. It follows that
$$\dim(V)\leq\dim S(W)$$
In our case it means that the dimension of an irreducible space is at most $\dim S(k[G])$. So again it is bounded by a fixed cardinal.