Why is the collection of isomorphism classes of principal G bundles a set?

Question

Given a universal $G$-bundle $EG\longrightarrow{BG}$ it is well known that the set of isomorphism classes of principal $G$-bundles over some space $B$ denote it by $\mathcal{B}(G,B)$ is in bijection with the set of homotopy classes of maps $[B,BG]$, my first question would be: why is $\mathcal{B}(G,B)$ a set?
Is there another way to prove that is indeed a set, or just rather to accept that because of $\mathcal{B}(G,B)\cong[B,BG]$ then it must be a set.
In addition to this, $\mathcal{B}(G,BG)$ would only consists of one isomorphism class right? This would follow from the fact that if $E\longrightarrow{BG}$ is some principal $G$-bundle over $BG$, because of the universality of $EG$ there would only exists a morphism $E\longrightarrow{EG}$ and because they are principal $G$-bundles over the same base space they must be isomorphic.
Does this make sense or did I make a mistake? Thank you for any help and comments.

Answer 1

All $G-$bundles $E$ over some space $X$ have the same cardinality, namely the cardinality of $X \times G$, thus a $G-$bundle $E \rightarrow X$ is isomorphic to $X \times G \rightarrow X$ but where $X \times G$ may have some other topology than the product one, thus the collection of $G-$bundles over $X$ is smaller than or equal in size to the set of topologies on $X \times G$ which in turn has a smaller size than $\mathcal P (\mathcal P (X \times G))$ where $\mathcal P(S)$ denotes the powerset of $S$.

Answer 2

Noel Lundström's answer addresses the first question. As for the second question, your claim is false. We have $\mathcal{B}(BG, G) = [BG, BG]$. If this were a singleton, then $BG$ (and hence $G$) would be contractible: any two maps $BG \to BG$ would be homotopic, in particular $\operatorname{id}_{BG}$ and a constant map.

Your mistake occurs here: "because they are principal $G$-bundles over the same base space they must be isomorphic". That is not true. For example, over $BG$ we have the universal bundle $EG \to BG$, but also the trivial bundle $BG\times G \to BG$. As $EG \to BG$ is the universal bundle, there is map $f : BG \to BG$ such that $f^*(EG) \cong BG\times G$, e.g. a constant map. This doesn't imply that the bundles are isomorphic though as an isomorphism of principal $G$-bundles must cover the identity.