Typical Terminology: A basis $\mathcal{B}$ for a topology on a set $X$ is a set of subsets of $X$ such that (i) for all $x\in X$ there is some $U\in\mathcal{B}$ such that $x\in U$, and (ii) if $x\in U\cap V$ for some $U,V\in\mathcal{B}$, then there is some $W\in\mathcal{B}$ such that $x\in W\subseteq U\cap V$. The topology generated by such a basis is the set of unions of subsets of $\mathcal{B}$.
A topology $\mathcal{T}$ on $X$ is a collection of subsets of $X$ with $X$ and $\emptyset$ as elements, that is closed under unions and finite intersections. The $\mathcal{T}$-open subsets of $X$ are the elements of $\mathcal{T}$.
A topology $\mathcal{T}$ is generated by a basis $\mathcal{B}$ iff each $\mathcal{B}$-set is $\mathcal{T}$-open and for each $\mathcal{T}$-open $U$ and each $x\in U$ there is some $V\in\mathcal{B}$ such that $x\in V\subseteq U$.
The First Obstacle: The definitions above are problematic when one wishes to imbue a proper class $\mathbf{M}$ (such as the ordinals) with a topology in the setting of ZF(C). We can't really say that such a topology "exists" (or even easily describe it) in that context, as it would be a "class of classes" of which some are proper.
One remedy I've encountered for this is to describe a "basis class"--that is, a class $\mathbf{B}$ of subsets of $\mathbf{M}$ having the properties of a basis as described above. At that point, we would then say a subclass $\mathbf{U}$ of $\mathbf{M}$ is open iff for each $x\in\mathbf{U}$ there is some $V\in\mathbf{B}$ such that $x\in V\subseteq\mathbf{U}$.
The Second Obstacle: The aforementioned remedy doesn't allow one to imbue a proper class with the indiscrete topology, for example (not that one would want to, per se, but the ability to do so would be nice). Of course, one could still "topologize" a proper class indiscretely by convention with no real difficulty, but not by means of the aforementioned remedy. This suggests that there may be "topologies" on proper classes that couldn't be defined either by convention or by the aforementioned remedy.
My Proposed Remedy: First, we require that a "basis class" $\mathbf{B}$ satisfy only the property (ii) in the above definition of basis. At that point, we would then say a subclass $\mathbf{U}$ of $\mathbf{M}$ is open iff one of the following occurs: (a) $\mathbf{U}=\emptyset$, (b) $\mathbf{U}=\mathbf{M}$, or (c) for each $x\in\mathbf{U}$ there is some $V\in\mathbf{B}$ such that $x\in V\subseteq\mathbf{U}$.
The Question: It seems to me that such a convention uniquely determines a "topology" on $\mathbf{M}$ in the same fashion as the remedy I previously encountered, while simultaneously giving greater freedom in defining said "topology"--in fact, this allows our basis class to be a set (and not a proper class) in some cases, where otherwise this wouldn't be possible. I don't think I've made any errors or ZF(C)-illegal moves here, but I'd appreciate feedback, especially if my suggested approach is problematic in a way that the first-mentioned remedy isn't. Thoughts? Has such an approach been suggested before?
