# Relative merits, in ZF(C), of definitions of “topological basis”.

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?

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Weird! I was just looking at this thread... then you edited it. –  Asaf Karagila Nov 3 '12 at 22:49
Ha! Gotta love the timing. –  Cameron Buie Nov 3 '12 at 22:51

Your first problem is actually more acute: in ZF(C) classes are merely definable collections. In simple topological spaces a basis may be described nicely, but what happens at those crazy not-even-$T_0$ with huge cardinality? In those cases we often say "Well, there is a set with such and such properties..." but that too is not going to work with classes, again because we cannot quantify a subclass.
My suggestion is to start with an inaccessible cardinal $\kappa$ and consider the classes you want to talk about as subsets of $V_\kappa$ (preferably definable subsets so you could pull the same stunts on proper classes later).
When you work in $V_\kappa$ all the quantifiers are over sets, so everything is legal again. Of course the downside is that you need to assume the existence of a large cardinal - but that's not as bad as it sounds anyway.
The nice thing about your approach is that it is simple to consider the ordinals as a subset of $V_\kappa$--they're just $\kappa$! –  Cameron Buie May 19 '12 at 16:09
@Cameron: Judging by the post on the meta, I already figured that you are trying to apply this to the ordinals (as well you tried to apply this to the ordinal in that other post). Try with an inaccessible cardinal, you don't really have to use anything but the fact that $\kappa=\mathbf{Ord}^{V_\kappa}$, and this too is just for getting the hang of the idea, after you have done it successfully for the inaccessible case you can try and apply it to the "entire class of ordinals" case. –  Asaf Karagila May 19 '12 at 18:37