Is there a simple way to know whether the collection of all objects of a certain type is a set? Please forgive me if this question seems really unsubstantial and only serves to expose my ignorance of set theory. I have seen some examples where considering the set of all objects of certain kind leads to a contradiction when using naive set theory (When the collection of all objects of a certain type is not a set).
For example the collection of sets, ordinals, sets that don't contain themselves etc.
For which objects is the collection of all of those objects a set?
I realize this is ambiguous.
So concretely can we answer the following?


*

*Is the collection of all topologies a set?

*Is the collection of all groups a set?


Thank you kindly.
Regards.
 A: There is no "immediate way", and it will depend on some additional information.
For example, the collection of all finite ordinals is a set, because we can prove there is only one finite ordinal of each cardinality, and that this collection is bounded by the assumption that infinite sets exist (more specifically, that an infinite ordinal exists).
But if your "type" is cardinality invariant, e.g. every set of two elements has a group structure, then the only way that "all objects of this type" is a set is if only the empty set is that type of object. Otherwise you will have to include every singleton, or every countable set, or so on, and those are already collections which are proper classes.
This is the same as with "topologies" a set $\tau$ is a topology if $X=\bigcup\tau$ is such that $(X,\tau)$ is a structure satisfying certain axioms. Since we can translate a topology on $X$ to a topology on any set equipotent with $X$, even the collection of trivial topologies is already not a set.
There are ways, however, to overcome this issue. Specifically, if we are only interested in finite groups, then we can ask the group structure to be endowed to a particular finite set. Namely, $\{0,\ldots,n\}$ for some $n$, and then we can look only at groups which have those underlying sets. This would be a set, since each concrete set $G$ can only have set-many group structures (simply because a group structure is a binary operator, and those are elements of $\mathcal P(G\times G\times G)$).
Or if you are only interested in separable metric spaces, then you can show that those are either finite, countable, or have size $2^{\aleph_0}$, so you can only look at those separable metric spaces whose underlying set is a subset of $\mathcal P(\Bbb N)$, and you will know that it is enough for that.
