homeomorphism of topological spaces is an equivalence relation ?

Would it be ok to say that homeomorphism of topological spaces is an equivalence relation ?

I know that there isn't a base "set of all topological space" but since I encountered this phrase in several places, I believe it either can be made rigorous or it is just an expression that isn'T to be taken literally. What do you think ?

• Nothing prevents us from considering the set of topological spaces and since "homeomorphism property" is reflexive, symmetric and transitive then yes it is an equivalence relation. – palio Oct 14 '11 at 18:11
• @palio the problem is that there is no "set" of topological spaces. If you take all topological spaces together in a collection there are too many to be a set. This is the same problem as there is no "set of all sets". – Bill Cook Oct 14 '11 at 18:13
• @BillCook, but there is a category of topological spaces where the spaces are objects. Granted it's concrete... – alancalvitti Jan 16 '13 at 3:43

First, note that isomorphism is a map between two objects which preserves the structure. The more structure, the "less" isomorphisms you might have. Isomorphism is an equivalence relation: the identity map gives us reflexivity; the fact the inverse of an isomorphism is also an isomorphism gives symmetry; and by composing isomorphisms we have transitivity.

This means that "locally" isomorphism is an equivalence relation on a set. What does that mean locally? If you take a set collection of structures, then isomorphism classes are equivalence classes which are sets.

Example: If we have two countable sets, without any structure then any bijection is an isomorphism of sets. However, if we give an ordering we may have less maps, lastly if we insist on well-ordering the sets then there is a unique isomorphism (if it exists to begin with).

When we say that $X$ is a topological space, or rather $(X,\tau)$ is a topological space we endow the set $X$ with some structure. In this case, a family of subsets of $X$ which has some properties.

So what is an isomorphism between two topological spaces? Firstly it has to be a bijection (as any isomorphism), but it has to preserve open sets. It is an open map, as well as a continuous map (as we want the inverse map to be open as well).

If so, an isomorphism between topological spaces is exactly a homeomorphism.

Of course, if the topology allows us - we can ask for more. If the topology is metric, then we can ask for an isometric map - which is not only preserving open sets but also the distance. We can ask for differentiable or measurable homeomorphism if the topological structure allows us.

For your second point, here are three possible solutions:

Firstly if we agree that sets of same cardinality may have the same topologies - we can choose a representative for a set of each cardinality, and consider only topologies defined on this set. This allows us simply to take the set of all topologies defined on $X$, which is a set since it is a subset of $P(P(X))$.

For example, if we wish to work with finite dimensional real vector spaces, for the sake of argument we can assume that the underlying set is always the same. It is a set theoretical cheat, since it prevents us from having $\mathbb R\subseteq\mathbb R^2$. However the latter is already an abuse of notation since $\mathbb R^2$ is a set of pairs, while $\mathbb R$ is not.

Secondly, we can use Scott's trick (named after Dana Scott) which is to use the axiom of foundations (known as axiom of regularity in some places) and define the equivalence classes as sets in the following way:

$$[(X,\tau)] = \{(Y,\rho)\mid (Y,\rho)\cong(X,\tau)\land\operatorname{rank}(Y)\text{ is minimal}\}$$

That is, we use the fact that every set can be given a rank, and the collection of sets in a given rank is indeed a set. Now we can take all the topological spaces homeomorphic to $(X,\tau)$ whose rank is the least possible.

Lastly, we can stay with classes (or move to a set theory which allows classes, such as Von Neumann–Bernays–Gödel set theory). Classes are syntactical objects. They are defined by a function, perhaps parameterized. In this case, the equivalence class of topological spaces homeomorphic to $(X,\tau)$ can be defined using $(X,\tau)$ as a parameter.

• I actually had in mind a fourth idea, which I only remembered when Skolem posted his answer, so no point in adding that here, I think. :-) – Asaf Karagila Oct 15 '11 at 10:21

It is an equivalence on the class (or category) of topological spaces.

You are right that the collection of all topological spaces is "too big" to be a set. But you can speak of equivalence relations on classes or categories.

http://en.wikipedia.org/wiki/Class_%28set_theory%29

http://en.wikipedia.org/wiki/Category_theory

• Or you could restrict yourself to a "small category" of topological spaces, such as the category of all topological spaces on subsets of a particular set, essentially setting a maximum size for your topologies. – Thomas Andrews Oct 14 '11 at 18:10

One clever, if limited, solution is to simply say that once you are given a set $\mathcal{X}$ of topological spaces, homeomorphism between spaces in that set is an equivalence relation.

This works in some simple contexts where it may be akin to 'choosing' a particular "universal set" $X$ when doing some basic set theory with unions, intersections, and complements/differences. For example, one can think of equivalence relations on elements of $X = \mathbb{Z}$ without worrying about constructions like $\mathbb{Q}$, $\mathbb{Z}_n$, or $\mathbb{Z} \times \mathbb{Z}$. In particular, I think a student could find their way through a first course in point set topology using this approach.

However, in other contexts, this construction is obviously deficient: a fixed set $\mathcal{X}$ may not or cannot have certain desired closure properties. For example, if one is working with category theory, such a set cannot be closed under infinite products, infinite sums, etc...

Consider also closure under homeomorphisms (which might be the whole point, in this situation). This brings brings classes into the picture immediately: consider the (class) inclusion $x \mapsto \{x\}$. This shows there are as many one-element topologies as there are elements, so even a (homeomorphism-closed) collection of one-element spaces can't be a set. This is why it's so useful to choose representatives somehow, as mentioned in the answer above.