TITLE: homeomorphism of topological spaces is an equivalence relation ?
QUESTION [13 upvotes]: 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 ?

REPLY [4 votes]: 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.