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Is the class of all 2-countable manifolds a set?

I think so: each such space is a countable union of sets of cardinality $|\mathbb{R}^n|\!=\!|\mathbb{R}|$, i.e. a manifold has cardinality continuum, and there are less than $\mathbb{R}^\mathbb{R}$ ways of gluing it together.

Why then is this set still called a class?

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Every set is a class. Classes which are not sets are called proper classes. The word "class" is also used e.g. for the equivalence classes of an equivalence relation on a set, i.e. the maximal subsets where all elements are equivalent according to that relation. Of course subsets of a set are also sets. – celtschk Aug 11 '12 at 20:28
Depending on the source in which you read this phrase, "class" may not be being used here in the technical sense. – Qiaochu Yuan Aug 11 '12 at 20:30
@LeonLampret Let $X$ be any countable manifold. For each arbitrary set $a$ you can give a structure of 2-countable manifold to the set $X\times\{a\}$. Obviously all are different (although diffeomorphic). – azarel Aug 11 '12 at 20:31
I know, but when one uses the term class, usually one has in mind 'large classes', i.e. proper. E.g. I've never heard anyone say 'class of natural numbers', or 'class ofreal functions', but 'class of groups/rings/modules' is used often. – Leon Aug 11 '12 at 20:31
I don't think you can argue with cardinality here (for the reason pointed out by azarel). However, Whitney's theorem tells you that every $n$-manifold embeds into $\mathbb{R}^{2n+1}$ (even $\mathbb{R}^{2n}$), so there aren't that many different diffeomorphism types. – t.b. Aug 11 '12 at 20:33
up vote 8 down vote accepted

No, the class of all manifolds (2nd countable) that are is not quite a set, as it is a proper class. To see this, let us consider the "isomorphism class" of the connected zero dimensional manifolds. This is just a point. But their are as many singletons as their are sets. To each set $S$, we may associate a singleton, $\{S\}$, which may be regarded as a manifold. The upshot is that all of the UP TO ISOMORPHISM, their is only one such point. A similar thing may be said of 2nd countable manifolds. Their is a class worth of these things, but if you select one representitive from each equivalence class, you do get a set. One might say that the category of 2nd countable manifolds is not small, but ESSENTIALLY SMALL.

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The class you consider is not a set basically for the same reason that the class of all sets consisting of just one element is not a set. If it were then the mapping sending any set $S$ to the singleton set consisting of $S$ as its single element would be an injection from the class of all sets to the set of all singleton sets, making the former a set (which it's not).

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Your question, as it seems to me, is mainly a nomenclature issue.

Class, in mathematics, means usually something described by a formula: all things with a certain property.

Equivalence classes, for example, are called classes but could just as well be sets. In fact sets are also classes in the context of set theory. You can see this in other places as well (e.g. conjugacy class).

In set theory the Russell paradox (and various other paradoxes as well) showed that not every collection is a set. This is why the notion of a class was invented. It turns out that this notion is (at least philosophically) close to other notions of class in mathematics.

Formally, however, note that there are "normal functions; normal spaces; normal distributions; etc." but many of these are somewhat orthogonal and have nothing to do with one another. Classes of set theory are constructs in set theory, while conjugacy classes are constructs in group theory (used elsewhere as well, of course).

So to sum up, the word "class" is commonly used to describe a definable collection (definable from what? well, that depends on the particular context); and as Qiaochu remarked, in a very technical sense every set is a class anyway (in the set theoretical context of the word).

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