**Question**: Consider the proper class of all $n$-dimensional smooth manifolds. If we take the equivalence classes where two manifolds are identified if there exists a diffeomorphism between them, is this collection of equivalence classes **a set**? 


*Remark*: I do not assume my manifolds to be Hausdorff nor second countable. If the answer depends on those assertions (*Edit*: [it most definitely does](https://mathoverflow.net/questions/143947/is-the-class-of-n-dimensional-manifolds-essentially-small?noredirect=1#comment371342_143947)) I would like to hear about the difference. 


*Remark 2*: As [Omar pointed out](https://mathoverflow.net/questions/143947/is-the-class-of-n-dimensional-manifolds-essentially-small?noredirect=1#comment371340_143947), there may be a problem with the various long lines. To make the question slightly more tractable, I would be (mostly) satisfied if there is a statement even ignoring the smooth structure and consider the case of topological manifolds and homeomorphisms. 


**Motivation**: for something that I am working on I need to consider the collection of all $n$-dimensional smooth manifolds satisfying "property $X$". Unfortunately property $X$ is diffeomorphism invariant, so most definitely this collection is not a set, which invalidates many constructions (I want to build a manifold out of this collection; if the collection is a proper class then even on the set level the thing that I constructed will be a proper class, instead of a set) or at least forces me to rethink how this constructions ought to go. Fortunately for my argument it suffices that I have one object in each diffeomorphism class in my collection. 



Is the class of n-dimensional manifolds essentially small?
If we assume the Hausdorff separation axiom, connectedness, but not second countability then each such manifold has cardinality at most continuum. Therefore the collection of equivalence classes of such manifolds is a set and not a proper class. First of all, $U$ must be path connected since every connected locally path connected space is path connected. Fix $x\_{0}\in U$. Then every point of $U$ is reachable from $x\_{0}$ by a path. I now claim that there can be at most continuumly many choices of paths $f:I\rightarrow U$ with $f(0)=x\_{0}$. To each $x\in U$, we associate an open set $U\_{x}$ locally homeomorphic to $\mathbb{R}^{n}$. Let $T$ be the following tree of height $\omega\_{1}$. The nodes of $T$ will branch into continuumly many parts at each point of $T$, so the cardinality of $T$ is at most continuum. The objects of $T$ are tuples $((r\_{\alpha})\_{\alpha\leq\lambda},(y\_{\alpha})\_{\alpha\leq\lambda},(f\_{\alpha}))$ where $(r\_{\alpha})\_{\alpha\leq\lambda}$ is an increasing sequence in $[0,1]$, $(y\_{\alpha})\_{\alpha\leq\lambda}$ is a sequence of points in $U$ where $y\_{\alpha+1}\in U\_{y\_{\alpha}}$ and $f\_{\alpha}:[r\_{\alpha},r\_{\alpha+1}]\rightarrow U\_{y\_{\alpha}}$ is continuous. Now for each element $((r\_{\alpha})\_{\alpha\leq\lambda},(y\_{\alpha})\_{\alpha\leq\lambda},(f\_{\alpha}))$, we associate a path $\bigcup\_{\alpha}f\_{\alpha}$ if possible (there could be discontinuities at limit ordinals). Clearly every path is determined by an element of this tree, so there could be at most continuumly many paths. Therefore $U$ has cardinality at most continuum.


