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What is an example of a compact Hausdorff space that cannot be given the structure of a

(i) differential manifold

(ii) topological manifold?

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closed as off-topic by Michael Albanese, Jon Mark Perry, Claude Leibovici, Shaun, RedMushroom Feb 13 at 11:06

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Please specify if you include manifolds with boundary. If not, $[0,1]$ is the most basic example. – Martin Brandenburg Feb 12 at 14:26
    
Also, is connectedness required? If not, $\{0,1\}$ will do. I suppose you should state your definitions of these objects, since there are several in common use. – MPW Feb 12 at 14:32
    
@MPW Your example is still a compact Hausdorff 0-manifold, isn't it? – JoeyBF Feb 12 at 15:48
    
@JoeyBF: Yes, I should have said "if so" rather than "if not". My bad. The point is that it violates the connectivity requirement. – MPW Feb 12 at 16:21
    
For us to be certain. I am looking for the manifolds that are without boundary and that are connected. So $[0,1]$ is a very good example. – Abo Kutis-Felan Feb 12 at 16:28
up vote 10 down vote accepted

The union of the $x$-axis and $y$-axis in $\mathbb{R}^2$, intersected with the closed unit disk in $\mathbb{R}^2$.

enter image description here

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Is this compact? – Abo Kutis-Felan Feb 12 at 14:18
    
Hi @Abo, sorry yes, it wasn't compact as I initially stated, but that condition can be satisfied with a slight modification. – Amitesh Datta Feb 12 at 14:20
    
A fan or comb captures this idea – MPW Feb 12 at 14:22
    
Perhaps a more intuitive description of this space is to say the union of two line segments sharing a non-end-point. – leftaroundabout Feb 12 at 16:08
    
Hi @leftaroundabout, yes, or simply a "plus sign"/"cross". I've added an image to hopefully illustrate the point, although it's a bit thicker than I would like ... – Amitesh Datta Feb 12 at 17:14

The Cantor set is a nice example.

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9  
For some strange values of "nice"... – Zhen Lin Feb 12 at 14:18
3  
For example, "nice" $=$ "looks locally the same at every point", as a manifold does. @ZhenLin – Lee Mosher Feb 12 at 14:21
1  
I'd really add that comment to your answer -- in a way it's a more insidious counterexample than the cross. – djechlin Feb 13 at 4:33

The cartesian product $[0,1]^{\Bbb R}$ is obviously compact and Hausdroff, but isn't locally homeomorphic to any $\Bbb R^n$. The really interesting (and hard) case is a topological manifold without differentiable structure.

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Why $[0,1]^{\mathbb{R}}$? Actually $[0,1]^I$ for every non-empty set $I$ works. And if we are only interested in examples, $\#I=1$ suffices. – Martin Brandenburg Feb 13 at 8:41
    
@MartinBrandenburg, true. But the case of manifolds vith borders/corners was already mentioned by... you. – Martín-Blas Pérez Pinilla Feb 13 at 14:57

Any set that has an accumulation point and an isolated point

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You mean subspace of $\mathbb{R}$? There is no notion of an isolated/accumulation point for a set. – Martin Brandenburg Feb 13 at 8:42
    
What? Of course I intend for this set to be endowed with a topology. Specifically, "Any compact Hausdorff space that has an isolated point and an accumulation point". – MPW Feb 13 at 11:17
    
So why are you writing "set"? – Martin Brandenburg Feb 13 at 13:02
    
Enough of this. That the set carries a compact Hausdorff topology is implicit in the terminology I used and the context of the question. – MPW Feb 13 at 14:14

The interval $[0,1]$ is a compact Hausdorff space which doesn't carry the structure of a manifold without boundary. (Of course, it carries the structure of a manifold with boundary.)

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