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I don't think I understand what exactly a manifold is until I watched Milnor's video:

"Def $X$ is a smooth manifold of dim $m$ if each $x\in X$ has a neighborhood diffeomorphic to a open subset of $\mathbb R^m$."

So changing "diffeo" to "homeo" we get the definition of manifold?

By this definition, a manifold of dim $m$ must be also a manifold of dim $m+1$?

We can also change the $\mathbb R^m$ to any space like Banach to make a Banach manifold? But $\mathbb R^m$ is field and Banach is a space, how could you freely interchange the field and space?


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  • $\begingroup$ Why would a manifold of dimension $m$ also be a manifold of dimension $m+1$? $\endgroup$
    – littleO
    Jun 30 '17 at 5:11
  • $\begingroup$ @littleO because if the neighborhood is diffeomorphic to a subset $S$ of $\mathbb R^m$, then $S$ is also a subset of $\mathbb R^{m+1}$ $\endgroup$
    – High GPA
    Jun 30 '17 at 5:13
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    $\begingroup$ But if $S \subset \mathbb R^m \subset \mathbb R^{m+1}$ then $S$ is not an open subset of $\mathbb R^{m+1}$. $\endgroup$ Jun 30 '17 at 5:18
  • $\begingroup$ @AnthonyCarapetis Thank you! I finally understand. Shall I delete this beginner's question? $\endgroup$
    – High GPA
    Jun 30 '17 at 5:19
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A topological $m$-manifold is a Hausdorff space locally homeomorphic to $\Bbb R^m$.

$\Bbb R^m$ is a vector space, not a field.

If $m\ne n$ no nonempty $m$-manifold is an $n$-manifold. This is because no open subset of $\Bbb R^m$ is homeomorphic to an open subset of $\Bbb R^n$ (Brouwer's "invariance of domain" theorem).

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    $\begingroup$ Just to emphasize: This is a surprisingly nontrivial result, whereas in the case of smooth manifolds, linear algebra will tell you that an open set in $\Bbb R^n$ cannot be diffeomorphic to an open set in $\Bbb R^m$ when $m\ne n$. $\endgroup$ Jul 2 '17 at 0:33

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