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Encountered the term parametrizable for the first time:

The support of $\omega$ is contained inside a single parametrizable open subset $W$ of $X$.

So I am just curious, what kind of sets are not parametrizable? The intersection of $\mathbb{R}^2$ and the Weierstrass function?

1LiterTears
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    I would interpret "parametrizable open subset" of a manifold $X$ to mean "coordinate patch of $X$", i.e. an open set of $X$ that (with its smooth structure from $X$) is diffeomorphic to an open subset of $\mathbb{R}^n$. – Zev Chonoles Aug 07 '13 at 02:36
  • Thanks @ZevChonoles, that is how I understand it. But I am really curious about is that what kind of open subset is not diffeomorphic to an open subset of $\mathbb{R}^n$? – 1LiterTears Aug 07 '13 at 03:13
  • I mean locally not diffeomorphic to an open subset of $\mathbb{R}^n$, that's why Weierstrass function came to my mind. @ZevChonoles – 1LiterTears Aug 07 '13 at 04:31

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If the context of this is manifold theory, the open subset $W$ should be a coordinate neighborhood: it admits a diffeomorphism $\phi_W:W\to \mathbb{R}^n$. That is, $W$ can be given a parametrization by $\mathbb{R}^n$.

So sets that aren't parametrizable are sets that aren't diffeomorphic to $\mathbb{R}^n$. For example, a manifold with nontrivial fundamental group is not parametrizable (although it is locally parametrizable). Another example, any neighborhood of the cusp at $0$ in the double-cone given by $x^2 + y^2 = z^2$ is not parametrizable.

Neal
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