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I'm trying to see whether $\mathbb{Z}$ or the cantor set $C$ are submanifolds or $\mathbb{R}$.

Actually, I thought that $\mathbb{Z}$ was not a submanifold. As every subset of $\mathbb{Z}$ is countable while every open subset of $\mathbb{R}$ is uncountable, so there should be no diffeomorphism between them?

However, a quick research has yielded that this can't be true since $\mathbb{Z}$ is in fact a submanifold. But I didn't find proof for that.

Can anyone help me with that? Besides, is the cantor set a submanifold, too? By what argument?

[edit] Definition: $M$ is a $k$-dimensional submanifold of $\mathbb{R}^n$ iff for every $a \in M$ there are open subsets $U, V \subset \mathbb{R}^n$, such that $a \in U$, and a diffeomorphism $\phi: U \rightarrow V$ such that $\phi(M \cap U)=(\mathbb{R}^k \times \{0\}) \cap V.$

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    $\begingroup$ It may be important to specify whether you mean an immersed or embedded submanifold; see en.wikipedia.org/wiki/Submanifold Either way, submanifolds do not need to be diffeomorphic to the manifold containing them, and generally are not. $\endgroup$
    – Ben Grossmann
    Jun 21, 2013 at 3:22
  • $\begingroup$ Hum, the definition I know goes like this: $M$ is a submanifold of $\mathbb{R}$ iff for every $a \in M$ there are open subsets $U, V \subset \mathbb{R}$, such that $a \in U$, and a diffeomorphism $\phi: U \rightarrow V$ such that $\phi(M \cap U)=(\mathbb{R} \times \{0\}) \cap V.$ Sorry, should have included it right from the start. $\endgroup$
    – Amarus
    Jun 21, 2013 at 3:41
  • $\begingroup$ Ah, so $\mathbb Z$ must be diffeomorphic to its image in $\mathbb R$, but not $\mathbb R$ itself. $\endgroup$
    – Ben Grossmann
    Jun 21, 2013 at 3:43
  • $\begingroup$ Yes, but if I understand this definition correctly, for $\mathbb{Z}$ to be a $1$-dimensional submanifold of $\mathbb{R}$, it should be open, but we know it is not... (I included the general definition for higher dimensions in the opening post.) $\endgroup$
    – Amarus
    Jun 21, 2013 at 3:53

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$\mathbb Z$ is a $0$-dimensional submanifold because every point has a neighborhood (obtained by intersecting with an open set in $\mathbb R$) that is diffeo to (an open set in) $\mathbb R^0$.

Can the same be said of the Cantor set?

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  • $\begingroup$ Sorry, I don't understand what you mean. $\mathbb{R}^0=\{0\}$? About the cantor set: I have no idea. $\endgroup$
    – Amarus
    Jun 21, 2013 at 5:32
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    $\begingroup$ Yes, $\mathbb R^0=\{0\}$. So a $0$-dimensional submanifold consists of isolated points; i.e., each point $p$ has a neighborhood in the big manifold consisting of $p$ alone. $\endgroup$
    – Ted Shifrin
    Jun 21, 2013 at 12:22
  • $\begingroup$ Ah, so $\mathbb{Z}$ is a $0$-dimensional submanifold, but the cantor set is not because it has no isolated points. $\endgroup$
    – Amarus
    Jun 25, 2013 at 17:09
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    $\begingroup$ True. One non-isolated point is enough to mess it up. E.g., $\{0,1,1/2,1/3,1/4,\dots\}\subset\mathbb R$ is not a $0$-dimensional manifold. $\endgroup$
    – Ted Shifrin
    Jun 25, 2013 at 17:40

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