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Let $M\subseteq\mathbb R^d$ for some $d\in\mathbb N$, $1\le k_1<k_2\le d$ and $\alpha\in\mathbb N_0\cup\{\infty\}$.

If for all $x\in M$, there a $C^\alpha$-diffeomorphism $\psi$ from an ($\mathbb R^d$)-open neighborhood $U$ of $x$ onto an open subset of $\mathbb R^d$ with$^1$ $$\psi(U\cap M)=\psi(U)\cap\left(\mathbb R^{k_1}\times\{0\}\right),\tag1$$ then we say that $M$ is a $k_1$-dimensional $C^\alpha$-submanifold.

If $M$ is an arbitrary subset of $\mathbb R^d$ and $x_i\in M$, it should clearly be possible$^2$ that there is a $C^\alpha$-diffeomorphism $\psi_i$ from an ($\mathbb R^d$)-open neighborhood $U_i$ of $x_i$ onto an open subset of $\mathbb R^d$ with $$\psi_i(U_i\cap M)=\psi_i(U_i)\cap\left(\mathbb R^{k_{\color{red}i}}\times\{0\}\right).\tag2$$ So, in order for the dimensionality notion for a submanifold to make sense, I guess we can somehow show that $\psi_1$ and $\psi_2$ as above cannot simultaneously exist, if $M$ is a $C^\alpha$-submanifold (i.e. if it is a $k_1$-dimensional $C^\alpha$-submanifold, only $\psi_1$ can exist). How can we prove that

$^1$ $0\in\mathbb R^{d-k_1}$ and hence $\mathbb R^{k_1}\times\{0\}\subseteq\mathbb R^d$.

$^2$ In fact, assuming $x_1\ne x_2$ and $U_1\cap U_2=\emptyset$, we just need to define $M:=U_1\cup U_2$.

0xbadf00d
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  • You can't prove it unless $M$ is assumed to be connected. – Berci Jul 06 '20 at 17:51
  • @Berci From my intuition, that's plausible. So, in general, $M$ can be a $k_1$-dimensional submanifold and a $k_2$-dimensional submanifold simultaneously? – 0xbadf00d Jul 06 '20 at 18:18
  • Well, it can be a disjoint union of those, e.g. a line plus a plane, or a circle plus a sphere in the space. – Berci Jul 06 '20 at 18:23
  • @Berci I'm not sure if you got me right. Let me rephrase the question: Is it possible that $M$ is a $k_1$-dimensional submanifold and a $k_2$-dimensional submanifold? Or, equivalently, if $M$ is a $k_1$-dimensional submanifold, is it possible that the diffeomorphism $\psi_2$ in the question exists? – 0xbadf00d Jul 06 '20 at 18:28
  • No, manifolds of different dimension are not even homeomorphic. For example, see this discussion: https://math.stackexchange.com/questions/396660/topological-manifolds-dimension – mxnoqwerty Jul 07 '20 at 07:45

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