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First note, that this question is related to this question, but this time a different definition and a proof is required.

I define a submanifold of $\mathbb{R}^n$ as follows:

A subset $M$ of $\mathbb{R}^n$ is said to be an m-dimensional $C^q$ submanifold of $\mathbb{R}^n$ if, for every $x_0\in M$, there is in $\mathbb{R}^n$ an open neighborhood $U$ of $x_0$, an open set $V$ in $\mathbb{R}^n$, and a $\varphi\in\mathrm{Diff}^q(U,V)$ such that $\varphi(U\cap M)=V\cap (\mathbb{R}^m\times {0})$.

Is if the union of two disjoint submanifolds (of the same dimension) $M_1, M_2 \subset \mathbb{R}^n$, satisfying $M_1\cap M_2=\emptyset$, again a submanifold?

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  • $\begingroup$ Can you specify exactly what you mean by disjoint union? If you mean some kind of abstract disjoint union (as described in Najib's answer) then the answer is yes; but If you mean the union of two submanifolds $M,N\subset \mathbb R^n$ with $M \cap N = \emptyset$, then I think the answer is no. $\endgroup$ Feb 19, 2018 at 11:31

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To be perfectly clear, this answer is about the "union of two disjoint manifolds", where we are considering $M_1,M_2 \subset \mathbb R^n$ as embedded submanifolds that do not intersect and taking the set union $M_1 \cup M_2.$ I claim this union is not necessarily an embedded submanifold (i.e. a manifold in the sense you have defined.)

Let $M_1 = (0,2)\times\{0\}$ and $M_2=\{1\}\times(-1,0),$ i.e. two line segments arranged to form the shape of the letter T. Since the intervals are open, these sets are disjoint, and they are both clearly 1-dimensional submanifolds of $\mathbb R^2.$ (For example, for $M_1$ you can simply take $\varphi = \mathrm{id}, V=(0,2)\times\mathbb R.$)

The union $M_1 \cup M_2$ is not a manifold because of the point $x_0=(1,0) \in M_1$ where the two line segments (almost) meet. If $\varphi \in \mathrm{Diff}^q(U,V)$ for some neighbourhood $U$ of $x_0$, then $M_1\cap U, \bar M_2 \cap U$ must be mapped to differentiable curves intersecting at $\varphi(x_0).$ The fact that they meet at a non-zero angle in $\mathbb R^2$ along with the invertibility of the Jacobian $D\varphi|_{x_0}$ tells us that these curves must meet at a non-zero angle at $\varphi(x_0) \in V$, and thus cannot lie in the one-dimensional subspace $\mathbb R \times \{0\}.$ Since the curves belong to $\varphi(U \cap (M_1 \cup M_2))$, this rules out $M_1 \cup M_2$ being a manifold of dimension 0 or 1.

Since every neighbourhood of $x_0$ intersects the complement $\mathbb R^2 \setminus (M_1 \cup M_2)$, $M_1 \cup M_2$ cannot be a manifold of dimension 2 either.

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  • $\begingroup$ Hey Anthony, fancy seeing you here! Remember me? I was your room mate at AMSI. How's the PhD coming? $\endgroup$ Feb 19, 2018 at 12:20
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    $\begingroup$ @TheoBendit: Hi, of course I remember! My thesis was approved last week. I think we're connected on facebook if you want to chat, probably a more appropriate venue than the comments here. :P $\endgroup$ Feb 19, 2018 at 23:06
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Apparently OP meant "union of two disjoint submanifolds" and not "disjoint union of two submanifolds". I'll leave this answer here for posterity.

Yes, it's true. A quick way to see this is as follows. Suppose $M$ and $N$ are both $m$-dimensional submanifolds of $\mathbb{R}^n$ with the same $n$ (if not, say $M \subset \mathbb{R}^n$ and $N \subset \mathbb{R}^{n'}$, then you can just embed both in $\mathbb{R}^{\max(n,n')}$). Then $M \sqcup N$ is the same as the $m$-dimension submanifold of $\mathbb{R}^{n+1}$ given by $M \times \{0\} \cup N \times \{1\}$. Given a chart $U$ around, say, $x \in M$, then you find a chart $U \times (-1/2,1/2)$ around $(x,0) \in M \times \{0\}$, similarly for $N$.

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  • $\begingroup$ I specified the term disjoint union in my edit. $\endgroup$ Feb 19, 2018 at 11:36

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