Let $M_1,M_2\subset \mathbb{R}^n $be k-dimensional manifolds (without boundary) and $M=M_1\cup M_2$.
Prove that if $\bar{M_1}\cap M_2=\bar{M_2}\cap M_1=\varnothing$ Then M is a k-dimensional manifold (without boundary)
My work: I proved that if $x\in M_1$, then $x\in\mathbb{R}^n$\ $\bar{M_2}$ then because $\mathbb{R}^n$\ $\bar{M_2}$ is an open set, there is a neighbourhood of x with $\delta>0 \ s.t \ B(x,\delta)\cap \bar{M_2}=\varnothing$. In the same way we prove the same for $x\in M_2$.
So I wanna say that from that we conclude that there are no problematic points for us to deal with but I'm not sure how to go about it.
Any help?
EDIT: I added the definition of a Manifold without boundaries if that helps.
The definition of a $k$-manifold we are given is a set $M\subset\mathbb{R}^n$ such that the following equivalent conditions hold for each $x\in M$:
- There exists a mapping $f:\mathbb{R}^n\to\mathbb{R}^{n-k}$, continuously differentiable near $x_0$, such that $(Df)_{x_0}=A:\mathbb{R}^n\to\mathbb{R}^{n-k}$ is onto and $$ x\in M \iff f(x)=f(x_0)\text{ for all $x$ near $x_0$}$$
- There exists a local diffeomorphism $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ near $x_0$, such that $$ x\in M \iff \varphi(x)\in \mathbb{R}^k\times\{0_{n-k}\}\text{ for all $x$ near $x_0$}$$
- There exists a permutation $(i_1,\dots,i_n)$ of ${1,\dots,n}$ and a mapping $g:\mathbb{R}^k\to\mathbb{R}^{n-k}$, continuously differentiable near $(x_{0,i_1},\dots,x_{0,i_n})$ such that $$ x\in M \iff g(x_{i_1},\dots, x_{i_k})=(x_{i_{k+1}},\dots,x_{i_n})\text{ for all $x$ near $x_0$}$$
You have the idea of the proof. To end it neatly, you can use any of your definitions. For instance, with the second definition, you can define $\phi_1$ as the restriction of $\phi$ on $B(x,\delta)$. It is easy to see that $\phi_1$ has exactly the desired property (as $x\in M \mbox{ iff } x\in M_1$ for $x \in B(x,\delta)$).