While $M_1\cap M_2=\emptyset$, $M_i$ a manifold, show $M=M_1\cup M_2$ is not necessarily a manifold.
Another question, prior to this one, was to show the union is a manifold, where the conditions were slightly different: one manifold was to be disjoint from the other's closure. So I believe I should be looking at intervals that are open, semi-open, closed, etc, within the coordinates in a space of bigger dimension. I am just afraid I am not too familiar with manifolds. One definition states that $M$ is a $k$-manifold near $x_0\in \Bbb{R}^n$ if there exists a permutation $(i_1,...,i_n)$ of $\{1,...,n\}$ and a mapping $g$, continuously differentiable near $({x_0}_{i_1},...,{x_0}_{i_k})$ such that: $$x\in M \iff g (x_{i_1},...,x_{i_k})=(x_{i_{n-k}},...,x_{i_n})$$.
So far, I looked at $M=$ $x$-axis in $\Bbb{R}^2$, with the identity permutation and $g(x)=0$. For all $(x,y)$ on the $x$ axis, $M$, $y=0$, and therefore $g(x)=0=y=x_2$. I am not sure about how legal it is and about differentiability. Is there a way to proceed? I would really appreciate reference to my attempts.