Consider the set $\mathcal{M}_{1} = \{ \ (x,y,z) \in \mathbb{R}^{3}\ | \ y = -1 \ \}$, this is a plane. Also consider the set $\mathcal{M}_{2} = \{ \ (x,y,z) \in \mathbb{R}^{3}\ | \ x=y=0 \ \}$, which is a line.
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My question: Is $\mathcal{M}_{1} \cup \mathcal{M}_{2}$ a topological submanifold of $\mathbb{R}^{3}$?
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I know that separately, $\mathcal{M}_{1}$ is a 2-D submanifold, and $\mathcal{M}_{2}$ is a 1-D submanifold. But since the dimensions of these manifolds are not the same, I am thinking that $\mathcal{M}_{1} \cup \mathcal{M}_{2}$ cannot be a proper manifold, since the dimension of a manifold should be the same everywhere....am I correct in thinking this?