$\bigcup_{i \in I} M_i$ is a manifold when $\forall j \in I, M_j \cap \overline{\bigcup_{i \neq j} M_i}= \emptyset$

Question

Let $M_i, i \in I$ be a family of $d$-dimensional manifolds in $\mathbb{R}^p$ so that $$\forall j \in I, M_j \cap \overline{\bigcup_{i \neq j} M_i}= \emptyset$$

How can I show that $\bigcup_{i \in I} M_i$ is a manifold?

Answer 1

A subspace $N \subset \mathbb{R}^p$ is a $d$-dimensional submanifold if for each $y \in N$ there exists a submanifold chart for $N$ around $y$, the latter being a homeomorphism $h : V \to V'$ between open $V, V' \subset \mathbb{R}^p$ such that $y \in V$ and $h(V \cap N) = V' \cap \mathbb{R}^d \times \{ 0 \}$. You can also define smooth $d$-submanifolds by requiring the submanifold charts to be diffeomorphisms. Note that if $W \subset \mathbb{R}^p$ is an open neighborhood of $y$ such that $W \subset V$, then the restriction $h_W : W \to h(W)$ is also a submanifold chart for $N$ around $y$.

Now consider $M = \bigcup_i M_i$. We shall show that each $M_j$ is open in $M$. This implies that $M$ is a $d$-dimensional submanifold:

Each $x \in M$ is contained in some $M_j$. Let $h : V \to V'$ be submanifold chart for $M_j$ around $x$. Since $M_j$ is open in $M$, there exists an open $U \subset \mathbb{R}^p$ such that $U \cap M = M_j$ (subspace topology!). Then $h_{V \cap U}$ is a submanifold chart for $M_j$ around $x$. But now $(V \cap U) \cap M = V \cap (U \cap M) = V \cap M_j$ so that $h_{V \cap U}$ is a submanifold chart for $M$ around $x$.

Let us now verify that $M_j$ is open in $M$. Denoting by $\overline{\phantom{X}}$ closure in $\mathbb{R}^p$, we know that

$(\ast)$ $M_j \cap \overline{\bigcup_{i \ne j} M_i} = \emptyset$, i.e. $\overline{\bigcup_{i \ne j} M_i} \subset \mathbb{R}^p \setminus M_j$.

But $\overline{\bigcup_{i \ne j} M_i}^M = (\overline{\bigcup_{i \ne j} M_i}) \cap M$, where $\overline{\phantom{X}}^M$ denotes closure in the subspace $M$. Hence $\overline{\bigcup_{i \ne j} M_i}^M \subset (\mathbb{R}^p \setminus M_j) \cap M = M \setminus M_j = \bigcup_{i \ne j} M_i$. For the last equation note that the $M_i$ are pairwise disjoint by $(\ast)$.

We conclude that $\overline{\bigcup_{i \ne j} M_i}^M = \bigcup_{i \ne j} M_i$, thus $M_j = M \setminus \bigcup_{i \ne j} M_i$ is open in $M$.