Is every embedded smooth submanifold either open or closed?

Question

One often wants to show that a smooth embedded submanifold is open or closed.

Is it sufficient to show such a submanifold is not open to imply it is closed and vice versa?

Answer 1

We consider an $n$-manifold $M$ without boundary.

1. A submanifold $N \subset M$ is open if and only if $\dim N = n$.

If $N$ is a $k$-dimensional submanifold of $M$, then for each $x \in N$ there is a chart $\phi : U \to V \subset \mathbb R^n$ on $M$ with $p \in U$, where $U$ is open in $M$, such that $\phi(U \cap N) = V \cap (\mathbb R^{k} \times \{0\})$ with $0 \in \mathbb R^{n-k}$. Thus if $k < n$, then $N$ cannot be open in $M$ because otherwise $V \cap (\mathbb R^{k} \times \{0\})$ would be open in $\mathbb R^n$ which is not true. If $k = n$, then $V \cap (\mathbb R^{k} \times \{0\}) = V$ and hence $U \cap N = U$, therefore $U \subset N$.

2. If $k < n$, then $N$ may be closed or non-closed in $M$.

For example, if $N$ is compact, then it is closed. Another example is $\mathbb R^{k} \times \{0\}$ which is closed in $\mathbb R^n$.

However, each closed $k$-dimensional submanifold $N$ contains a subset $N' \subset N$ which is open in $N$ but not closed in $N$. Then $N'$ is also a $k$-dimensional submanifold of $M$, but it is not closed in $M$.

To find such an $N'$, let $\psi : U \to V$ be a chart on $N$. For $x \in V$ there exists $r > 0$ such the closed ball $B^c_r(x)$ is contained in $V$. Then the preimage $N' =\psi^{-1}(B_r^o(x))$ of the open ball $B_r^o(x)$ is an open subset of $U$ (and thus also an open subset of $N$) which is not closed in $N$ and therefore not closed in $M$.