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An open set in an $n$-manifold is clearly a submanifold of the same dimension as its containing manifold (see open manifolds).

Now, given an $n$-manifold $M$, is it true that a set, to be the underlying set of a submanifold of $M$ with dimension $n$, must be open?

Bruno Stonek
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    This depends on whether you allow manifolds with boundary. – Justin Campbell Nov 22 '11 at 15:00
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    The OP is clearly learning the basics of differential geometry. Let's try not to make him more confused with subtleties... – PseudoNeo Nov 22 '11 at 15:15
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    Converse: https://math.stackexchange.com/questions/2144470/open-submanifold-having-the-same-dimension-of-a-manifold –  Jul 21 '19 at 09:19

3 Answers3

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Yes.

In general, $N \subset M$ is a submanifold if you can find for every $x \in N$ an open neighbourhood $U \subset M$ of $x$, an open neighbourhood $V \subset \mathbb R^{\dim M}$ of $0$ and a diffeomorphism $\Phi : U \to V$ such that $\Phi(x) = 0$ and $\Phi(N) = (\mathbb R^{\dim N} \oplus 0) \cap V$. (It is legit to talk about diffeomorphisms because $M$ is a manifold).

Now, if $\dim N = \dim M$, the condition implies $\Phi(N) = V$. In other words, the open neighbourhood $U$ (the domain of the chart) must lie in $N$. So, $N$ contains an open neighbourhood of each of its points, and it is therefore open.

PseudoNeo
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Let $f$ be embedding of submanifold $N$ into $M$. I only need to prove $f(N)$ is open. Now consider $Df(p)$ from tangent space of $N$ at $p$ to tangent space of $M$ at $f(p)$.

Now $Df(p)$ is one-to-one (by defintion of submanifold). But $\dim{N}= \dim{M }$. Using the Rank Nullity theorem we get that $Df(p)$ is one-to-one and onto. Hence $f$ is local diffeomorphism at each point.

Hence f is diffeomorphism.(although proof of this also goes in same way as @PseudoNeo gave)

In particular $f$ homeomorphism. Hence $f(N)$ open.

Josef K.
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Sushil
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I believe other answers assume a differentiable structure. Your statement also holds for topological manifolds, though it does require the invariance of domain.

Suppose $N \subset M$ is a submanifold with $\dim N = \dim M = n$. For each $x \in N$, choose open neighborhoods $U$ of $x$ in $M$ and $V$ of $x$ in $N$ that are both homeomorphic to $\mathbb{R}^n$. Let

$$ f: U \xrightarrow{\sim} \mathbb{R}^n, \qquad g: V \xrightarrow{\sim} \mathbb{R}^n $$

be these homeomorphisms (note that $V$ is open in the subspace topology on $N$).

Now consider

$$ f \circ g^{-1} : g(U \cap V) \;\longrightarrow\; \mathbb{R}^n. $$

Since $U \cap V$ is open in $N$, its image $g(U\cap V)$ is open in $\mathbb{R}^n$. By invariance of domain, $f(U\cap V)$ is then open in $\mathbb{R}^n$, which means $U\cap V$ is open in $M$. Because this holds for every $x\in N$, it follows that $N$ is open in $M$.

Jerrome Chu
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