I'd like to prove the fact that for an $n-$manifold $M\subseteq \mathbb{R}^{n}$ it's the same thing to say that $M$ is open or is the closure of an open set of $\mathbb{R}^n$.

I think the case where $M$ has no boundary is easier since $M = \bigcup\limits_{p_i \in M}\varphi_i(U_i)$ which is open (since union of open sets), because $\varphi_{i} (U_{i})$ is a local diffeomorphism between and open set $U_i$ of $\mathbb{R}^n$ and $V_{p_{i}} \cap M$ open neighboorhood of $p_{i} \in M$.

What about the case where I've got local diffeomorphism $\varphi_i(U_i \cap \left\lbrace x_{n} \geq 0 \right\rbrace) = V_{p_i} \cap M$ ?

Any help or direct proof would be appreciated.


By invariance of domain, if $f: U \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$ is injective, then $f(U)$ is open in $\mathbb{R}^n$, so for a manifold without boundary this is true, since it os the union of inverses of charts. Now, if $M$ has a boundary this is false, just consider $[0,1) \subset \mathbb{R}$.


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