# $M \subset \mathbb{R}^n$ $n-$manifold is an open set or the closure of an open set in $\mathbb{R}^n$

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 $$[0,1) \subset \mathbb{R}$$.