# Question Regarding Relatively Open Sets of a Submanifold

I am reading a text in introductory differential geometry and it says " If $$M$$ is a submanifold of $$\mathbb{R}^n$$, and $$N$$ is open relative to $$M$$, then it follows easily from the definitions (of a submanifold) that $$N$$ is a submanifold of $$\mathbb{R}^n$$, with same dimension as $$M$$.

I don't see how it follows so easily. I think that since $$M$$ is a submanifold of $$\mathbb{R}^n$$, then there exists a smooth $$F$$ such that $$F:U \subset \mathbb{R}^n \rightarrow \mathbb{R}^k$$, where $$k = n-m$$. In other words, $$F^{-1}(c) = U \cap M$$. If we were to consider the restriction of $$F$$ to the set $$U \cap M$$, then by definition of relativity open (i.e. $$\exists O$$ open such that $$O \cap M$$ is relatively open), the map $$F$$ restricted to $$U \cap M$$ maps to $$\mathbb{R}^k$$, and the image is contained in a neighborhood of $$c$$. This restriction is also smooth, so the fact that a relatively open subset is a submanifold follows.

Is this all that I need? I feel I am getting mixed up somewhere.

Consider a smooth atlas on $$M$$, say ($$\varphi_{\alpha}, U_{\alpha})$$, where every $$U_{\alpha}$$ is open. Now for any $$\alpha$$ such that $$U_{\alpha} \cap N \neq \emptyset$$ consider $$\overline{\varphi_{\alpha}}: U_{\alpha} \cap N \rightarrow \mathbb{R}^k$$ defined by taking the restriction of $$\varphi_{\alpha}$$ over $$U_{\alpha} \cap N$$.

The new atlas ($$\overline{\varphi_{\alpha}}, U_{\alpha} \cap N$$) is a smooth atlas on $$N$$ (the restriction of smooth maps to open sets is smooth, and ($$U_{\alpha} \cap N$$)$$_{\alpha}$$ is a partition of $$N$$).

• I see, that's what I figured but wasn't quite sure how to say it.. thanks. – LordVader007 Feb 14 at 8:12