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Theorem 22.3 (Smooth invariance of domain). Let $U \subset\mathbb{R}^n$ be an open subset, $S \subset\mathbb{R}^n$ an arbitrary subset, and $f : U \rightarrow S$ a diffeomorphism. Then $S$ is open in $\mathbb{R}^n$.

I can't understand why the set $S$ is not automatically open in $\mathbb{R}^n$. The mapping is a diffemorphism，which means it is continuous in both directions，so $S$ is open.

cmk
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Toney Leung
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  • Smooth invariance of domain : Let $U\subseteq \mathbb{R}^n$ be an open set and $S\subseteq \mathbb{R}^n$ be an arbitrary set. If there exists a diffeomorphism $f:U\rightarrow S$, then $S$ is an open subset of $\mathbb{R}^n$. This diffeomorphism does not say about open/closedness inside $\mathbb{R}^n$. – Praphulla Koushik Sep 22 '19 at 15:45

1 Answers1

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All you know a priori, is that open sets $V$ of $U$ satisfy: $ f(V)$ is open in $S$, not that $f(V)$ is open in $\mathbb R^n .$ So, $f(U)=S$ is open $\textit{in S}$. The claim is then that $f(U)=S$ is actually open in $\mathbb R^n$, which is not the same thing and is not automatic. It requires proof.

Matematleta
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