Lemma 5.2.798494B0CCC31

Proof. If $X$ is empty, take $N=0$. Otherwise let $n$ be the dimension of $X$. For each $x\in X$ choose a holomorphic chart $\kappa _ x\colon U_ x\to \mathbb C^ n=\mathbb R^{2n}$ with $x\in U_ x$. By proposition 5.2.6, applied to the compact set $\{ x\} $ and the open set $U_ x$, there is a smooth function $\chi _ x\colon X\to [0,1]$ with compact support in $U_ x$ which equals one on an open neighborhood $W_ x$ of $x$. Since $X$ is compact, finitely many of the sets $W_ x$ cover $X$; we denote the corresponding charts, cutoff functions and neighborhoods by $\kappa _ i\colon U_ i\to \mathbb R^{2n}$, $\chi _ i$ and $W_ i$, $1\le i\le k$. The map $\chi _ i\kappa _ i$, defined as the product on $U_ i$ and as $0$ on $X\setminus U_ i$, is smooth: it is smooth on $U_ i$, it vanishes on $X\setminus \operatorname{Supp}\chi _ i$, which is open because the compact set $\operatorname{Supp}\chi _ i$ is closed in the Hausdorff space $X$, and these two open sets cover $X$ because $\operatorname{Supp}\chi _ i\subset U_ i$. Put $N=k+2nk$ and

a smooth map. It is injective. Indeed, let $F(x)=F(y)$, and choose $i$ with $x\in W_ i$. Then $\chi _ i(y)=\chi _ i(x)=1$, so $y\in \operatorname{Supp}\chi _ i\subset U_ i$, and $\chi _ i\kappa _ i(x)=\chi _ i\kappa _ i(y)$ gives $\kappa _ i(x)=\kappa _ i(y)$; since $\kappa _ i$ is injective on $U_ i$, we obtain $x=y$.

If $C\subset X$ is closed, then $C$ is compact, so $F(C)$ is compact and hence closed in $\mathbb R^ N$. Thus the continuous bijection $F\colon X\to F(X)$ maps closed sets to closed sets, so its inverse is continuous, and $F$ is a homeomorphism onto $F(X)$. Hence $d$, the Euclidean metric of $F(X)$ transported to $X$ by $F$, is a metric on $X$ that induces its topology. $\square$