What shown below is a reference from "Analysis on manifolds" by James R. Munkres
So I want discuss more carrefully why it is sufficent to prove the statement using the sup metric: infact we know that topology induced by the euclidean metric is equivalent to the topology induced by the sup metric and so clearly the statement holds, right? Finally I don't understand why if $\epsilon:=\text{min}\{f(x):x\in X\}$ then the $\epsilon$-neighborhood of $X$ is contained in $U$. So could someone help me, please?
Best Answer
No, the fact that the topologies are the same does not show that proving the theorem for one metric automatically proves it for the other. It works here because of the relationship between the two metrics involved. Let $N_E(X,\epsilon)$ be the Euclidean $\epsilon$-nbhd of $X$, and let $N_S(X,\epsilon)$ be the sup metric $\epsilon$-nbhd of $X$. Then $N_E(X,\epsilon)\subseteq N_S(X,\epsilon)$, so if $N_S(X,\epsilon)\subseteq U$, then automatically $N_E(X,\epsilon)\subseteq U$ as well.
For your second question, suppose that $\epsilon=\min\{f(x):x\in X\}$, and let $y\in\Bbb R^n\setminus U$. Then for each $x\in X$ we have $d_S(x,y)\ge f(x)\ge\epsilon$, so $y\notin B_S(x,\epsilon)$. And since $x\in X$ was arbitrary, $y\notin\bigcup_{x\in X}B_S(x,\epsilon)=N_S(X,\epsilon)$. That is $\Bbb R^n\setminus U$ is disjoint from $N_S(X,\epsilon)$, and hence $N_S(X,\epsilon)\subseteq U$.