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What shown below is a reference from "Analysis on manifolds" by James R. Munkres

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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?

Antonio Maria Di Mauro
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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 $N_S(X,\epsilon)\subseteq U$.

Brian M. Scott
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  • Okay, I understood the secon point. However why $N_E(X,\epsilon)\subseteq N_S(X,\epsilon)$? could you explain? – Antonio Maria Di Mauro May 01 '20 at 18:06
  • Then if the two topologies are equal what you said it seems strange to me: indeed if two topology are equivalent then I can prove any statement using which of them I prefer, is this incorrect? – Antonio Maria Di Mauro May 01 '20 at 18:07
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    @AntonioMariaDiMauro: Just check that $d_E(x,y)\le d_S(x,y)$ for all $x,y\in\Bbb R^n$: $$\sqrt{(x_1-y_1)^2+\ldots+(x_n-y_n)^2}\le\max{|x_1-y_1+\ldots|x_n-y_n|};.$$ I’m not sure what you’re asking in your second comment. – Brian M. Scott May 01 '20 at 18:08
  • With the secon commment I want say that the topology induced by the sup metric is the same of the topology induced by euclidean metric and so if I prove a statement with the topology of sup metric then it is true even for the topology of euclidean metric and vice versa, is this uncorrect? – Antonio Maria Di Mauro May 01 '20 at 18:11
  • However I think you wanted write $$\sqrt{(x_1-y_1)^2+\ldots+(x_n-y_n)^2}\le\sqrt{n}\max{|x_1-y_1|+\ldots|x_n-y_n|}$$ right? Indeed I know that $d_S(x,y)\le d(x,y)\le\sqrt{n} d_S(x,y)$. – Antonio Maria Di Mauro May 01 '20 at 18:14
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    @AntonioMariaDiMauro: It is incorrect: whether proving something for one metric also proves it for an equivalent metric depends on the metrics and on what is being proved. And what I meant to write was: $$\sqrt{(x_1-y_1)^2+\ldots+(x_n-y_n)^2}\color{red}{\ge}\max{|x_1-y_1|,\ldots,|x_n-y_n|};.$$ That is what guarantees that the $\epsilon$-ball in the Euclidean metric is a subset of the $\epsilon$-ball in the sup metric. – Brian M. Scott May 01 '20 at 18:17
  • Okay, so if $d_E(x,y)\ge d_S(x,y)$ then why $N_E(X,\epsilon)\subseteq N_S(X,\epsilon)$? – Antonio Maria Di Mauro May 01 '20 at 18:22
  • @AntonioMariaDiMauro: because if something is $\epsilon$ far away in the sup metric, then it is $\epsilon$ at least $\epsilon$ far away in the Euclidean metric and possibly even further away. – Brian M. Scott May 01 '20 at 18:29
  • Sorry, I'm slightly confused. If $d_S(x,y)<\epsilon$ then generally it is not $d_E(x,y)<\epsilon$ and so it seems to me that $N_E(X,\epsilon)\nsubseteq N_S(X,\epsilon)$. Could you explain better? Excuse me, forgive my confusion. – Antonio Maria Di Mauro May 01 '20 at 18:32
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    @AntonioMariaDiMauro: No. $d_E(x,y)\color{red}{\ge}d_S(x,y)$, so if $d_S(x,y)\ge\epsilon$, then $d_E(x,y)\ge\epsilon$. That is the fact that is used in the proof. Equivalently, if $d_{\color{red}E}(x,y)\color{red}{\le}\epsilon$, then $d_S(x,y)\le\epsilon$. – Brian M. Scott May 01 '20 at 18:35
  • Okay, and from this how can I conclude the statement? – Antonio Maria Di Mauro May 01 '20 at 18:36
  • @AntonioMariaDiMauro: Which statement? – Brian M. Scott May 01 '20 at 18:37
  • $N_E(X,\epsilon)\subseteq N_S(X,\epsilon)$ – Antonio Maria Di Mauro May 01 '20 at 18:40
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    @AntonioMariaDiMauro: It’s an immediate consequence: if $x\notin N_S(X,\epsilon)$, then $d_S(x,X)\ge\epsilon$, so $d_E(x,X)\ge\epsilon$, and therefore $x\notin N_E(X,\epsilon)$. – Brian M. Scott May 01 '20 at 18:43
  • Okay, all clear. My error was this: if $B_E(x,\epsilon)$ is a ball in the euclidean topology and $B_S(x,\epsilon)$ is a ball in the sup topology and if $d_S(x,y)\le d_E(x,y)$ then $N_S(X,\epsilon):=\bigcup_{x\in X}B_S(x,\epsilon)\subseteq\bigcup_{x\in X}B_E(x,\epsilon)$. Thanks too much!!! – Antonio Maria Di Mauro May 01 '20 at 18:45
  • @AntonioMariaDiMauro: Yes, it’s easy to get the inclusion going the wrong way. You’re welcome! – Brian M. Scott May 01 '20 at 18:47
  • Hi professor Scott, could I ask your assistance here? – Antonio Maria Di Mauro May 02 '20 at 18:54