When are the sup and euclidean metric interchangeable in analysis on $\mathbb{R}^{n}$ In Analysis on Manifolds, Munkres makes the statement that for most purposes, the sup metric, which is $$max\{|x_{i} - y_{i}| \space i \in \{1 \dots n\}\}$$ and the euclidean metric are "equivalent." In other words, they can be interchanged in theorems like this:

I cannot find an instance when interchanging the metrics in a theorem is unacceptable -- I suspect that this has to do with the fact that one can place, within any neighborhood of one metric in $\mathbb{R^{n}}$, another neighborhood of the second metric.
Is my hunch correct -- that since neighborhoods formed using either metric can be placed within each other, all theorems involving a metric can be formulated using either metric? At least with respect to analysis on $\mathbb{R^{n}}$?
 A: If you are talking about the exponent $\alpha$ of a Hölder condition on a function, then the metric matters: the Hölder exponent of a function can change when you change the metric. Even worse, it might satify a Hölder condition with respect to one metric but no Hölder condition at all with respect to another. 
The same issue holds for a Lipschitz condition on a function as well.
A: Yes -  actually, you have in $\mathbb{R}^n$ that all the norms are equivalent, that means, exists  $a, b>0$ such that
$$ a|x|_1\leq |x|_2\leq b|x|_1 $$
This is the same as your neighborhood image - for every sphere (neighborhood using euclidean norm) you can place inside a cube (neighborhood using sup norm). This makes a lot (sadly, not all) of the theorems have very similar proofs when you change the norm.  

For example, lets call $|\cdot|_1$ the sup norm, and $|\cdot |_2$ the euclidean norm. If we know that

4.7-Sup: Let $X$ be a compact subspace of $\mathbb{R}^m$; let $f:X\to \mathbb{R^n}$ be continuous. Given $\varepsilon>0$, there is a $\delta>0$ such that whenever $x, y\in X$,
$$|x-y|_1<\delta \text{ implies } |f(x)-f(y)|_1<\varepsilon $$

Using that, we can prove the following theorem

4.7-Euc: Let $X$ be a compact subspace of $\mathbb{R}^m$; let $f:X\to \mathbb{R^n}$ be continuous. Given $\varepsilon>0$, there is a $\delta>0$ such that whenever $x, y\in X$,
$$|x-y|_2<\delta \text{ implies } |f(x)-f(y)|_2<\varepsilon $$

Proof: Recall that exists some (fixed) $a, b>0$ such that, for all $p\in \mathbb{R}^m$
$$a|p|_1\leq |p|_2 \leq b|p|_1$$
Let $\varepsilon>0$. We know that exists a $\eta>0$ such that, for all $x, y \in \mathbb{R^m}$
$$ |x-y|_1<\eta \text{ implies } |f(x)-f(y)|_1<\frac{\varepsilon}{b} $$
Now, if, $\eta> \frac{1}{b}|x-y|_2$, then $\eta>\frac{1}{a}|x-y|_2\geq |x-y|_1$, and then implies $\varepsilon>b|f(x)-f(y)|_1\geq |f(x)-f(y)|_1$. And calling $\delta=\eta b >0$, we have 
$$ |x-y|_2<\delta \text{ implies } |f(x)-f(y)|_1<\varepsilon$$
So, 4.7-Sup implies 4.7-Euc. 
A: The Euclidean norm is strictly convex, and the sup norm isn't.  This is relevant when unique best approximations are relevant.  For example, given a closed convex subset $X$ of $\mathbb R^n$, and a point $p\in\mathbb R^n$, there is a unique $x\in X$ with smallest Euclidean distance to $p$.  This is not true with the sup norm; e.g., in $\mathbb R^2$, if $X$ is the closed unit "ball" in the sup norm, and $p=(2,0)$, then the smallest distance from $p$ to $X$ is $1$, but this minimum is achieved for all of the infinitely many points $(1,t)\in X$.  
