Question on proof that euclidean and square metrics induce the same topology on $\mathbb{R}^n$ There are a few things I'm trying to understand regarding Theorem 20.3 in Munkres, which states that $\left(\mathbb{R}^n, d_E\right)$ ($d_E$ is the Euclidean metric) and $\left(\mathbb{R}^n, d_{\infty}\right)$ (the square metric) induce the same topology.

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*I understand the open balls in the square metric are in fact "open squares." I'm trying to understand where this comes about geometrically. I tried to restrict to the $n=2$ case, but I haven't had any success with sketching it. Could someone give me some intuition on this? Are they only "open squares" in $\mathbb{R}^2$, or in $\mathbb{R}^n$ as well? I assume in $\mathbb{R}^n$, it's in effect an open $k$-cell.


*I'm going to denote an open ball with respect to the Euclidean metric by $B_r^{E}$ and an open ball with respect to the square metric (am I using the correct terminology by saying "with respect to" here?) by $B_r^{\infty}$. The proof argues that $B_r^{E} \subset B_R^{\infty}$ and $B_{r/\sqrt{n}}^{\infty} \subset B_r^{E}$. I don't have any intuition on why this is the case. The standard hint I've seen is to restrict to the $n=2$ case, but I don't know how that generalizes to the $n$-dimensional case.
I'd appreciate any help on understanding these, specifically the associated geometry.
 A: *

*To sketch $B^{\infty}_1(0)$ in $n=3$ (it will be harder in higher dimensions), draw the planes $x_1=\pm 1$, $x_2=\pm 1$ and $x_3=\pm 1$ and the interior of the ball is the bounded region cut out by these. The same is true in $n=2$. Just draw $x_1=\pm 1$ and $x_2 = \pm 1$ and you can see they break the plane into several parts and the ball is the bounded region.


*We have two inequalities about the $E$ and $\infty$ norms
$$\|{x}\|_{E}=\left(\sum_{i=1}^n \vert x_1\vert^2\right)^{1/2}\leq \left(\sup_i \vert x_i\vert^2\sum_{i=1}^n 1\right)^{1/2} = \sup_i\vert x_i\vert \sqrt{n}=\sqrt{n}\|x\|_{\infty}$$
and
$$\|x\|_{\infty}=\sup_{i}\vert x_i\vert =(\sup_i\vert x_i\vert^2)^{1/2}\leq\left(\sum_{i=1}^n\vert x_i\vert^2\right)^{1/2}=\|x\|_{E}$$
The first shows that $d_{\infty}(x,y)=\|x-y\|_{\infty} < r \implies d_{E}(x,y)=\|x-y\|_{E} < \sqrt{n}r$ or $$B^{\infty}_r\subset B^{E}_{\sqrt{n}r}$$
The second shows that $d_E(x,y)=\|x-y\|_E < r \implies d_{\infty}(x,y)=\|x-y\|_{\infty} < r$ or $$B^E_r\subset B^{\infty}_r$$
A: 1. To see that open balls with respect to $d_∞$ are really cubes, consider the unit ball $d_∞(\vec x, \vec 0) = 1$. Expanding the definition of $d_∞$, this is $\max_i|x_i| = 1$, which means that all the components of $\vec x$ satisfy $-1 ≤ x_i ≤ +1$, with at least one of them being on the edge, $x_j = ±1$. In other words, if we pick one of the coordinates to be fixed at $±1$, the others vary and describe a hypersquare region of sidelength $2$ — this is a face of the hypercube. The whole locus is the shell of a cube of (inner) radius of $1$.
So the open ball $B_R^∞$ described by $d_∞(\vec x, \vec 0) < R$ is the interior of a cube.
2. To see that $B_R^E \subset B_R^∞$, just convince yourself that (the interior of) a ball of radius $R$ fits inside (the interior of) a cube of inner radius $R$ — in any number of dimensions. More formally, this amounts to showing $\|\vec x\|_E < 1 \implies \|\vec x\|_∞ < 1$, or $\sqrt{\sum_ix_i^2} < 1 \implies \max_i|x_i| < 1$.
For $B_{R/\sqrt n}^∞ \subset B_R^E$, think about the largest cube that fits inside a sphere of radius $R$. A bit of thought shows us that this interior cube has an inner radius $R/\sqrt 2$ in the case of a 2D square, and an inner radius of $R/\sqrt 3$ in 3D. The general case $R/\sqrt n$ can be derived using Pythagoras’ theorem to relate the hypercube’s inner and outer radii.
(It is correct to talk about a ball “with respect to” some norm, since the definition of a ball $B_R = \{d(\vec x, \vec 0) < R \mid \vec x ∈ \mathbb R^n\}$ indeed depends on a metric $d$.)
A: Munkres has a named trick for the square metric in the Topology 2nd Edition. His trick is confusing for readers. The square metric in his mathematic term is actually the classical Chebyshev distance for the two dimensions. So it is explicitly a square region for the two dimensions. Please see the following links for the definition and the picture.
wikipedia:
https://en.wikipedia.org/wiki/Chebyshev_distance
Chebyshev norm:
https://en.wikipedia.org/wiki/Uniform_norm
real_analysis-Page-104-and-108:
http://tomlr.free.fr/Math%E9matiques/Math%20Complete/Analysis/Basic%20Elements%20of%20Real%20Analysis%20-%20M.%20Protter.pdf
