I am reading the wonderful work of Munkres on Topology. I have crossed paths with a really interesting exercise, in section §20 of Ch.2. It reads:
In $\mathbb{R}^n$ define: $$ d'({\bf x},{\bf y})=|x_i - y_i|+ \dots + |x_n - y_n|$$ Show that $d'$ is a metric that induces the usual topology on $\mathbb{R}^n$.
It is a truly remarkable result, as I thought any two different metrics give rise to different topologies. It almost looks like, even if there are infinite ways to construct a topology on infinite sets, most of our spatial intuition relies on a few interesting topologies. I now wish to show my attempt, then make a claim that could possibly shorten the proof. (This proof generalises pretty easily to $l_p$ metrics with $p\in\mathbb{N}$, as well.)
The proof
Let us denote the Manhattan (the $d'$ metric) and the Euclidean metric by $d_1$ and $d_2$, and the topologies induced by the two metrics by $\mathcal{T_1}$ and $\mathcal{T_2}$. First of all, let's note that for any ${\bf x},{\bf y}\in\mathbb{R}^n$ the following inequality holds: $$ n \cdot d_2({\bf x},{\bf y}) \ge d_1({\bf x},{\bf y}) \ge d_2({\bf x},{\bf y})$$ Let $U\in\mathcal{T_2}$, let ${\bf x}\in U$. Then we can find an $\epsilon$-ball $B_{d_2}({\bf x}, \epsilon)$ contained in $U$. Let ${\bf y}\in B_{d_2}({\bf x}, \epsilon)$. Now consider the $\delta$-ball $B_{d_1}({\bf y}, \delta)$ with: $$ \delta=\epsilon - d_2({\bf x},{\bf y})$$ Let ${\bf z}\in\mathbb{R}^n$. We wish to show that if ${\bf z}\in B_{d_1}({\bf y}, \delta)$, then it is contained in $B_{d_2}({\bf x}, \epsilon)$. $$ d_1({\bf z},{\bf y}) < \epsilon - d_2({\bf x},{\bf y}) \\ \implies d_2({\bf z},{\bf x}) \le d_2({\bf z},{\bf y}) + d_2({\bf x},{\bf y}) \le d_1({\bf z},{\bf y}) + d_2({\bf x},{\bf y}) < \epsilon \\ \implies d_2({\bf z},{\bf x}) < \epsilon $$ Then $B_{d_1}({\bf y}, \delta) \subseteq B_{d_2}({\bf x}, \epsilon) \subseteq U$ for any ${\bf y}\in B_{d_2}({\bf x}, \epsilon)$.
Conversely, let $U\in\mathcal{T_1}$, let ${\bf x}\in U$. Then we can find an $\epsilon$-ball $B_{d_1}({\bf x}, \epsilon)$ contained in $U$. Let ${\bf y}\in B_{d_1}({\bf x}, \epsilon)$. Now consider the $\delta$-ball $B_{d_2}({\bf y}, \delta)$ with: $$ \delta= \frac{\epsilon - d_1({\bf x},{\bf y})}{n}$$ Let ${\bf z}\in\mathbb{R}^n$. We wish to show that if ${\bf z}\in B_{d_2}({\bf y}, \delta)$, then it is contained in $B_{d_1}({\bf x}, \epsilon)$. $$ d_2({\bf z},{\bf y}) < \frac{\epsilon - d_1({\bf x},{\bf y})}{n} \\ \implies d_1({\bf z},{\bf x}) \le d_1({\bf z},{\bf y}) + d_1({\bf x},{\bf y}) \le nd_2({\bf z},{\bf y}) + d_1({\bf x},{\bf y}) < \epsilon \\ \implies d_1({\bf z},{\bf x}) < \epsilon $$ Then $B_{d_2}({\bf y}, \delta) \subseteq B_{d_1}({\bf x}, \epsilon) \subseteq U$ for any ${\bf y}\in B_{d_1}({\bf x}, \epsilon)$.
The claim
I have read somewhere it is sufficient to show that every open ball in one topology contains an open ball of the second topology with the same center. Is it true? Will it suffice?
I feel like my proof does that, in some way. The idea behind requiring the two balls to have the same center is to show that in any point of an open set of a topology one can find an open set of the other topology containing the point and contained in the open set, which is also the base for my proof.
