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Let $\mathbb{R}^2$, where $p \in \mathbb{R}^2$ is the ordered pair $(p_1, p_2)$, have the taxicab metric, i.e. for $p,q \in \mathbb{R}^2, d(p,q) = |p_1 – q_1|\ +\ |p_2 – q_2|$. Consider $\mathbb{R}^2$ with the topology induced by $d$.

The following metrics, $g(p,q)$, induces an equivalent topology on $\mathbb{R}^2$: $g(p,q) = \max\{|p_1 – q_1|,|p_2 – q_2|\}$ - (Maximum Metric)
$g(p,q) = \sqrt{(p_1 – q_1)^2+(p_2 – q_2)^2}$ - (Euclidean Metric)
$g(p,q)= \begin{cases} 0, & \text{if $p=q$} \\[2ex] |p_1 – q_1|\ +\ |p_2 – q_2|+1, & \text{if $p\neq q$} \end{cases} $

Are these correct? Are there any other well-known "standard" metrics that induce an equivalent topology on $\mathbb{R}^2$?

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Your third metric does not induce the same topology. For example, the set $U_\frac{1}{2}(0) = \{0\}$ is open in the last topology, but not in the other topologies. More generally, every set is open in your last topology - this is called the discrete topology.

The topology that is induced by the taxicab metric, the maximum metric and the Euclidean metric is called Euclidean (or standard) topology on $\mathbb{R}^2$. There are many other metric that induce this topology, for example the $l^p$-metrics for $p > 0$: $$d_p(x, y) = (|x_1 - y_1|^p + |x_2 - y_2|^p)^{\min(1, p^{-1})}$$

This class contains your metrics as special cases. $p = 1$ is the taxicab metric, $p = 2$ is the Euclidean metric and in the limit $p \to \infty$ you get the maximum metric.

There are many more metrics that induce the euclidean topology. For example, a large class of sets can induce the euclidean topology via the Minkowski functional.

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  • $\begingroup$ Thanks. What about $g(p,q) = min\{|p1 – q1|+|p2 – q2|,1\}$ as an alternative to my third? $\endgroup$
    – Learner
    Dec 20 '16 at 13:15
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    $\begingroup$ No, each $p > 0$ yields a different metric. See also here en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions $\endgroup$
    – Dominik
    Dec 20 '16 at 13:18
  • $\begingroup$ Interesting - might be above my studies so far - but looks like my min{...} example (edited it into my 1st comment) would work as it seems like the $L^p$ space goes from min to max? $\endgroup$
    – Learner
    Dec 20 '16 at 13:25
  • $\begingroup$ I don't quite understand your question. Note that for any metric $d$ the function $q(x, y) = \min\{d(x, y), c\}$ is also a metric, if $c > 0$. $\endgroup$
    – Dominik
    Dec 20 '16 at 13:27
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A footnote to the other A's: Some other ways to generate equivalent metrics:

(I).If $d$ is a metric on $X$ and $r>0$ then $e(p,q)= \min (r, d(p,q))$ is an equivalent metric. Observe that for $0<s<r/2$ the set of open $d$-balls of radius $s$ coincides with the set of open $e$-balls of radius $s.$

(II). If $d$ is a metric on $X$ then $e(p,q)=d(p,q)/(1+d(p,q))$ is an equivalent metric.

(III). Theorem. Let $(X,d)$ and $(Y,e)$ be metric spaces. Let $f:X\to Y$ be continuous. Then for $p,q\in X,$ the function $d'(p,q)=d(p,q)+e(f(p),f(q))$ is a metric on $X$ equivalent to $d.$

We prove this by showing, for $p\in X$ and $r>0,$ that there exist positive $s,s'$ such that $B_{d'}(p,s')\subset B_d(p,r)$ and $B_d(p,s)\subset B_{d'}(p,r).$ (The first inequality is immediate with $s'=r.$ The second requires the use of the continuity of $f.$)

Thus for any metric $d$ on $\mathbb R^2$ that generates the Cartesian topology, and any continuous $f:\mathbb R^2 \to \mathbb R$ (with the usual topology on $\mathbb R$) we can let $e(p,q)=d(p,q)+|f(p)-f(q)|.$

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