Equivalent metrics may assign different numerical distances while producing the same convergent sequences and the same topology. The focus keyword equivalent metrics is useful because it permits a more convenient metric to replace another when studying continuity or convergence. This lesson compares standard Euclidean metrics, bounded metric transforms, and non-equivalent metrics on function spaces. :::definition[Equivalent Metrics] Let $d_1$ and $d_2$ be metrics on a nonempty set $X$. They are called equivalent if for every sequence $\{x_n\}$ and every $x\in X$, $d_1(x_n,x)\to0$ if and only if $d_2(x_n,x)\to0$. ::: :::remark The metrics $d_1$ and $d_2$ are equivalent if and only if the identity mappings $(X,d_1)\to(X,d_2)$ and $(X,d_2)\to(X,d_1)$ are both continuous. ::: :::theorem[Sufficient Condition for Equivalent Metrics] Let $d_1$ and $d_2$ be metrics on $X$. If there exists $K>0$ such that $(1/K)d_2(x,y)\leq d_1(x,y)\leq Kd_2(x,y)$ for all $x,y\in X$, then $d_1$ and $d_2$ are equivalent. ::: :::proof Given the two-sided inequality. To prove equivalence. If $d_1(x_n,x)\to0$, then $d_2(x_n,x)\leq Kd_1(x_n,x)\to0$. Conversely, if $d_2(x_n,x)\to0$, then $d_1(x_n,x)\leq Kd_2(x_n,x)\to0$. Hence, the metrics are equivalent. $\square$ ::: :::example[Standard Metrics on Euclidean Space] On $\mathbb R^n$, the metrics $d_1(x,y)=\sum|x_i-y_i|$, $d_2(x,y)=(\sum|x_i-y_i|^2)^{1/2}$, and $d_\infty(x,y)=\max_i|x_i-y_i|$ are equivalent. ::: :::proof For real numbers $a_i$, $(\sum a_i^2)^{1/2}\leq\sum|a_i|\leq n\max_i|a_i|\leq n(\sum a_i^2)^{1/2}$. Applying this to $a_i=x_i-y_i$ gives the required metric comparisons. $\square$ ::: :::example[Bounded Transform of a Metric] For a metric $d$, the formula $\rho(x,y)=d(x,y)/(1+d(x,y))$ defines a bounded metric equivalent to $d$. Convergence to zero in either metric is equivalent because $d=\rho/(1-\rho)$ when $\rho<1$. ::: :::example[Uniform and Integral Metrics] On $C[0,1]$, $d(f,g)=\sup|f-g|$ and $e(f,g)=\int_0^1|f-g|$ are not equivalent. For $f_n(x)=x^n$, $e(f_n,0)=1/(n+1)\to0$, but $d(f_n,0)=1$. ::: :::solved-problem For $x,y\in\mathbb R^n$, define $d_p(x,y)=(\sum_{i=1}^n|x_i-y_i|^p)^{1/p}$. Show that $\lim_{p\to\infty}d_p(x,y)=d_\infty(x,y)$. ::: :::solution Let $M=d_\infty(x,y)$. Then $M\leq d_p(x,y)\leq n^{1/p}M$. Since $n^{1/p}\to1$, the squeeze theorem gives $d_p(x,y)\to M=d_\infty(x,y)$. ::: :::faq[Frequently Asked Questions] Q: Do equivalent metrics have the same numerical distances? A: No. They have the same convergence and topology. Q: Does $d/(1+d)$ preserve convergence? A: Yes. It is equivalent to $d$. :::