prove that a function is continuous with respect to metric

Question

Prove that the multiplication function $m : \mathbb{R}^2 → \mathbb{R}$, defined by $m(x, y) = xy$, is continuous with respect to the Euclidean metric $d_2$ on $\mathbb{R}^2$ and the usual (also Euclidean) metric on $\mathbb{R}$. HINT: pass to an equivalent metric.

a function $f: X \rightarrow Y$ is $(d_X, d_Y)$ continuous if for all $a \in X, \epsilon > 0$, there exists $\delta > 0$ such that $$d_X (x, a) < \delta \Rightarrow d_Y (f(x), f(a)) < \epsilon$$

so since $d_2$ is strongly equivalent to $d_\infty$, i want to prove that $d_\infty (p, q) < \delta \Rightarrow d_2 (m(p), m(q)) < \epsilon$ for points $p=(a, b)$ and $q=(x, y)$, but i don't know how to start

Answer 1

If $\{x_{n}\}$ and $\{y_{n}\}$ are two sequences of real numbers converging to x ,y $\in \mathbb{R} $ respectively then the sequence $\{x_{n}y_{n}\}$ converges to xy. Now one can use the sequential criterion for continuity to prove that m is continuos .

Answer 2

We have \begin{align*} d_{1}(m(x,y),m(a,b))&=d_{1}(xy,ab)\\ &=|xy-ab|\\ &=|xy-xb+xb-ab|\\ &\leq|x||y-b|+|b||x-a|\\ &\leq(|x|+|b|)(|x-a|+|y-b|)\\ &\leq(|x|+|b|)\left(\sqrt{|x-a|^{2}+|y-b|^{2}}+\sqrt{|x-a|^{2}+|y-b|^{2}}\right)\\ &=2(|x|+|b|)d_{2}((x,y),(a,b))\\ &\leq 2(|x-a|+|a|+|b|)d_{2}((x,y),(a,b))\\ &\leq 2(d_{2}((x,y),(a,b))+|a|+|b|)d_{2}((x,y),(a,b)). \end{align*} Given $\epsilon>0$, choose $\delta=\min\{1,\epsilon/(2(1+|a|+|b|))\}$, then it is routine to check that $d_{2}((x,y),(a,b))<\delta$ implies that $d_{1}(m(x,y),m(a,b))<\epsilon$.