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Let $X=\Bbb R^2$ and $x = (x_1, x_2), y = (y_1, y_2)$ and $d(x,y)=\sqrt {|x_1−x_2|+|y_1−y_2|}$ be a metric on $X\times X$. Is the metric space $(X,d)$ complete?

  • If $X_i=(x_i,y_i)$ is cauchy in $(X,d_2)$ then note that ${x_i},\ {y_i}$ are Cauchy wrt $|\ |$ So we have limits $x_\infty,\ y_\infty$ Note that $(x_\infty,y_\infty)$ is limit of $X_i$ – HK Lee Nov 03 '16 at 07:52
  • Can you confirm that the metric is $d_2(x,y) = \sqrt{|x_1-x_2| + |y_1 - y_2|}$ and not $d_2(x,y) = \sqrt{|x_1-x_2|^2 + |y_1 - y_2|^2}$? – Stefan4024 Nov 03 '16 at 08:48
  • @Stefan4024 you are right. In fact, I would ask for usual $d_2$ metric. Sorry for this. But as I see, the method of solution is same :) Thank you very much for your solution. –  Nov 03 '16 at 09:14

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Let $X_i = (x_i,y_i)$ be a Cauchy sequence in $\mathbb{R}^2$. Then note that both of the sequences $\{x_i\}$, $\{y_i\}$ are Cauchy, as $|x_i - x_j| \le ||X_i - X_j||$. But we know that $\mathbb{R}$ is complete so both $\{x_i\}$, $\{y_i\}$ converge to let say $a,b$, respectively. Now because of this $\exists N_1 \in \mathbb{N}$ s.t. $ n > N_1 \implies |x_{n} - a| < \frac{\epsilon^2}{2}$ and $\exists N_2 \in \mathbb{N}$ s.t. $ n > N_2 \implies |y_{n} - b| < \frac{\epsilon^2}{2}$. Now choose $N = \max\{N_1,N_2\}$, then we have that: $$ n>N \implies ||X_n - (a,b))|| = \sqrt{|x_i - a| + |y_i - b|} < \sqrt{\frac{\epsilon^2}{2} + \frac{\epsilon^2}{2}} = \epsilon$$

Therefore $X_i$ converges to $(a,b)$ and hence $\mathbb{R}^2$ is complete.

Stefan4024
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