# Using the limit to prove the equivalence between two metrics

Two metrics $d$ and $p$ in some set $X$ are said to be equivalent if for any sequence $x_k \in X$ the following equivalence holds

$$\lim_{k \to \infty}d\left(x_k,x\right)=0 \Longleftrightarrow \lim_{k \to \infty}p\left(x_k,x\right)=0$$

How I can show that

1. $d(x,y)=\sqrt{\left(x_1-y_1 \right)^2+\left(x_2-y_2 \right)^2 }$
2. $p(x,y)=\left|x_1-y_1 \right|+\left|x_2-y_2\right|$

are equivalent?

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It is possible that you are confusing the meaning of subscripts here. Let me try to give hints anyway:

For two vectors $x = (x_1, x_2)$ and $y = (y_1, y_2)$ in $\mathbb{R}^2$ let:

$d(x, y) = \sqrt{ |x_1 - y_1|^2 + |x_2 - y_2|^2 }$

and

$p(x, y) = |x_1 - y_1| + |x_2 - y_2|$.

Let $x^{(k)} \in \mathbb{R}^2$, $k = 1, 2, \ldots$, be a sequence, and $x \in \mathbb{R}^2$.

Assume $d(x^{(k)}, x) \to 0$ as $k \to \infty$. Then what does this imply for the components $x_1^{(k)}$ and $x_1$? and also $x_2^{(k)}$ and $x_2$?

So, what can be said about $p(x^{(k)}, x)$ as $k \to \infty$?

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it's not absolute value under the root square so I don't think your answer works – Motasem M. Al-wazir Jan 13 '14 at 22:53
and I don't think it makes any difference; do you? – user66081 Jan 14 '14 at 0:27
it makes a lot of different – Motasem M. Al-wazir Jan 14 '14 at 10:20
what difference does it make then? – user66081 Jan 14 '14 at 13:37

It is also interesting to look at at things this way. Take any open ball with radius $\epsilon$ around a point $(x,y)$ in the metric $d$. Can you find an open ball in the metric $p$ which fits in this ball? Take for instance a ball in $p$ with radius $\epsilon/4$. This fits nicely in the original ball. Can you do the same thing the other way around?

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you mean that two metric are equivalent if they have the same topology. I know this way but I have to solve it like in the question above . – Motasem M. Al-wazir Jan 13 '14 at 22:58
You can use this for your proof. Take a sequence $x_k$ such that it converges in $d$. The argument in my answer proves that the sequence trivially converges in $p$. The other implication is analogous. – Leo Jan 13 '14 at 23:02
converges in d can be to anything not only 0. right or not? – Motasem M. Al-wazir Jan 14 '14 at 10:23