# Limit preserving metrics.

I need to prove that given a sequence of points $\{a_n\}$ in $(\Bbb R^k,d_m)$ that converges to $a$, then it converges to the same limit in both $(\Bbb R^k,d_e)$ and $(\Bbb R^k,d_t)$ (and conversely), where

$$d_m(x,a)=\max_{1\leq i \leq k} \{|x_i-a_i|\} \;;\;\text{ the max metric}$$

$$d_e(x,a)=\sqrt{\sum_{1\leq i \leq k}(x_i-a_i)^2} \;;\;\text{ the Euclidean metric}$$

$$d_t(x,a)={\sum_{1\leq i \leq k}|x_i-a_i|} \;;\;\text{ the so-called taxicab metric}$$

There is a nice characterization of a point $a$ being a limit of a sequence $\{ a_n\}$ which is

Given a sequence of points $\{a_n\}$ in a metric space $(X,d)$ then $\lim a_n = a$ if and only if for every neighborhood $V$ of $a$, only but finitely many of the $a_n$ are not contained in $V$, or equivalently $a_n \in V$ for almost all of the $a_n$.

More formally, one can say that for every nbhd $V$ of $a$ there is an $N$ such that whenever $n>N$, $a_n \in V$.

This said, one can prove:

THEOREM Let $\{a_n\}$ be a sequence of points in $\Bbb R^k$. Then if $\lim a_n=a$ under either one of the metrics $d_t,d_e,d_m$, it follows $\lim a_n=a$ for the other two.

(NEW) PROOF We have that

$$\tag a d_m(a,x) \leq d_e(a,x) \leq \sqrt{k}\cdot d_m(a,x)$$ $$\tag b d_m(a,x) \leq d_t(a,x) \leq {k}\cdot d_m(a,x)$$

Let $\lim a_n=a$ in $(\Bbb R^k,d_m)$. Let $T$,$E$ be neighborhoods of $a$ in $(\Bbb R^k,d_t)$ and $(\Bbb R^k,d_e)$ respectively. Then they contain an open $t$-ball and an open $e$-ball about $a$ for a $\delta>0$. $(a)$ and $(b)$ guarantee that each of these balls contain an open $m$-ball about $a$. But this ball $B_m$ is a neighborhood of $a$ in $(\Bbb R^k,d_m)$ so $a_n\in B_m$ whenever $n>N$ for a some $N$. This in turn means that $a_n \in B_t$ and $a_n \in B_e$ whenever $n>N$ so $\lim a_n=a$ in both the two other metrics.

The other two cases follow analogously.

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It seems a bit off. First, at the outset, you should state which case you are proving. I surmise you are starting by assuming that you have convergence in the $m$-metric. Now you have to prove you have convergence in, say, the $e$-metric. So you need to start with an arbitrary $e$-nhood of $a$ (not with an arbitrary $m$-nhood of $a$) and show it contains a tail of the sequence. To do this, you can use your inequalities to show such an nhood contains an $m$-ball centered at $a$, and thus a tail of the sequence. – David Mitra Jul 16 '12 at 2:34
@DavidMitra That was precisely what was haunting me. You're right. Thanks. – Pedro Tamaroff Jul 16 '12 at 2:37
@DavidMitra I've rewritten the proof. – Pedro Tamaroff Jul 16 '12 at 3:06

Optionally, you could structure the proof by preceding it with a definition and a lemma.

Definition: Two metrics $d$ and $d'$ are comparable if there exists a constant $C$ such that $d/C\le d'\le Cd$.

Lemma. Two comparable metrics have the same convergent sequences.

This way the neighborhoods and sequences appear only in the lemma and the rest becomes a matter of estimating metrics up to multiplicative constants.

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Does it necessarily have to be the case we take $C$ and $C^{-1}$? I Think $$K\cdot d(x,y) \leq d'(x,y) \leq C d(x,y)$$ is good enough. Peeking some pages ahead, I read that the identity mapping defines a topological equivalence between to metric spaces with the same underlying set and metrics $d'$ and $d$ if there exist $K$ and $K'$ such that \eqalign{ & d'\left( {x,y} \right) \leqslant K \cdot d\left( {x,y} \right) \cr & d\left( {x,y} \right) \leqslant K' \cdot d'\left( {x,y} \right) \cr} – Pedro Tamaroff Jul 16 '12 at 3:26
@Peter Using C and 1/C is just a device to reduce the number of constants flying around in the proof. It makes no difference otherwise. If your inequalities hold then so does mine, with C being the maximum of K and K'. – user31373 Jul 16 '12 at 3:33
@PeterTamaroff, you are correct, but there is no change in results given by finding optimal constants. Furthermore, if there are constants $K, K'$ as your book writes, there is also a $1/C,C$ as Leonid writes. More generally, two comparable metrics have the same Cauchy sequences. If the metric space is complete in either metric, you get the same results as for $\mathbb R^n.$ – Will Jagy Jul 16 '12 at 3:34
@WillJagy Right. Both you and Leonid are right. Thank you. – Pedro Tamaroff Jul 16 '12 at 3:55
We are filled with rightness. – Will Jagy Jul 16 '12 at 4:19