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.

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. 
