# Applaying equivalence of norms to show a sequence is a Cauchy sequence

Let $\|\cdot\|$ be any norm on $\mathbb R^n$ prove that a sequence $x \in \mathbb R^n$ is a Cauchy sequence under $\|\cdot\|_2$ if and only if it is a Cauchy sequence under any $\|\cdot\|$.

I tried following this link in wiki and this post with a very similar question.

I want to use the system from the post and say for a Cauchy sequence $|x_m-x_n|<\varepsilon$ and therefor $\|x_m-x_n\|_1<\varepsilon$, and then use $c\|x_m-x_n\|_2\leq \|x_m-x_n\|_1\leq d\|x_m-x_n\|_2$.

Can i say that if $|x_m-x_n|<\varepsilon$ (from wiki) then the norm is also smaller then $\varepsilon$ that is: $\|x_m-x_n\|<\varepsilon$.

Can I apply the same system as used in the post in this case? if not what will be a good way to prove this?

• Is $(x_k)_{k=1}^\infty$ a sequence in $\mathbb R^n$ ? If so, the expression $\lvert x_m-x_n \rvert < \epsilon$ doesn't make sense. Furthermore, I don't understand what your question is. – user42761 May 13 '16 at 18:22
• @Epsilon $|x_m-x_n| < \varepsilon$ is just like $|a_m-a_n| < \varepsilon$ in the link from wiki . – havakok May 13 '16 at 18:40
• Also @Epsilon I am trying to ask if i can deduce the same thing for the norm that is: $\|x_m-x_n\|<\varepsilon$. – havakok May 13 '16 at 18:41

Hint: Show that if $E$ is a vector space finite dimensional then all norms in $E$ are equivalents, i.e., if $\|\cdot\|_1$ and $\|\cdot\|_2$ are norms in $E$ then there are $c_1,c_2 >0$ constants such that
$$c_1 \|x\|_1\le\|x\|_2\le c_2\|x\|_1$$
for every $x\in E.$
• If $(x_k)_k\subset \Bbb{R}^n$ is a Cauchy sequence is $(\Bbb{R}^n,\|\cdot\|_1)$, then for all $\varepsilon >0$, we have $\|x_l -x_s\|_1< \varepsilon$, for $s,l>k_0\in \Bbb{N}$. By inequality $\frac{1}{c_2}\|x_l -x_s\|_2\le \|x_l -x_s\|_1<\varepsilon$, $s,l>k_0\in \Bbb{N}$. Then $(x_k)_k\subset \Bbb{R}^n$ is a Cauchy sequence is $(\Bbb{R}^n,\|\cdot\|_2)$. The other sense is analogue. – Irddo May 14 '16 at 16:05