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Let $V$ be a real vector space. Suppose that $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are two norms on $V$ which are equivalent.

I suspect the following to be true.

Let $(x_n)_{n=0}^\infty$ be a sequence in $V$. Then $(x_n)_n$ is a Cauchy sequence w.r.t. $\Vert \cdot \Vert_1$ if and only if $(x_n)_n$ is a Cauchy sequence w.r.t. $\Vert \cdot \Vert_2$.

Is it true?

I'm looking for a good way to explain why "equivalence" of norms is the right way of "comparing" norms without mentioning anything from topology besides "convergence".

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Yes, it is true. Have you tried to prove it? It's pretty simple. –  Nate Eldredge Oct 19 '11 at 18:05
    
What does it mean for the spaces when the norms are equivalent? –  Jonas Teuwen Oct 19 '11 at 18:10
    
Being a Cauchy sequence or not is not a purely topological concept. You need an uniform structure for that. But if the norms are equivalent, the topology they generate is the same. The convergent sequences are the same in both norms. –  André Caldas Oct 19 '11 at 23:21
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up vote 2 down vote accepted

Yes this is true. If $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent, then by definition there exist $C > 0, c > 0$ such that $$c\|x\|_1 \le \|x\|_2 \le C\|x\|_1$$ for all $x \in V$.

Suppose that $(x_n) \subset V$ is Cauchy in the norm $\|\cdot\|_2$ and let $\epsilon > 0$. Then there exists an $N > 0$ such that $$\|x_n - x_m\|_2 \le c\epsilon$$ whenever $n,m > N$. Thus $$\|x_n - x_m\|_1 \le \epsilon$$ whenever $n,m > N$. So $(x_n)$ is Cauchy in the norm $\|\cdot\|_1$. The proof of the reverse implication works the same way.

Basically, equivalent norms induce the same open sets, so that as topological spaces $(V, \|\cdot\|_1)$ and $(V, \|\cdot\|_2)$ are the same. In other words, they induce equivalent metrics $d_1$ and $d_2$ on $V$ so that elements of $V$ are "close" with respect to $d_1$ iff they are "close" with respect to $d_2$.

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