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Let $S=\{\{a_n\}_{n\in\mathbb{N}}: a_n\in\mathbb{R}, \sum|a_n|<\infty\}$. Determine if the metrics $d(\{a_n\},\{b_n\})=\sum|a_n-b_n|$ and $\rho(\{a_n\},\{b_n\})=\sup|a_n-b_n|$ are equivalent metrics (equivalent in sense of convergence of sequences).

I proved that both are metrics and are well defined. Also I proved that if $\{\{x_n\}_k\}_{k\in\mathbb{N}}$ is a sequence in $S$ such that converges in $d$ then converges in $\rho$ (this part is trivial because $\rho\leq d$). I need to determine if converse is true or false.

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Just a remark on notation/terminology: your set $S$ is usually denoted by $\ell^1$ or $l^1$ and is called the Banach space of (absolutely) summable sequences. The metric $d$ is called the $\ell^1$-metric and the metric $\rho$ is called the $\ell^\infty$-metric (or $\sup$-metric). See e.g. here. –  t.b. Aug 10 '12 at 7:24
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up vote 5 down vote accepted

It is not true,

Define $a_n$ a sequence of sequences as follow $$a_1 = (1,0,0, \cdots,0, \cdots $$ $$a_2 = (\frac{1}{2},\frac{1}{2},0,0,0 \cdots,0, \cdots $$ $$a_3 = (\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0 \cdots,0, \cdots $$ $$\cdots$$ $$a_n =(\frac{1}{n},\frac{1}{n},\cdots, \frac{1}{n},\frac{1}{n},0,0 \cdots,0, \cdots$$ where in the previous $\frac{1}{n}$ are $n$ times. So $ a_{n}$ is a sequence where the first $n$ ordinates are $ \frac{1}{n}$and the other $0$. Then take $a_{n}$ and $b=0$ where $b=(0,0, \cdots ,0, \cdots)$.
So $$ \rho( a_{n} , b )= \frac{1}{n} \rightarrow 0$$ that means $a_n$ converges to $b$ in respect to $ \rho$

$$d( a_n , b )= \sum _{i=1}^{n} \frac{1}{n} =1$$ That means that $a_n$ (a sequence of sequences) does not converge to $b$ (the zero sequence) in respect to $d$

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I think that all its ok but $\rho\to 1$. In fact $\rho=1$ for all $m$. –  Gastón Burrull Aug 9 '12 at 17:39
    
Is that good now ? –  clark Aug 9 '12 at 17:49
    
$\rho$ is not 1/n is 1 and converges to 1 and $d$ is a finite sum with $m$ terms instead of $n$ and diverges for $m\to \infty$ –  Gastón Burrull Aug 9 '12 at 17:51
    
I will try to write it to be more clear of what I am trying to say. –  clark Aug 9 '12 at 17:58
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Even better: it is true now. ;) –  tomasz Aug 9 '12 at 18:15
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