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Let $l=\{(a_n)_{n \in ℕ}|a_n \in \Bbb R \text{ and } \sup|a_n|<\infty\}$ If $a=(a_n)$ and $b=(b_n)$ are in $l$, define a metric on $l$ by

$$d(a,b)=\sup_{n \in \Bbb N}|a_n-b_n|$$

Prove that $d$ is a metric:

(1) $d(a,a)=\sup|a_n-a_n|=\sup|0|=0$ for all $a_n in l$

(2) Suppose $d(a,b)=\sup|a_n-b_n|$ Then, $\sup|a_n-b_n|=\sup|-(-a_n+b_n)|=\sup|-1||b_n-a_n|=\sup|b_n-a_n|=d(b,a)$ for all $a_n,b_n$ in $l$.

(3) Let $a,b,c$ be in $l$. Then


Is this correct?


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For completeness, might want to show that $\sup|a_n-b_n|$ exists. For (2), surely there is no need to show that $|a_n-b_n|=|b_n-a_n|$. Your proof is fine. In the $d(a,a)$ should be for all $(a_n)$ in $l$. – André Nicolas Oct 18 '12 at 18:36
By completeness, do you mean write that both a_n and b_n have a least upper bound, and hence their difference also has a least upper bound? – Alti Oct 18 '12 at 18:50
FYI, your space $l$ is usually denoted by $\ell_\infty$ ($\ell_\infty$ with LaTeX) – kahen Oct 18 '12 at 18:53
Yes. If you are going to be as detailed as you are, might as well show that the proposed metric is well-defined. to illustrate the point, suppose we "defined" the metric by $d(a,b)=\min|a_n-b_n|$. All that you wrote, with $\min$ replacing $\sup$, would be correct, but it would not be a metric. – André Nicolas Oct 18 '12 at 18:53
Good point, thank you. – Alti Oct 18 '12 at 18:57

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