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Let $l^1$ denote the space of all absolutely summable sequences, i.e., the space of all sequences ${a_n}$ such that $\sum_{n=1}^{\infty} |a_n|$ converges. Define,

$d_1({a_n},{b_n})$ = $\sum_{n=1}^{\infty} |a_n - b_n|$ for all ${a_n},{b_n}\in l^1$

Show that:
(1) $d_1$ is a metric in $l^1$.
(2) Let 0 be the zero sequence (0,0,...). Describe the close unit ball $B$ with center 0 in ($l^1, d_1$).
(3) Show that $B$ is not sequentially compact.

I asked this question a few days ago and figured out part (1), but I'm now stuck on (3) and, to some degree, (2).
I think the close unit ball $B$ is the set of all $a_n \in l^1$ such that $\sum_{n=1}^{\infty} |a_n| \leq 1$.
However, based on this definition, I'm having trouble finding an infinite sequence which doesn't have a convergent subsequence. Any help is very appreciated.

1 Answers 1

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$(1, 0, 0, 0, 0, \ldots), (0, 1, 0, 0, 0, \ldots), (0, 0, 1, 0, 0, \ldots), \ldots$

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    Beat me to it. Also, the OP's description of the closed unit ball is correct – 2017-02-15
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    @Callus Surely there exists a convergent subsequence for all of these examples - just take a subsequence which avoids the 1? – 2017-02-15
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    These are not examples, this is ONE example. Note that the elements of the space are sequences of numbers, but for checking sequential compactness, you need a sequence of elements of the space, that is, a sequence of sequences. – 2017-02-15
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    @Callus AH I think I understand...so you could choose a sequence of sequences $a_n$ such that the nth term is a 1? I think I'm understanding where this is going but I'm a bit confused definition-wise. – 2017-02-15
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    @mizichael Exactly. This is a sequence of points in the closed unit ball, each of which is at a distance of $2$ from every other point in the sequence, and therefore which has no convergent subsequence. – 2017-02-15
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    You are correct. – 2017-02-15
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    @Callus thank you both very much! I have a much deeper understanding now. – 2017-02-15