0
$\begingroup$

(I'm sorry if someone has already asked a similar question, I couldn't find anything from my search).

The question is here.

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'm not exactly sure what the question is asking...I know a metric space is a space in which distances between all members are well-defined, but I'm not sure what it would mean for $d_1$ to be a metric in $l^1$...

$\endgroup$
  • $\begingroup$ Please type your problem in your post. $\endgroup$ – Tim Thayer Feb 12 '17 at 19:50
  • $\begingroup$ @TimThayer Done. Sorry, still getting familiar with the code. $\endgroup$ – mizichael Feb 12 '17 at 20:02
  • $\begingroup$ Your characterization of a "metric space" is incomplete. Do you know the definition of a "metric" on a set? $\endgroup$ – Lee Mosher Feb 13 '17 at 21:27
  • $\begingroup$ @LeeMosher do you mean: 1) $d(x,y) > 0$ 2) $d(x,y) = 0$ implies $x=y$, etc.? $\endgroup$ – mizichael Feb 14 '17 at 21:59
  • $\begingroup$ Yes, particularly since those things (and the etc.) are what you have to prove. $\endgroup$ – Lee Mosher Feb 14 '17 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.