Do the bounded sequences in any metric space form a complete metric space?

Question

I know that

the set of all bounded sequences over $\mathbb R$ is complete w.r.t. sup norm.
Similarly the set of all bounded sequences over $\mathbb C$ is complete w.r.t. sup norm.

Does this result hold for the set of all bounded sequences over abitary metric space?

And a minor thing. Among the above two banach spaces which is said to be $l_\infty?$

Take a help from this link http://math.stackexchange.com/questions/296805/is-the-set-of-all-bounded-sequences-complete — Srijan, May 11 '13 at 04:05

TCL · Accepted Answer · 2013-05-11T14:51:00.423

3

Let $(M,d)$ be a complete metric space. Then the set of all bounded sequences $\ell_\infty(M)$ in $M$ form a complete metric space with the distance $D$ defined by $$D(s,t)=\sup_k d(s_k, t_k)$$ for any bounded sequences $s=(s_k), t=(t_k)$. It is straightforward to check that this is indeed a metric, and that it is complete (using the completeness of $M$.)

Since $M$ is embedded in the space $\ell_\infty(M)$ by $x\to (x,x,x,x,\cdots)$, the converse is also true, i.e. if $\ell_\infty(M)$ is complete, so is $M$.

Either one can be said to be $\ell_\infty$.

edited May 11 '13 at 14:51

answered May 11 '13 at 04:11

TCL

14,262

The converse is also true, right? (If the space of bounded sequences with the supremum norm is complete, then the metric space is complete.) – Pete L. Clark May 11 '13 at 04:20
Yes. Just consider constant sequences. – TCL May 11 '13 at 04:22
Perhaps you'd consider adding this to your answer, with a little more detail? – Pete L. Clark May 11 '13 at 04:34
Thanks for the suggestion. It is too late now. Maybe tomorrow. – TCL May 11 '13 at 05:08
what does constant sequence signifies to prove the converse @TCL – Myshkin May 17 '13 at 10:55

Do the bounded sequences in any metric space form a complete metric space?

1 Answers1