4
$\begingroup$

which of the following metric spaces are separable?

  1. $C[0,1]$ with usual 'sup norm' metric.

  2. the space $l_1$ of all absolutely convergent real sequences, with the metric $$d_1(a_i,b_i)=\sum_{1}^{\infty}|a_i-b_i|$$

  3. The space $l_{\infty}$ of all bounded real sequences with the metric $$d_{\infty}(a_i,b_i)=\sup|a_i-b_i|$$

Well, 1 is separable as polynomials are dense in $C[0,1]$ so I can construct a set of polynomial with rational coefficients that is going to be a countable dense set for $C[0,1]$

I have no idea about 2,3 .

Well, along with this question I just want to ask The closed unit ball is compact with respect to $l_1$ metric? I guess no, because Sequence $e_1=(1,0,\dots),\dots e_n=(0,0,\dots,1(nth place),0,0\dots)$ this seqquence has no convergent subsequence so not sequentially compact. Am I right?

$\endgroup$

2 Answers 2

9
$\begingroup$

Another possibility for (3): You have $F=\{0,1\}^{\mathbb{N}} \subset \ell_{\infty}$ and moreover for all $x,y \in F$, if $x \neq y$ then $d(x,y) =1$. So $\{B(x,1),x \in F\}$ is a family of disjoint open sets whereas $F$ is uncountable.

$\endgroup$
3
  • $\begingroup$ thank you for nice answer. $\endgroup$
    – Myshkin
    Commented Jul 12, 2012 at 20:41
  • $\begingroup$ Very nice.${}{}$ $\endgroup$ Commented Jul 12, 2012 at 20:43
  • $\begingroup$ @Seirios Lovely! +1 $\endgroup$
    – WhySee
    Commented Jan 10, 2018 at 13:04
5
$\begingroup$

HINTS:

For (2), notice that if $x\in\ell_1$, then the terms of $x$ converge to $0$. Consider sequences that have rational terms that are eventually $0$.

For (3), let $D=\{x_n:n\in\Bbb N\}$ be a countable subset of $\ell_\infty$. For each $n\in\Bbb N$ choose $y(n)\in[0,1]$ so that $|y(n)-x_n(n)|\ge\frac12$. Let $y=\langle y(n):n\in\Bbb N\rangle$; is $y$ in the closure of $D$?

$\endgroup$
4
  • 1
    $\begingroup$ could you tell me why if $x=(x_n)\in l_1\Rightarrow x_n\to 0$ $\endgroup$
    – Myshkin
    Commented Jun 1, 2013 at 6:39
  • $\begingroup$ @TaxiDriver: If $x\in\ell_1$, then $\|x\|_1=\sum_n|x_n|<\infty$. $\endgroup$ Commented Jun 1, 2013 at 15:27
  • 1
    $\begingroup$ well if a series is convergent then n'th term must converges to $0$ so $x_n\to 0$, thank you! $\endgroup$
    – Myshkin
    Commented Jun 1, 2013 at 15:33
  • $\begingroup$ @TaxiDriver: Exactly. $\endgroup$ Commented Jun 1, 2013 at 15:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .