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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?

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up vote 4 down vote accepted


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 $D$?

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could you tell me why if $x=(x_n)\in l_1\Rightarrow x_n\to 0$ – La Belle Noiseuse Jun 1 '13 at 6:39
@TaxiDriver: If $x\in\ell_1$, then $\|x\|_1=\sum_n|x_n|<\infty$. – Brian M. Scott Jun 1 '13 at 15:27
well if a series is convergent then n'th term must converges to $0$ so $x_n\to 0$, thank you! – La Belle Noiseuse Jun 1 '13 at 15:33
@TaxiDriver: Exactly. – Brian M. Scott Jun 1 '13 at 15:33

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.

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thank you for nice answer. – La Belle Noiseuse Jul 12 '12 at 20:41
Very nice.${}{}$ – Brian M. Scott Jul 12 '12 at 20:43

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