Good day,
I have the following question
"Prove that for $1 \leq p < \infty$, $l_p(I,K)$ is separable if and only if $I$ is countable, and $l_{\infty}(I,K)$ is separable if and only if $I$ is finite."
From the lecture I know:
- A topological space is called separable iff it contains a countable, dense subset.
- $l_\infty(I,K)=\{f:I \to K: ~ ||f||_\infty = \sup_{x \in X} |f(x)| < \infty\}$, $l_p(I,K)=\{f:I \to K: ~ ||f||_p < \infty\}$ where $K$ is a field
- $||~ ||_p$ is a norm on $l_p(I,K)$ for $p \in [1,\infty]$
- $l_p(I,K)$ is Banach space for $p \in [1,\infty]$
- Lemma: A normed space $(X,||~||)$ is separable iff there is a countable set S such that $X=\overline{span(S)}$.
- $span(S):=\{ \sum_{i=1}^n a_i x_i : ~ n \in \mathbb{N}, x_i \in S, a_i \in \mathbb{R} \}$
I think have to use the Lemma to prove the first equivalence. But I have no approach how to prove this. Can someone please help me?
Thanks a lot, Marvin
For finite $p$, suppose let $D$ be a countable dense set (containing $0$) of $K$ (which, as Daniel Fischer remarked in his comment) must exist, as $l_p(\{p\})$ for a singleton set $I$ is isomorphic to $K$.
Then all sequences from $D$ that eventually $0$ can be shown to be dense in $l_p(I,K)$
If $I$ is not countable, you can easily find $|I|$ many vectors in $l_p(I,K)$ that are $1$ apart in distance. This kills separability (why?).
For $l_\infty(I,K)$ we can always find $2^{|I|}$ many vectors that are 1 apart, and this is uncountable for $I$ non-finite.
The finite case should be easy to do for $l_\infty(I,K)$, I shoudl think. The obvious candidate (use $D$ again!) works.