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I want to prove that $\ell^p$ is separable. I was given the following hint: consider a rational sequence in $\ell^p$ with all terms $0$ except for finitely many.

So consider the set of sequences defined as

$$S=\{(x_k)_k\in \mathbb{Q}:x_1 = q_n, x_k=0\mbox{ for $k\ge 1$}\}$$

Then $S$ is countable. But is it dense? We need to show that for every sequence $(y_k)_k$ in $\ell^p$ and every $\epsilon>0$ there exists a sequence $(x_k)_k$ in $S$ such that

$$\sum_k |y_k-x_k|^p<\epsilon$$

Is this correct?

But if this is what we need, then, obviously, there exists an $\epsilon >0$ and a sequence in $X$ such that $\sum_k |y_k-x_k|^p<\epsilon$, for any sequence in $S$.

So what is it that I do not understand? Would appreciate a clarification.

Edit: Or do we just need one point in $(x_k)_k$ be in the $\epsilon$-range of at least one point of $(y_k)_k$? I.e. to satisfy

$$|x_k-y_l|^p<\epsilon$$

Ali
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sequence
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1 Answers1

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In $\ell_p$: $$d(a_n,b_n)=\sqrt[p]{\sum_{n \in \Bbb N} |a_n-b_n|^p}$$

  • So, to prove that there is a dense subset in $\ell_p$, do I need to prove that $$\left(\sum_k |y_k-x_k|^p\right)^{1/p}<\epsilon?$$ – sequence Oct 06 '17 at 02:49
  • @sequence You need to prove for all $\varepsilon>0$ that there is some blah blah blah such that that is true, yes –  Oct 06 '17 at 02:50
  • Then I guess $S$ is not dense in $X$ and the hint was wrong? @ZacharySelk – sequence Oct 06 '17 at 02:51
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    @sequence No, the hint should work. See for example https://math.stackexchange.com/questions/487319/prove-that-sequence-space-ell-p-mathbb-r-is-separable –  Oct 06 '17 at 02:53