Is $C_0$ dense in $l^{p}$ with $1\leq p\leq \infty$ where $C_0=\{ (x_n): x_n\rightarrow 0, x_n\in R\}$. Well I think that if $p<\infty$ is true because by definition if i take $y=(y_n)\in l^p$ then $\sum (y_n)^p <\infty$ so $(y_n)^p \rightarrow 0$ imply $y_n \rightarrow 0$ then I can choose the same $y_n \in C_0\cap l^\infty$ such that $y_n \rightarrow y_n$ and this is the definition of density, for all $z$ in the big set exist one succession $z_n$ in the small set such that $z_n\rightarrow z$. But I don't know how to do with $p=\infty$. Please somebody can you help me? Thank you
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Martin Sleziak
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weymar andres
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2How can it be? Take the sequence $e=(1,1,...)$. For any $x \in C_0$ you must have $|e-c|_\infty = 1$. – copper.hat Jun 23 '20 at 01:14
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So is no dense for $p=\infty$ right? And my attept of proof for the other $p$ is good?thank you – weymar andres Jun 23 '20 at 01:16
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1Yes, $l_p \subset C_0$ for all $p \in [1,\infty)$. – copper.hat Jun 23 '20 at 01:29
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$C_0$ is not dense in $l^{\infty}$. Actually it is a closed subset in $l^{\infty}$ and it does not have the constant sequence $1$ which is in $l^{\infty}$.
To show that it is closed, suppose that $((a_n)^{(k)})$ is a sequence of $C_0$ points congerges to $(x_n)$ in $l^{\infty}$. Then for every positive $\epsilon$, we can choose $k$ large enough so that $|(a_n)^{(k)}-x_n|<\epsilon$ for all $n$. Since $(a_n)^{(k)} \in C_0$, we can choose $N$ large enough so that $|(a_n)^{(k)}|<\epsilon$ whenever $n>N$. But then for each $n >N$ we have $|x_n|<2\epsilon$. Since $\epsilon$ were arbitrary, we have $x_n \in C_0$ which shows the desired closedness.
Thanks for a good question!
seoneo
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Thank you so much, and im.confuse in my argument, is good my attempt please? – weymar andres Jun 23 '20 at 01:25
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Thank you, and why $(1,\ldots,1,\ldots)$ is not in $l^{\infty}$ , well i m not sure because the definition is that every element in the component is bounded? That is? Thank you – weymar andres Jun 23 '20 at 01:33
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@weymarandres $(1)_n$ is in $l^\infty$ but not in $C_0$ as you can see. – seoneo Jun 23 '20 at 01:35
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Some proofs for the closedness of $c_0$ in $\ell_\infty$ can also be found here: Prove the set of sequences $c_0$ which converge to zero in $l_{\infty}$ is closed. and in other posts linked there. – Martin Sleziak Jun 23 '20 at 03:34