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Is $\bigcup_{p<\infty}\ell_p=c_0$?

At least one inclusion obvious: every p-summable sequence converges to zero.

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4 Answers 4

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Consider $$ x(n)=\frac{1}{\log(n+1)} $$ It is easy to check that $x\in c_0\setminus \left(\bigcup_{1\leq p<\infty}\ell_p\right)$. Indeed $$ \lim\limits_{n\to \infty} x(n) = \lim\limits_{n\to \infty}\frac{1}{\log (n+1)}=0 $$ so $x\in c_0$. Now for fixed $p\in [1,+\infty)$ there exist $N\in \mathbb{N}$ such that for all $n>N$ we have $\log^p (n+1)<n$, so $$ \Vert x\Vert_p= \left(\sum\limits_{n=1}^\infty \frac{1}{\log^p (n+1)}\right)^{1/p}= \left(\sum\limits_{n=1}^N \frac{1}{\log^p (n+1)}+ \sum\limits_{n=N+1}^\infty \frac{1}{\log^p (n+1)}\right)^{1/p}> $$ $$ \left(\sum\limits_{n=1}^N \frac{1}{\log^p (n+1)}+ \sum\limits_{n=N+1}^\infty \frac{1}{n}\right)^{1/p}=+\infty $$ the last equality holds since the series $$ \sum\limits_{n=N+1}^\infty \frac{1}{n} $$ diverges.

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Why is it the case that for a fixed $p \in [1, +\infty)$ there exists $N \in \mathbb{N}$ such that for all $n > N$ we have $\log^p (n+1) < n$? –  Harry Williams Nov 10 '12 at 4:48
    
Because $$\lim\limits_{n\to\infty}\frac{\log^p(n+1)}{n}=0$$ –  Norbert Nov 10 '12 at 8:19
    
Could you give me a clue as to how you'd prove this? –  Harry Williams Nov 10 '12 at 14:22
    
Using l'Hopitale rule prove that $$\lim\limits_{n\to\infty}\frac{\log(n+1)}{n^{1/p}}=0$$ –  Norbert Nov 10 '12 at 14:34
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Prove: If $V$ is a separable Banach space, then $V$ cannot be written as a countable union of proper linear subspaces.

Because of inclusions among the $\ell_p$ spaces, your union is, in fact, equal to a countable union.

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$\frac{1}{\ln n}$ converges to 0 but $\sum \frac{1}{\ln^p n} = \infty$ for any $p \geq 1$.

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$$ \lim_{n\to\infty}\frac{1}{\log(n+1)}=0\text{, but }\sum_{n=1}^{\infty}\frac{1}{(\log(n+1))^p}=\infty\quad\forall p>0. $$

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