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Is this true or false: $$\bigcup_{1 \leq p < q} \ell^p = \ell^q?$$ I've tried to give an counter example, but I did not get any one.

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Probably not what you are looking for but one COULD use some functional analysis to answer the question: If $\ell^q = \bigcup_{p<q} \ell^p =\bigcup_{n\in\mathbb N} \ell^{q-1/n}$ Baire's theorem implies that some $\ell^{q-1/n}$ would be of second category in $\ell^q$ and then the open mapping theorem for the inclusion $\ell^{q-1/n} \hookrightarrow \ell^q$ would imply the surjectivity of this inclusion. So, all you need to know is that $\ell^p \neq \ell^q$ for $p<q$.

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    $\begingroup$ Killing a mosquito with a thermonuclear device. Nice. $\endgroup$ Aug 31, 2013 at 19:57
  • $\begingroup$ I could be wrong, but don't those sets you are taking the union of have to be closed with respect to $l^q$ norm to apply Baire? $\endgroup$
    – Evan
    Aug 31, 2013 at 21:13
  • $\begingroup$ Ok davide addresses how to correct for this in the other thread. $\endgroup$
    – Evan
    Aug 31, 2013 at 21:53
  • $\begingroup$ @Evan There's nothing "to correct for". We don't need to suppose anything closed. What we need is 1) $\ell^q$ is complete, 2) we can write the union as a countable union. Since $\ell^q$ is a complete metric space, it is of the second category in itself, and a countable union of sets of the first category is of the first category. So if the union were the entire space, at least one of the $\ell^p,\,p<q$ would be of the second category. But the open mapping theorem asserts that a continuous linear map whose range is of the second category is onto if the domain is an $F$-space. $\endgroup$ Sep 1, 2013 at 0:28
  • $\begingroup$ @Evan Simple only if you know the theorems. The used form of the open mapping theorem isn't everyday knowledge. $\endgroup$ Sep 1, 2013 at 1:53
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Consider $$x_n = \frac{1}{(n+2)^{1/q}(\ln (n+2))^{1/q} \ln (\ln (n+2))}, \quad n =1,2, \ldots$$ Then $\{x_n\}_1^{\infty} \in \ell^q$, but $\{x_n\}_1^{\infty} \notin \ell^p$ for $p<q$.

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For $q=\infty$, anything you can think of likely works as a counterexample.

Otherwise, use the fact that $\sum \frac{1}{n (\log n)^a} < \infty$ for $a>1$ to build counterexamples.

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