I have to prove that the space

$\ell^p(J)$ defined as the set of all functions

$\psi: J\rightarrow \mathbb{F}$

s.t. $\psi$ is null except in a contable subset of $J$


$||\psi||_p :=\bigg(\sum_{t\in J}|\psi(t)|^p\bigg)^{1/p}<\infty$

is Banach (complete).

Well, given a Cauchy sequence $(\psi_n)_n\subset \ell^p(J)$ I defined $\psi$ such that each $\psi(t)$ is defined as $\lim \psi_n(t)$ (each of these sequences $\psi_n(t)$ is Cauchy in $\mathbb{F}$).

I proved that $\psi_n\rightarrow \psi$ and $||\psi||_p<\infty$. However, I could not prove that $\psi$ is null except in a contable subset of $J$.

Many thanks for any help.


Hint: if $\psi(t) \ne 0$, then some $\psi_n(t) \ne 0$.

  • $\begingroup$ Yes, for $n$ bigg... But combining various $n$'s biggs, I could not prove yet the result. Many thanks for the help. $\endgroup$ – Na'omi Apr 23 at 16:38
  • $\begingroup$ Maybe because intersection/union of contable sets are contable? $\endgroup$ – Na'omi Apr 23 at 16:41
  • $\begingroup$ A secondary doubt: a non-contable sum can be finite? $\endgroup$ – Na'omi Apr 23 at 16:42
  • 1
    $\begingroup$ The union of countably many countable sets is countable. $\endgroup$ – Robert Israel Apr 23 at 17:50
  • $\begingroup$ $ \psi (t) \neq 0 \implies \psi_n (t) \neq 0 $ to $ n $ big enough. Thus, where $ J_n $ is the subset of $ J $ with zero measure out of which $ \psi_n$ cancels, the points on which $ \psi $ does not vanish are contained in the enumerable union $ \cup J_n $, also measure null... Is it correct? Many thanks. $\endgroup$ – Na'omi Apr 23 at 18:10

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