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For an arbitrary set $I$, not necessarily countable, is the space $\ell^1(I)$ consist of functions on $I$ whose support is always countable? Here "support" doesn't refer to the closure, but only those elements in $i\in I$ such that $f(i)\neq 0$, where $f\in \ell^1(I)$.

This is because, for an uncountable set $I$, the norm $$\|f\|_1:=\sup_{F\in \mathcal{F}}\left\{\sum_{j\in F} |f(i_j)|\right\}< +\infty,$$ where $\mathcal{F}$ is the set/family of all finite subsets of $I$, forces fucntion $f$ only takes nonzero values on at most countably many points $i_j\in I$.

Is that correct? I know $\ell^1(I)$ is equivalent to $L^1(I)$ through the counting measure.

user760
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    Yes, if $f:I\to \mathbb R$ has uncountable support, then for some $n$ there are uncountably many values $i\in I$ such that $|f(i)>\frac1n.$ Then show that $f\notin \ell^1(I).$ – Thomas Andrews Oct 12 '22 at 21:37
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    Basically, the set of values ${i\in I\mid |f(i)|>1/n}$ must be finite for each $n.$ or $f\notin\ell^1.$ And the support is then the (countable) union of these finite sets. – Thomas Andrews Oct 12 '22 at 21:40
  • @AnneBauval Thanks for the link. Yes it's connected. I just try to confirm functions in $\ell^1(I)$ are like that. – user760 Oct 12 '22 at 21:45
  • It was much more than connected. What was still missing? – Anne Bauval Oct 13 '22 at 08:05

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Yes, if $f\in \ell^1(I)$, then $f$ has countable support. Indeed, note that

$$\{i \in I: f(i)\ne 0\}= \bigcup_{n=1}^\infty \{i\in I: |f(i)|> n^{-1}\}.$$ The set $\{i \in I: |f(i)| > n^{-1}\}$ is finite for every $n$, otherwise you could pick an infinite subset $I_0$ with the property that $|f(i)| > n^{-1}$ for all $i\in I_0$ and then $$\|f\|_1=\sum_{i \in I}|f(i)|\ge \sum_{i \in I_0}|f(i)| \ge \sum_{i \in I_0} n^{-1}=\infty.$$ Hence, the support of $f$ is a countable union of finite sets, and thus a countable set.

J. De Ro
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