0
$\begingroup$

Let $X$ be any set (could be uncountable). Let $f:X \to \Bbb R$ (real numbers) be any function. If the series $\sum f(x)$ is absolutely convergent for all $x \in X$, then the set $A=\{x\in X: f(x)\not=0\}$ is at most countable.

I am having 2 cases: a) if $X$ is countable. b) if $X$ is uncountable.

I am completely fine with a. But I am getting stock with b

What I did, I distinguish 2 cases; first/ If $f(x)=0$ for all $x\in X$ (since $f$ is any function), then $A=\emptyset$ which is countable.

Second/if there is $x$ such that $f(x)=0$, then $|A|\le|X|$. Now we have $X= A \cup \{x\}$ which makes $A$ uncountable (here I could not continue)

I would appreciate any help or hint. Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ It is unclear what $\displaystyle \sum_{x \in X} |f(x)|$ would even mean if we had $|f(x)| > 0$ for an uncountable quantity of $x \in X$. The way sums are defined, you need a bijection between $\mathbb{N}$ and the set to be summed. $\endgroup$
    – Kaj Hansen
    Oct 18 '17 at 7:15
  • 2
    $\begingroup$ @KajHansen There is a general notion of a sum $$\sum_{x \in X} f(x)$$ for arbitrary $X$ and $f : X \to \mathbb{R}$. Let $\mathcal{F} \subseteq \mathcal{P}(X)$ be the family of finite subsets of $X$ ordered by inclusion. Then $\mathcal{F}$ is a directed set and we define $$\sum_{x \in X} f(x) = \lim_{F \in \mathcal{F}} \sum_{x \in F} f(x).$$ See limits of nets. $\endgroup$
    – Adayah
    Oct 18 '17 at 7:18
  • $\begingroup$ Thanks for that information @Adayah ! I was unaware of this $\endgroup$
    – Kaj Hansen
    Oct 18 '17 at 7:24
4
$\begingroup$

No cases are necessary. Let

$$\sum_{x \in X} |f(x)| = L.$$

Then for each $n$ the set $A_n = \{ x \in X : |f(x)| \geqslant \frac{L}{n} \}$ is finite, because it has at most $n$ elements. Now just notice that

$$A = \bigcup_{n=1}^{\infty} A_n$$

is countable as a countable union of finite sets.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.