Hi I am having certain doubts about the sum of uncountable numbers

In my class on functional analysis we proved that if $\sum_{\alpha \in A} X_{\alpha}$ converges then $X_{\alpha}$ is not equal to zero at most a countably many times. I know that there is a similar question on the Math Stack exchange website but I am having serious troubles understanding the logic behind the proof in the class

The proof given by @Benji makes complete sense.

The sum of an uncountable number of positive numbers

We had a very similar proof in the class which was partly discussed in the comment section .

We used the definition that A sequence $\sum_{\alpha \in A} X_{\alpha}$ is convergent iff $\sup_{F \subset A} \sum_{\alpha \in F}X_{\alpha} < \infty$ where $F$ refers to a finite subset of $A$

The proof is as follows:

Proof: By Contradiction. We assume that $\sum_{\alpha \in A} X_{\alpha}$ converges and $\{ \alpha |X_{\alpha}>0 \}$ is uncountable

we defined sets $S_n=\{\alpha |X_\alpha >\frac{1}{n}\}$ but the we have $\bigcup_{n \in \mathbb{N}} S_n=\{\alpha |X_\alpha >0\}$ But by assumption the RHS is uncountable and since LHS is countable union of sets , then it implies that $\exists N \in \mathbb{N}$ such that $S_N$ is uncountable

(Please note in the following whenever I use $F$ , it is a finite subset of A)

Now the prof concludes that $\sup_{F \subset A} \sum_{\alpha \in F}X_{\alpha} = \infty$ And from the definition in bond it follows that $\sum _{\alpha \in A}X_{\alpha}$ does not converge and we get a contradiction

Now if I consider the set $S_N=\{\alpha_1,\alpha_2,\dots \dots\}$ defined as before i.e $X_{\alpha_j}>\frac{1}{N}$ $\forall \alpha_j \in S_N$ and consider a generic $F=\{\alpha_1,\alpha_2,\dots ,\alpha_k \}$ Then

$\sum_{\alpha \in F \subset S_N} X_{\alpha}>\frac{k}{N}$ for any k as cardinality of F needs to be finite

So therefore I have that $\sup_{F \in \mathcal{F}} \sum_{\alpha \in F} X_{\alpha} <\sup_{k \in \mathbb{N}} \frac{k}{N}=\infty$

where $\mathcal{F}$ is the set of all finite subsets of $A$

**I don't understand why does $S_N$ need to be uncountable for this thing thing to work. I mean wouldn't it work as long as it is infinite(countably)? ** Please let me know if this question is not clear.


The argument would work if $S_N$ was countably infinite, but the point is that this is the only way to get that $S_N$ is infinite. If we have assumed $\{\alpha : X^{\alpha} >0\}$ countable then each $S_n$ could easily be finite.

But your issue is correct: $S_N$ needs only be infinite, nor necessarily uncountable.


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