The sum of an uncountable number of positive numbers

Claim : If $(x_\alpha)_{\alpha\in A}$ is a collection of actual numbers $x_\alpha\in [0,\infty]$ such that $\sum_{\alpha\in A}x_\alpha<\infty$ , after that $x_\alpha=0$ for almost at the majority of countably several $\alpha\in A$ ( $A$ need not be countable).

Evidence : Let $\sum_{\alpha\in A}x_\alpha=M<\infty$ . Take into consideration $S_n=\{\alpha\in A \mid x_\alpha>1/n\}$ .

After that $M\geq\sum_{\alpha\in S_n}x_\alpha>\sum_{\alpha\in S_n}1/n=\frac{N}{n}$ , where $N\in\mathbb{N}\cup\{\infty\}$ is the variety of components in $S_n$ .

Hence $S_n$ contends the majority of $Mn$ components.

Therefore $\{\alpha\in A \mid x_\alpha>0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of limited collections. $\square$

First, is my evidence deal with? Second, exist extra concise/elegant evidence?

174
2022-06-04 17:29:49
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Answers: 2

There is also one question directly relating to this question, that is, how to define the sum of uncountably many numbers (not necessarily positive numbers). The difficulty lies in the fact that there could not be any order of this summation, since there are uncountably many of terms. So, when we talk about the sum of $x_\alpha$, namely, $\sum_{\alpha\in A}x_\alpha$, we are actually saying the following,

For every countable subset of $I\subset A$ with arbitrarily given order, the sequence $(x_\alpha)_{\alpha\in I}$ should be convergent. In other words, the sequence $(x_\alpha)_{\alpha\in I}$ should be absolutely convergent.

A proper definition is given in Paul Halmos' book, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, as follows:

$x=\sum_{\alpha\in A}x_\alpha$ means that for any positive number $\varepsilon$ there is some finite set $I_0$ such that for any finite set (or more generally, countable set) $I\supset I_0$, we have $|x-\sum_{\alpha\in I}x_\alpha|<\varepsilon$.

Note that, the set $\{1,-\frac{1}{2},\frac{1}{3},\cdots,(-1)^{k-1}\frac{1}{k},\cdots, 0,\cdots\}$, where in the end there are uncountably many $0$'s is not convergent any more. But the sequence $\{1,-\frac{1}{2},\cdots\}$ is convergent in the standard sense.

Now we invoke Zorn's lemma, on all countable subsets $I$, with respect to which, the sequence $x_\alpha$ is absolutely convergent, with the inclusion as the order. Note that for any $I_1\subset I_2\subset I_3\subset \cdots$ a chain of countable subsets of $A$, the set $I^*=\bigcup_iI_i$ is also a countable subset of $A$ and by the definition, the sequence with index in $I^*$ is also absolutely convergent. By Zorn's lemma, there should exists a maximal countable subset $I_{max}$. This means that any number $x_\alpha$ with $\alpha\notin I_{max}$ should be $0$, otherwise we can construct another strictly larger countable subset on which the number sequence is absolutely convergent.

24
2022-07-02 20:39:33
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Just so the inquiry obtains a solution : of course , your evidence is proper and also is just one of numerous wordings of the fastest evidence that I recognize.

41
2022-06-04 18:07:49
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