Dragonfall
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Does there exist a converging uncountable sum of strictly positive reals?
The discussion revolves around the existence of a converging uncountable sum of strictly positive real numbers. Participants explore definitions and implications of uncountable sums, questioning whether such sums can converge and how they might be defined in contrast to countable sums and integrals.
Participants do not reach a consensus on the existence of converging uncountable sums of positive reals. Multiple competing views remain regarding definitions and implications of such sums.
There are limitations in the definitions provided, particularly regarding the assumptions underlying uncountable sums and the conditions for convergence. The discussion also highlights the potential confusion surrounding ordinal notation and its implications for countability.
First you will have to define what you mean by "uncountable sum"! I know a definition for finite sums and I know a definition for countable sums (the limit of the partial, finite, sums), but I do not know any definition for an uncountable sum except, possibly the integral that bpet suggested.Dragonfall said:Does there exist a converging uncountable sum of strictly positive reals?
g_edgar said:Definition Let [itex]S[/itex] be an index set. Let [itex]a \colon S \to \mathbb{R}[/itex] be a real function on [itex]S[/itex]. Let [itex]V[/itex] be a real number. Then we say [itex]V = \sum_{s\in S} a(s)[/itex] iff for every [itex]\epsilon > 0[/itex] there is a finite set [itex]A_\epsilon \subseteq S[/itex] such that for all finite sets [itex]A[/itex] , if [itex]A_\epsilon \subseteq A \subseteq S[/itex] we have [itex]\left|V - \sum_{s \in A} a(s)\right| < \epsilon[/itex] .
Dragonfall said:So does there exist sequences [tex]x_i[/tex] indexed by ordinals [tex]D\geq\epsilon_0[/tex] such that [tex]\sum_{i\in D}x_i[/tex] is finite, and that each x_i is positive?
Dragonfall said:Yes I was mistaken on the notation, it should be [tex]\omega_1[/tex].
You have yet to say why. You asserting it true doesn't constitute a proof.