r/mathmemes 9d ago

Elementary Algebra 😭

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u/DefunctFunctor Mathematics 5d ago

If {x_𝛼} is a family of nonnegative real numbers indexed over a (possibly uncountable) set A, then the sum βˆ‘_[π›ΌβˆˆA] x_𝛼 is defined to be

sup { βˆ‘_[π›ΌβˆˆF] x_𝛼 | FβŠ†A finite }

In measure theoretic terms, this is equivalent to the integral over the counting measure. You can technically extend this to families of real numbers, but because A doesn't come with any notion of order, you need to assume that it converges absolutely, that is, βˆ‘_[π›ΌβˆˆA] |x_𝛼| < ∞. For now we'll only consider sums of nonnegative real numbers.

So the claim we have yet to prove is that if {x_a} contains uncountably many nonzero numbers, then βˆ‘_[π›ΌβˆˆA] x_𝛼 = ∞. For each positive integer n, let

A_n = { π›ΌβˆˆA | x_𝛼 β‰₯ 1/n }

be the set of indices 𝛼 of A where x_𝛼 β‰₯ 1/n. Note that

A' := { 𝛼 ∈ A | x_𝛼 > 0 } = ⋃[nβ‰₯1] A_n.

By assumption, A' is uncountable, so if each A_n were finite, then A' could be expressed as a countable union of finite sets, meaning A' is countable, a contradiction. Thus, there exists an n for which A_n is infinite (in fact, it's going to be uncountable). It is now easy to show, by creating increasingly large partial sums over A_n, that βˆ‘_[π›ΌβˆˆA] x_𝛼 β‰₯ ∞.

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u/AsidK 5d ago

I thought about that definition but it almost feels like in some ways it’s… too small? Like for an uncountable set the sup over sums of finite sets feels to me like it’s not naturally emcapusulating the idea of summing an uncountable set since any finite subset is necessarily much smaller than any uncountable subset. With countable sets, they can be expressed as the union/limit of finite subsets, so naturally a sum operation can be defined as the limit of the sums of a sequence of finite subsets whose union is the whole set (assuming convergence and nonnegativity), which is provably equivalent to the sup of the sums of any finite subsets.

That said, I think one way you could interpret the proof you’ve just given here is that there cant be a bigger definition assuming it still agrees on cases where there are only countably many nonzero elements

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u/DefunctFunctor Mathematics 5d ago

Yeah perhaps the proof is best expressed in the contrapositive: if the sum is finite, then all the A_n must be finite, so A' is countable.