I’m extremeley confused by this, can someone explain or point to a proof of this? How do we get from an uncountable original set to a countable set of non-zero results?
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Like I get that if sum of f(x) is absolutely convergent, it means it’s going down fast enough, but it doesn’t mean any of those values are literally zero, does it? So why would we expect that filtering an uncountable set by zero would somehow give us a countable set?
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I think I’m probably reading this completely wrong somewhere but I don’t see where
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the result is correct. the proof is very cute: consider the subset X_n of X of all values at which |f(x)| > 1/n. if the sum is absolutely convergent then each of these subsets must be *finite*, because the absolute sum must be at least |X_n| / n
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the union of the X_n is the set of all values at which |f(x)| is positive, and since it's a countable union of finite sets it's countable. so we conclude that |f(x)| = 0 for all but countably many values of x
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Thanks!
>because the absolute sum must be at least |X_n| / n
Where does this come from?
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ah I get it, it’s cause we set them all to be > 1/n so the sum of them all is at least their count * 1/n
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yeah i elided that a bit but it's the key step. we are basically using the fact that the real numbers are archimedean:
