IMT Exercise 0.0.1

If {(x_{\alpha})_{\alpha\in A}} is a collection of numbers {x_{\alpha}\in [0,+\infty]} such that {\sum_{\alpha\in A} x_{\alpha} <\infty}, show that {x_{\alpha}=0} for all but at most countably many {\alpha\in A}, even if {A} itself is uncountable.

{\sum_{\alpha\in A} x_{\alpha} <\infty} and so none of the {x_{\alpha}} is equal to {\infty}, i.e. {x_{\alpha}\in [0,+\infty)}.

Let the sum {\sum_{\alpha\in A} x_{\alpha}=M}.

Define {A_0=\{\alpha:\alpha\in A, x_{\alpha}=0\}}. However many elements there are indexed by {A_0} they make no contribution to the sum.

Let {B=A\setminus A_0}, then {\sum_{\alpha\in B} x_{\alpha}=M}.

Let {\epsilon} be the smallest number in {B}, then the set {B} is finite and {\left\vert{B}\right\vert} is at most {M/\epsilon}. {\Box}

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