## 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 ${M/\epsilon}$. ${\Box}$