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Let $I,J$ be finite subsets of $\Bbb N$ and $(a_i):\Bbb N\to \Bbb R$

I define $\sum_{i\in I} a_i:=\sum_{k=1}^N a\_i_k$ where $N=|I|$ and $(i_k):\{1,....N\}\to I$ is strictly increasing and onto $I$.

Are the following propositions right? and how can I prove them:

  1. "$\sum_{i\in I\setminus J} a_i=\sum_{i\in I} a_i-\sum_{i\in I\cap J} a_i$"

  2. "$\sum_{i\in I\cup J} a_i=\sum_{i\in I} a_i+\sum_{i\in J\setminus I} a_i$"

  3. "$\sum_{i\in I} a_i=\sum_{i\in I\cap J} a_i+\sum_{i\in I\setminus J} a_i$"

J.-E. Pin
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  • The claim is definitely false. Say, $A={1}$ and $B={2}$. Your claim says that $a_1-a_2=a_1$. – Batominovski Apr 06 '20 at 07:14
  • thank you for your counter example. now i modified the proposition –  Apr 06 '20 at 07:41
  • @MartinR It isn't strictly a question about abstract algebra, but (1) it can trivially be turned into one, and (2) any valid answer will apply equally well to any commutative monoid in place of $\mathbb{R}.$ Therefore,I think the [abstract-algebra] tag was appropriate; but I won't presume to restore it. (You could say that inside this question there's an abstract algebra question struggling to get out.) – Calum Gilhooley Apr 06 '20 at 08:16

1 Answers1

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(1) and (3) are clearly equivalent. But actually the three assertions follow from this property:

if $E$ and $F$ are disjoint subsets of some set $U$, then $\sum_{i \in E \cup F} a_i = \sum_{i\in E} a_i + \sum_{i \in F} a_i$.

J.-E. Pin
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