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Suppose that $T_1,T_2\subset \mathbb{N}$ and $a_i\geq 0$. I need to prove that $$\sum \limits_{i\in T_1\cap T_2}a_i+\sum \limits_{i\in T_1\cup T_2}a_i=\sum \limits_{i\in T_1}a_i+\sum \limits_{i\in T_2}a_i.$$

I do not know how to prove this fact at all because I even don't know what is the definition of $\sum \limits_{i\in K}a_i$ for $K\subset \mathbb{N}$.

I will highly appreciate if someone will explain my questions, please!

EDIT: Also I assume that $\sum \limits_{I\in T_1}a_i<\infty,\sum \limits_{I\in T_2}a_i<\infty$ and as I said above $a_i\geq 0$.

peter.petrov
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RFZ
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  • For summation over non-negative numbers indexed by any infinite set $K$ (even uncountable ones),it is defined as a supremum of sum over all finite subsets. $$\sum_{i \in K} a_i \stackrel{def}{=} \sup\left{ \sum_{j \in J} a_j : J \subset K, |J| < \infty \right}$$ As long as you can verify the identity for finite sums, you can use the properties of supremum to derive corresponding identity for infinite sums. – achille hui May 18 '20 at 05:26
  • It occurs to me, rather belatedly (probably because I was musing on this very topic a few weeks ago, and become bogged down in elaborate commutative diagrams, so I decided to stop thinking about it for a while!), that whoever set the problem may have been taking the meaning of $\sum_{i \in K}a_i$ to be self-evident when $K$ is finite, and in the infinite case, may have been appealing to a basically trivial generalisation of the meaning of $\sum_{i=0}^\infty a_i$ or $\sum_{i=1}^\infty a_i,$ which is seldom written down anywhere, not even in Terence Tao, Analysis I, where I'd hoped to find it. – Calum Gilhooley May 18 '20 at 15:14
  • @CalumGilhooley, thanks a lot for your comments! After some thoughts I begin to realize that the definition which you have used in the previous topic where you defined $\sum \limits_{n\in K}a$ to be $\sum \limits_{n=1}^{\infty}a_n[n\in K]$ is not the best one. I do not think that this is correct definition. In order to solve this problem firstly we have to define this summation in correct way. – RFZ May 18 '20 at 15:27
  • I was planning to write an answer listing four (?) different definitions of $\sum_{i \in K}a_i,$ and proving them all equivalent, but I became diverted into writing a long rant about there being no accepted definition, so I'll have to take a break to calm down! I'm not sure if there are now four or five different possible definitions to prove equivalent. P.S. In my answer to your previous question, I carefully didn't define $\sum_{i \in K}a_i$ in that way, but merely stated that whatever the definition might be, it could be supposed to satisfy that identity. – Calum Gilhooley May 18 '20 at 15:32
  • @CalumGilhooley, so it turns out that the problem of the previous topic and this one are still open? It is quite sad. I really want to see the proofs and definitions :( – RFZ May 18 '20 at 15:34
  • Please don't lose heart! We've made definite progress. The question (i.e. the broader question of the definition of expressions like $\sum_{i \in K}a_i$ when $K$ is a possibly infinite subset of $\mathbb{N}$) is very much open, but that's not your fault, or mine. – Calum Gilhooley May 18 '20 at 15:38
  • @CalumGilhooley, By the way if we define $\sum \limits_{I\in K}a_i$ as achille hui did above could you show why it satisfies identity $\sum \limits_{n=1}^{\infty}a_n[n\in K]$. To be honest, I am so lost – RFZ May 18 '20 at 15:39
  • Yes, @achillehui's definition (which is an accepted one in a more general and abstract context) is one of those I was going to include in my list. On the other problem (which is more a problem in social psychology than in mathematics): you have been put in a double bind by wheoever set the question, so I repeat, please don' t think it is your fault. – Calum Gilhooley May 18 '20 at 15:41
  • Let us continue this discussion in chat. – RFZ May 18 '20 at 15:42

2 Answers2

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One definition for summations of non negative numbers over an arbitrary index set is $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$.

Note that it follows immediately that if $I \subset I'$ then $\sum_{n \in I} a_n \le \sum_{n \in I'} a_n$.

Now suppose $A \subset I$, then I claim that $\sum_{n \in I} a_n = \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Suppose $J_1 \subset A, J_2 \subset A^c$ are finite, then since $J_1 \cup J_2$ is finite we have $\sum_{n \in I} a_n \ge \sum_{n \in J_1 \cup J_2} a_n = \sum_{n \in J_1} a_n + \sum_{n \in J_2} a_n$, taking the $\sup$s on the right hand side gives $\sum_{n \in I} a_n \ge \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Similarly, if $J \subset I$ is finite, then $\sum_{n \in J} a_n = \sum_{n \in J\cap A} a_n + \sum_{n \in J \cap A^c} a_n \le \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Now, note that $T_1 \cup T_2 = T_1 \setminus T_2 \cup (T_1 \cap T_2) \cup T_2 \setminus T_1$, a disjoint union, hence $\sum_{n \in T_1 \cup T_2} a_n+ \sum_{n \in T_1 \cap T_2} a_n= \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n + \sum_{n \in T_2 \setminus T_1} a_n$ and $\sum_{n \in T_1 } a_n + \sum_{n \in T_2}a_n = \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n + \sum_{n \in T_2 \setminus T_1} a_n$.

Note that this result holds whether or not the sum is finite.

copper.hat
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  • Very nice answer! In the last equation I guess that you have used the following $\sum \limits_{n\in T_1 \backslash T_2}a_n+\sum \limits_{n\in T_1 \cap T_2}a_n+\sum \limits_{n\in T_2 \backslash T_1}a_n+\sum \limits_{n\in T_1 \cap T_2}a_n$ and the first two and last two terms are $\sum \limits_{n\in T_1}a_n$ and $\sum \limits_{n\in T_2}a_n$, respectively. Because $T_1=(T_1 \backslash T_2)\sqcup (T_1\cap T_2)$ and $T_2=(T_2 \backslash T_1)\sqcup (T_1\cap T_2)$. – RFZ May 18 '20 at 17:50
  • @ZFR: That is correct. – copper.hat May 18 '20 at 17:57
  • Brilliant answer! You know I remember I had an issue with this for a long time because I did not know the precise definition of $\sum \limits_{n\in I}a_n$. Can we apply this definition for arbitrary $a_n$ real? In my case all $a_n\geq 0$ because I am learning measure theory and $a_n$ in my case are mostly measures which are nonnegative. – RFZ May 18 '20 at 18:11
  • @ZFR: There are various definitions, in https://math.stackexchange.com/a/3680999/27978 I used one that mimics one of the standard Lebesgue integral definitions. – copper.hat May 18 '20 at 18:24
  • One pattern that repeats (as with integrals) is to deal with a finite number of non negative objects, then an arbitrary number of non negative objects, then an arbitrary number of objects. – copper.hat May 18 '20 at 18:28
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For the notation part: For example, let $K$ be the even numbers. Then $\sum\limits_{i\in K}a_i$ is just $a_2+a_4+a_6+....$

For the proof part:

Lemma: If $A$ and $B$ are disjoint and the terms are always positive and the sums converge: $\sum\limits_{i\in A\cup B}(a_i) = \sum\limits_{i\in A}a_i + \sum\limits_{i\in B}a_i$

Hence, since $T_2= T_1 \cap (T2-T1)$ with this to sets being disjoint: $\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2-T_1}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$

And again by the same lemma. combining the first and the second terms from the right side of the equality (remember that $T_1 \cup T_2 = T_1 \cup (T_2 - T_1)$):

$\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T1\cup T_2}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$.

Fernando Chu
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  • Hmm. Thank you! Could you also show the proof of the first question? – RFZ May 18 '20 at 04:23
  • Oh, my bad, thought you were just asking about the notation. I'm on it but are there no assumptions about convergence on the sum of the $a_i$s? – Fernando Chu May 18 '20 at 04:24
  • Look at my edit, please. So it turns out that that I can define $\sum \limits_{i\in K}a_i$ for $K\subset \mathbb{N}$ as $\sum \limits_{n=1}^{\infty}a_n[n\in K]$ , where $[]$ is Iverson bracket. Right? – RFZ May 18 '20 at 04:28
  • yup seems about right – Fernando Chu May 18 '20 at 04:30
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    then it follows that $[n\in T_1\cup T_2]+[n\in T_1\cap T_2]=[n\in T_1]+[n\in T_2]$ and our relation follows easily, right? Or maybe I am wrong? – RFZ May 18 '20 at 04:31
  • Also I am using the fact that $\sum \limits_{n=1}^{\infty}(a_n+b_n)=\sum \limits_{n=1}^{\infty}a_n+\sum \limits_{n=1}^{\infty}b_n$ for $a_n,b_n\geq 0$. This equality is true even series diverges. – RFZ May 18 '20 at 04:36
  • In your proof you are doing circular reasoning. I am not sure that you are correct. – RFZ May 18 '20 at 04:41
  • Yup, exactly like that. – Fernando Chu May 18 '20 at 04:41
  • Could you elaborate on the circular reasoning? – Fernando Chu May 18 '20 at 04:42
  • Giving the example where $K$ is the set of all even numbers doesn't define what $\sum_{i\in K}a_i$ means in general, any more than writing "$a_1 + a_2 + a_3 + \cdots$" would define what $\sum_{i=1}^\infty a_i$ means in general. The OP is surely not in any doubt as to the intuitive meaning of the notation; what is lacking is a mathematical definition. Similarly, all the hard work lies in proving the lemma $\sum_{i \in A \cup B}a_i = \sum_{i \in A}a_i + \sum_{i \in B}a_i,$ but you have only stated this lemma, not proved it. – Calum Gilhooley May 18 '20 at 13:06