Big rearrangement theorem/Big associative law

Question

Translated from German Großes Assoziativgesetz/Grpßer Umordnungssatz.

I am currently reading the textbook analysis 1 from Konrad Königsberger. It is page 70 I have Problems with.

I don't know the englisch expressions but the chapter I am currently reading is called "Summierbare Familien". Summable families.

Definitions

Let $I$ be a nonempty set and $a:I\rightarrow \mathbb{C}$ a set. $a$ or also $(a)_{i\in I}$ is called a Family and $a_i$ is defined as $a(i)$. The set of all finite subsets of $I$ is defined as $\mathscr{E}(I)$. For $J\in\mathscr{E}(I)$ define

$$a_J:=\sum_{i\in J}a_i\quad |a|_J:=\sum_{i\in J}|a_i|$$

$a_J$ is then the partialsum of the Family $a$ to the indexset $J$.

A Family $(a)_{i\in I}$ is called summable if there is a $s\in\mathbb{C}$ such that for every $\epsilon>0$ there is a finite indexset $I_{\epsilon}\subset I$ such that for all $J\subset \mathscr{E}(I)$ with $I_{\epsilon}\subseteq J$ we have

$$|s-a_J|\leq \epsilon$$

$s$ is then called the sum of the Family $a$ and also written as $\sum_{i\in I}a_i$.

Already proven facts

A family has not more than one sum

By Definition if $\pi:I\rightarrow I$ is a permutation then $\sum_{i\in I}a_{\pi(i)}=\sum_{i\in I}a_i$

Test for summability: A family $(a_i)_{i\in I}$ is summable if and only if the set $\{|a|_J \, |\,J\in\mathscr{E}(I) \}$ of partialsums from the family $|a|$ is bounded

Rearangementtheorem: A family $a=(a_i)_{i\in\mathbb{N}}$ is summable if and only if $\sum_{i=1}^{\infty}a_i$ converges absolutely. The respecitive sums are equal in this case

$$\sum_{i\in\mathbb{N}}a_i=\sum_{i=1}^{\infty}a_i$$

Theorem

The big rearrangement Theorem states that if $(a_i)_{i\in I}$ is summable and $I_k$, $k\in K$ are pairwise disjoint subsets of $I$ such that the union of those is $I$ again. Then every subfamily $(a_i)_{i\in I_k}$ is summable and the family $(s_k)_{k\in K}$ of the sums $s_k:=\sum_{i\in I_k}a_i$ is sumable and we have

$$\fbox{$\sum_{i\in I}a_i=\sum_{k\in K}s_k=\sum_{k\in K}\big(\sum_{i\in I_k}a_i\big)$}$$

Proof:

The summability of the subfamilies results from the summability test. The summablity of the family $(s_k)_{k\in K}$ also results from the summabilty test. Because if we take an arbitrary finite subset $\{k_1,...,k_n\}\subset K$ we can make the estimate

$$\sum_{v=1}^{n}|s_{k_{v}}|\leq \sum_{v=1}^{n}\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}=\sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}\leq \sup_{J\in\mathscr{E}(I)}\{|a|_J\}$$

To prove the Formula we set $S:=\sum_{i\in I}a_i$. Furthermore let $\epsilon>0$. We then have to find a finite indexset such that with $S_M:=\sum_{k\in M}s_k$ we have

$$\tag{*}|S_m-S|\leq \epsilon\quad \text{for all }M\in\mathscr{E}(K) \text{ with } K_\epsilon \subset M $$

According to the the deinition of $S$ there is a finite indexset $I_{\epsilon}\subset I$ such that $|a_J-S|\leq \epsilon/2$ for alll $J\in\mathscr{E}(I)$ with $I_\epsilon \subset J$. To $I_\epsilon$ we choose a finite set $K_\epsilon\subset K$ such that the union of $I_k,\, k\in K_{\epsilon}$ contains $I_\epsilon$. We now Show that $K_\epsilon$ satisfies $(*)$. If $m$ is the number of Elements in $M$, we choose for every subfamily $(a_i)_{i\in I_k}$ a dinite subset $I_{k,\epsilon}\subset I_k$ with

$$|a_{I_{k,\epsilon}}-s_k|\leq\frac{\epsilon}{2m}\quad\text{ and } \quad (I_k\cap I_{\epsilon})\subset I_{k,\epsilon}$$

For $|S_M-S|$ we get the estimate:

$$|\sum_{k\in M}s_k-S|\leq \sum_{k\in M}|s_k-a_{I_{k,\epsilon}}-S|\leq m\frac{\epsilon}{2m}+\frac{\epsilon}{2}=\epsilon;$$

for that we used that $\bigcup_{k\in M}I_{k,\epsilon}=:J$ is a union of disjoint sets hence $a_J=\sum_{k\in M}a_{I_{k,\epsilon}}$ and that $|a_J-S|\leq \epsilon/2$ because $J$ contains according to the construction $I_{\epsilon}$ $\square$

Question:

I could only understand the first line of the proof and I was really trying hard to understand this estimate:

$$\sum_{v=1}^{n}|s_{k_{v}}|\leq \sum_{v=1}^{n}\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}=\sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}\leq \sup_{J\in\mathscr{E}(I)}\{|a|_J\}$$

I am not sure about the other parts but I think I could somehow figure it out because all is jstified. If not I will have to open a new Question or eddit this Question again.

Why is

$$|s_{k_{v}}|\leq\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}$$

And also the equality, how can I prove it?

$$\sum_{v=1}^n\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}=\sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}$$

For the last inequality I have got an idea:

If I take an arbitrary $J_v\in \mathscr{E}(I_{k_{v}})$ and we look at $\sum_{v=1}^{n}|a|_{J_v}$ then since each $J_v$ is also an element of $\mathscr{E}(I)$ we can also express $\sum_{v=1}^{n}|a|_{J_v}$ as $\sum_{i\in J} |a_i|=|a|_J$ for some $J$. which is less or equal than $\sup_{J\in\mathscr{E}(I)}\{|a|_J\}$.

I don't know how I can argue such that also we can establish the case for $\sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}$ ?

Please help me and let me know if something is unclear.

Answer 1

Why is $$|s_{k_{v}}|\leq\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}$$

Basically its a form of the triangle inequality.

Let $\epsilon>0$. Let $I_\epsilon\subset I$ be a finite index set such that for all $J\subset \mathscr{E}(I_{k_v})$ with $I_{\epsilon}\subseteq J$ we have $|s_{k_v}-a_J|\leq \epsilon$. Then $|s_{k_v}|\leq |a_J|+\epsilon$ follows by inverse triangle inequality. By the triangle inequality, we also have $|a_J|\leq |a|_J$, since the latter is a finite sum. Now this implies $$|s_{k_{v}}|\leq\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}+\epsilon.$$ Since $\epsilon>0$ was aribtrary, the inequality $$|s_{k_{v}}|\leq\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}.$$ follows.

And also the equality, how can I prove it? $$\sum_{v=1}^n\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\}=\sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}$$

The right-hand side can be rewritten as $$ \sup_{J_1\in \mathscr{E}(I_{k_1})} \sup_{J_2\in\mathscr{E}(I_{k_2})} \cdots \sup_{J_n\in \mathscr{E}(I_{k_n})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}. $$ Now both terms contain $n$ supremums. It can be seen that $\sup_{J_1\in \mathscr{E}(I_{k_1})}$ can be exchanged with $\sum$: $$ \sum_{v=1}^n\sup_{J_v\in\mathscr{E}(I_{k_{v}})}\{|a|_{J_{v}}\} = \sup_{J_1\in \mathscr{E}(I_{k_1})} |a|_{J_1} + \sup_{J_2\in \mathscr{E}(I_{k_2})} |a|_{J_2} + \cdots \sup_{J_n\in \mathscr{E}(I_{k_n})} |a|_{J_n} = \sup_{J_1\in \mathscr{E}(I_{k_1})} \left( |a|_{J_1} + \sup_{J_2\in \mathscr{E}(I_{k_2})} |a|_{J_2} + \cdots \sup_{J_n\in \mathscr{E}(I_{k_n})} |a|_{J_n} \right). $$ If you repeat this process for $v=2,\dots n$ you can get the equality that you want.

For the last inequality I have got an idea: If I take an arbitrary $J_v\in \mathscr{E}(I_{k_{v}})$ and we look at $\sum_{v=1}^{n}|a|_{J_v}$ then since each $J_v$ is also an element of $\mathscr{E}(I)$ we can also express $\sum_{v=1}^{n}|a|_{J_v}$ as $\sum_{i\in J} |a_i|=|a|_J$ for some $J$. which is less or equal than $\sup_{J\in\mathscr{E}(I)}\{|a|_J\}$.

This is a good idea. It can be improved by starting with an arbitrary $(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})$. Then you have shown that $$ \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}\leq \sup_{J\in\mathscr{E}(I)}\{|a|_J\} $$ is true. Since $(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})$ was chosen arbitrary, you can also just take the supremum over those $(J_1,\dots,J_n)$ in the inequality. The result is then $$ \sup_{(J_1,...,J_n)\in\mathscr{E}(I_{k_{1}})\times … \times \mathscr{E}(I_{k_{n}})} \big\{\sum_{v=1}^n|a|_{J_{v}}\big\}\leq \sup_{J\in\mathscr{E}(I)}\{|a|_J\}. $$