Baby Rudin 3.54, Concerning the Definition of Rearrangement of a Series

Question

This theorem has already been asked about, however my specific question, to my knowledge has not. I hope it will be acceptable if I refer the reader of this question to either here or here for Rudin's proof, since it has already been written up.

Rudin says that for "sequences $(m_n)$, $(k_n)$" the following is "clearly a rearrangement of $\Sigma a_n$.'' $$ P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} - Q_{k_1+1} - \cdots - Q_{k_2} + \cdots $$

Can anyone tell me, is he saying that $\Sigma a_{n'}$ is a rearrangement of $\Sigma a_n$ where $(a_{n'})$ is defined $$ a_{1'} := \sum_{i=1}^{m_1} P_i \hspace{1mm}-\hspace{1mm} \sum_{j=1}^{k_1} Q_j, \quad\ldots\quad a_{n'} := \sum_{i=m_{n'}+1}^{m_{{n'}+1}} P_i \hspace{1mm}-\hspace{1mm} \sum_{i=k_{n'}+1}^{k_{{n'}+1}} Q_j, \quad\ldots $$ for any sequences of integers $(m_n)$ and $(k_n)$? Or is he saying that $\Sigma a_{n'}$ is a rearrangement of $\Sigma a_n$ where $(a_{n'})$ is defined $$ a_1 := P_1, \quad\ldots\quad a_{n'} = \begin{cases} P_{\ell+1} &\text{ if }\hspace{2mm} \ell\ni a_{n'-1} = P_\ell \wedge \ell\neq m_n\hspace{1mm}\forall n\in\{1,2,\ldots\} \\ Q_{\ell+1} &\text{ if }\hspace{2mm} \ell\ni a_{n'-1} = Q_\ell \wedge \ell\neq k_n\hspace{1mm}\forall n\in\{1,2,\ldots\} \\ Q_{k_\ell} &\text{ if }\hspace{2mm} \exists\ell\ni a_{n'-1} = P_{m_\ell} \\ P_{m_\ell + 1} &\text{ if }\hspace{2mm} \exists\ell\ni a_{n'-1} = Q_{m_\ell} \end{cases} $$

Moreover, this only makes sense if $(m_n)$ and $(k_n)$ are strictly increasing sequences of positive integers (although Rudin does not make this specification), is that correct?

Thanks to anyone who is willing to help.

Edit: My suspicion is that the latter of these two is correct, but I would be grateful for a second or third opinion.

Answer 1

Rudin is building a rearrangement of the $a_n$ terms via the recursion theorem:

Theorem 1: Let $X$ be a set with $x_0 \in X$ and $\psi: X \to X$ be any function. Then there exist one and only one function $\rho: \Bbb N \to X$ satisfyings

$\tag 1 \rho(0) = x_0$ $\tag 2 \forall n \in \Bbb N, \; \rho(n+1) = \psi(\rho(n))$

See, for example, the wikipedia article recursive definition.

You won't find recursion in the index. Apparently, for Rudin, building sequences via recursion is so routine it doesn't deserve mention. If you want to learn it, read his proofs!

You will notice that in the body of the proof you won't find any attempt to 'map things out' - the recursion machine implicitly takes care of things at 'run time'.

I suggest the serious student rework the proof for three cases:

Case 1: For all $n$, $\alpha_n = \alpha \in \Bbb R$ and $\beta_n = \beta \in \Bbb R$ and $\alpha = \beta$.

Case 2: For all $n$, $\alpha_n = \alpha \in \Bbb R$ and $\beta_n = \beta \in \Bbb R$ and $\alpha \lt \beta$.

Case 3: For all $n$, $\alpha_n = -n$ and $\beta_n = +n$.

Try working on Case 1 by first thinking it through and working out your ideas on some scrap paper.

Notice that in Rudin's proof he wants $\beta_1 \gt 1$. No doubt he needs that to build (seed) his machine. It is a red herring, but if you understand that you will be in business.

Finally, to really understand the theorem you can use Rudin's proof to get the main ideas and then build your own recursion machine(s).

Answer 2

Looks to me like it's close to the latter option. (As saulspatz commented, $(m_n)$ and $(k_n)$ are constructed to be strictly increasing, so that's not an issue.) The point here is that each of the terms $P_i$ is by definition one of the terms $a_n$, and each of the terms $-Q_j$ is by definition one of the terms of $a_n$, and they each correspond to a distinct term, so a sum of all the $P_i$s and $-Q_j$s is a rearrangement of $\sum a_n$.

The precise definition of the sequence $a'_n$ would be as follows, adopting the convention $m_0=k_0=0$:

$$ a'_n = \begin{cases}P_{n-k_j} &\text{if } m_j+k_j < n \leq m_{j+1}+k_j \\ -Q_{n-m_j} &\text{if } m_j+k_{j-1}<n\leq m_j+k_j \end{cases} $$

Answer 3

for what it's worth, I finally put together a pretty thorough reworking of Rudin's proof.

I'd be happy to receive feedback, but mostly I just thought I'd share in case anyone cares. Here's a link to the PDF, because I don't know how to make my LaTeX macros compatible with this website (sorry).