I have a question about the following part of a proof that a rearrangement of an absolutely convergent series converges to the same value as the original series.
Proof. Suppose that $\sum_{k=1}^{\infty}a_k$ is absolutely convergent and let $z: \mathbb{N} \to \mathbb{N}$ be a bijection. Pick $\epsilon > 0$. First find $N$ large such that for $n \geq N$ we have
$$\left|\sum_{k=1}^{\infty}|a_k| – \sum_{k=1}^n|a_k|\right| = \sum_{k=n+1}^{\infty}|a_k| < \epsilon.$$
Consider the maximum $M = \max\{z(1), z(2), z(3), \ldots , z(N)\}.$ Then for each $n \geq M$ the set
$$S_n = \{z(1), z(2), z(3), \ldots , z(n)\} \supset \{1, 2, 3, \ldots, N\}.$$
$\ldots$
Why is this final line true?
Yes, we can always go far enough through the terms of the rearranged sequence to reach a point where $a_1, \ldots, a_N$ have all shown up. But isn't this just because $z$ is a bijection, and so each term of the original sequence shows up eventually in the rearranged sequence.
Why does $n \geq M$ make this certain?
Can't we have for example with $N = 10$ that
$$\{z(1), \ldots, z(N)\} = \{101, 102, 103, \ldots, 110\}$$
where $M = 110$ and we have $n \geq M \geq N$.
Therefore,
$$\{z(1), \ldots, z(N), \ldots, z(M), \ldots, z(n)\} = \{101, \ldots, 110, \ldots, 210, 211\} \not \supset \{1, \ldots, 10\}. $$
Is this a faulty counterexample?
Thanks in advance.
Best Answer
I assume that they meant to write
$M=\max\{z^{-1}(1),z^{-1}(2),\ldots,z^{-1}(N)\}$.