The even and odd index maps and the alternating sequence: strictly increasing $e,o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0=1$, $s_{\sigma (k)}=-s_k$, which satisfies $\mathrm{\mid }{s}_k\mathrm{\mid }=1$, $s\circ e\equiv 1$ and $s\circ o\equiv -1$

Let $\sigma$ be the successor on $\mathbb{N}$ (The natural numbers $\mathbb{N}$ (von Neumann)). There are functions $e,o:\mathbb{N}\to \mathbb{N}$ and a sequence $(s_k)$ of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with the following properties.

1. The index maps. $e$ is the unique function with $e_0=0$ and $e_{\sigma (j)}=\sigma (\sigma (e_j))$, and $o$ the unique function with $o_0=\sigma (0)$ and $o_{\sigma (j)}=\sigma (\sigma (o_j))$. Both are strictly increasing.
2. The partition. $\mathbb{N}$ is the disjoint union of the ranges of $e$ and of $o$: every natural number is $e_i$ for exactly one $i$ or $o_i$ for exactly one $i$, and never both.
3. The alternating sequence. $(s_k)$ is the unique sequence of reals with $s_0=1,s_{\sigma (k)}=-s_k(k\in \mathbb{N})\mathrm{.}$
4. Its values. $\mathrm{\mid }{s}_k\mathrm{\mid }=1$ for every $k$, so $(s_k)$ is bounded; and $s_{e_j}=1,s_{o_j}=-1(j\in \mathbb{N}),$ that is $s\circ e$ is constantly $1$ and $s\circ o$ constantly $-1$.

This is the sequence usually written $s_k=(-1{)}^k$, with $e_j=2j$ and $o_j=2j+1$, presented by the recursions that its proofs actually use. It is collected here once because three separate items on this page and its companion need an alternating or interleaved witness, and rebuilding the recursion inside each of them is what this lemma exists to prevent.

• Why the recursion rather than $(-1{)}^k$. Written as a power, every one of the four claims would have to be unwound into the two recursion equations before it could be proved; written as a recursion, each is a two-line induction. The identification with $(-1{)}^k$ is available (Integer powers $a^m$) and is used nowhere.

• The parity statement is genuinely proved, not assumed. Claim 2 is where the work is: the covering half is the interleaved induction of step 1.6, which tracks $n$ and $\sigma (n)$ together because neither alone is preserved by the successor, and the disjointness half is settled by the sequence, since $s$ takes the value $1$ on one range and $-1$ on the other and $1\mathrm{\ne }-1$. Using the sequence to separate the two ranges is shorter than any direct parity argument and needs no arithmetic on $\mathbb{N}$ beyond the successor.

• What consumes this lemma. FALSE: a sequence with a convergent subsequence is bounded (the converse of Bolzano-Weierstrass) interleaves a constant sequence with an unbounded one along $e$ and $o$; The sequence $(-1{)}^k(1+1\mathrm{/}k)$ is bounded with subsequential limit set exactly $\{-1,1\}$ ↗ multiplies $s$ by a null perturbation to get a sequence with exactly two subsequential limits; and The sequence $1,1,2,1,3,1,4,\dots$ is unbounded and has a convergent subsequence ↗ is the witness for the first of those. The same sequence, built inline, refutes FALSE: every bounded sequence converges on the previous page; that item predates this lemma and is left as it stands.