Notebook 22: Merging Positions & Persistence
Merging Positions
The Ramsey decomposition needs a second finite-index equivalence, this time on the positions of an \(\omega\)-word. Fix a congruence \(\equiv\) on finite words. For positions \(k,k'\) and a later position \(m>\max\{k,k'\}\), say \(k\) and \(k'\) merge at \(m\), written \(k\mathrel{\widetilde{=}}_{\alpha,m}k'\), if the two segments reaching \(m\) agree: \[ \alpha(k,m)\ \equiv\ \alpha(k',m). \]
The stopping point \(m\) is ours to choose; the same two positions may fail to merge at one \(m\) and succeed at a later one. But persistence says: once merged, always merged.
We illustrate with the smallest possible congruence — parity of length \(\equiv_{\mathrm{par}}\) — on the word from the chapter figure, \(\alpha = b\,a\,b\,c\,c\,a\,b\,c\cdots\). Since \(\equiv_{\mathrm{par}}\) ignores symbols, \(\alpha(k,m)\) and \(\alpha(k',m)\) have lengths \(m-k\) and \(m-k'\); they share parity iff \(k\) and \(k'\) have the same parity as natural numbers. (With a richer congruence like \(\approx_{\mathcal A}\), merging becomes a genuine, symbol-dependent event.)
Widget: pick two positions
The default choice \(k=1\), \(k'=3\) reproduces the chapter’s figure: the segments \(\alpha(1,4)\) and \(\alpha(3,4)\) have lengths \(3\) and \(1\), both odd, so they merge at \(m=4\). For the parity congruence every even position merges with every even position and every odd with every odd — no real “sifting” is needed. The general Ramsey argument (Notebook 23) runs the same pigeonhole on a richer congruence, where finding an infinite pairwise-merging set takes genuine work.