Notebook 23: Ramsey Decomposition U·Vω

Even a non-ultimately-periodic ω-word splits into a prefix followed by infinitely many congruent blocks from one idempotent class. Watch the blocks abcⁱ all collapse to a single ≈𝒜-class V with V·V = V.

The Ramsey Decomposition

The last ingredient of complementation is a combinatorial fact: every \(\omega\)-word factors as \[ \alpha \in U\,V^{\omega}, \] a prefix from one class \(U\) followed by infinitely many blocks from a single idempotent class \(V\) (idempotent means \(V\cdot V=V\)). This holds even when \(\alpha\) is not ultimately periodic.

The chapter’s witness is the non-periodic word \[ \alpha = \underbrace{abc}_{v_1}\ \underbrace{abcc}_{v_2}\ \underbrace{abccc}_{v_3}\cdots,\qquad v_i = ab\,c^{\,i}. \] The number of trailing \(c\)’s grows forever, so \(\alpha\) is not ultimately periodic. Yet — because an extra \(c\) only loops at \(s_2\) — every block \(v_i\) has the same \(\approx_{\mathcal A}\)-profile. They all belong to one class \(V\), and \(V\) is idempotent. So \(\alpha=[\varepsilon]\,V^{\omega}\): prefix class \(U=[\varepsilon]\), then \(V\) forever.

Widget: the blocks collapse to one class

α = v₁ v₂ v₃ …  (vᵢ = ab cⁱ) — not ultimately periodic

Each block’s ≈𝒜-profile (all identical → one class V)

Idempotency: V · V = V

Why do the blocks collapse? An extra trailing \(c\) only loops at \(s_2\) (\(s_2\xrightarrow{c}s_2\)), which does not change the profile — every \(v_i=ab\,c^{i}\) takes \(s_0\to s_2\) and \(s_2\to s_2\), each visiting \(s_1\in F\), and has no path from \(s_1\). That is one class \(V\), and composing two such blocks yields the same shape, so \(V\) is idempotent. This is exactly the abstract decomposition the pigeonhole argument produces in general — here we simply read it off the automaton. Saturating on these class products is the final step: Notebook 24.