LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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The tail-complement automorphism fixes finitely supported names

Statement

Let $\dot{x}$ be an HS name in the tail-flip system, and choose $m < ω$ with $H_{m} \subseteq sym (\dot{x})$ . Let $Q \in Add (ω, ω)$ be a finite condition and let $r \geq m$ . There is $k_{0} < ω$ such that the automorphism which fixes every coordinate other than $r$ and flips $S_{r} (k)$ for every $k \geq k_{0}$ fixes both $Q$ and $\dot{x}$ , while sending $S_{r}$ to its complement modulo the finite initial segment $k_{0}$ .

Facts & Assumptions

Given: The HS name $\dot{x}$ , support bound $m$ , finite condition $Q$ , and coordinate $r \geq m$ .

[F1]

The tail-flip hereditary-symmetric model defines the ground-model bit-flip group, the subgroups $H_{m}$ , their action on coordinate Cohen reals, and the finite-support property of each HS name.

[F2]

Symmetry lemma for forcing automorphisms transports forced formulas and their names under the constructed automorphism.

Proof

technique · direct construction of a tail flip inside the supporting subgroup

1.1

The set $D = {k : (r, k) \in dom (Q)}$ is finite. Let $k_{0} = 0$ if $D = \emptyset$ , and otherwise let $k_{0} = 1 + \max D$ . Define $a \in (2^{ω \times ω})^{V}$ by $a (i, k) = 1$ exactly when $i = r$ and $k \geq k_{0}$ . By F1, $a$ induces an order automorphism $π_{a}$ of the forcing.

F1construct

2.1

No point of $dom (Q)$ belongs to the support of $a$ , so $π_{a} Q = Q$ . Because $r \geq m$ , the flip $a$ lies in $H_{m}$ . The support hypothesis therefore gives $π_{a} \dot{x} = \dot{x}$ . F2 then transports any forced formula containing $\dot{x}$ while leaving both its condition and that name fixed.

F1F2step 1.1

2.2

We have $π_{a} S_{i} = S_{i}$ for $i \neq r$ , while $π_{a} S_{r} = S_{r} △ {k : k \geq k_{0}}$ . Thus membership is reversed at every $k \geq k_{0}$ and preserved below $k_{0}$ , so the following exact symmetric-difference identity holds.

F1step 1.1

$π_{a} S_{r} △ (ω ∖ S_{r}) = k_{0} .$

The right side is the finite von Neumann initial segment.

3.1

The flip set is an infinite tail; only its intersection with the finite domain of $Q$ had to be empty. Replacing it by a finite flip would preserve membership in every free ultrafilter under finite modification and would not give step 2.2. The construction takes the maximum of one finite set and uses no Choice.

step 1.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Solomon Feferman, Some applications of the notions of forcing and generic sets, proof of Theorem 4.12, printed pp. 343–344 (standard reference, not scraped)
Eleftherios Tachtsis, On the Existence of Free Ultrafilters on omega and on Russell-sets in ZF, analogous tail-flip automorphism, pp. 5–7 (standard reference, not scraped)