Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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First-difference order on a binary tree

Statement

Let T=23, ordered by proper initial segment, and order the two successors at every nontop node by 0<1. The first-difference order on its eight terminal maximal branches is

000<001<010<011<100<101<110<111.

This finite calculation is ccc but is neither dense nor without endpoints, and it has nonempty separable open intervals. It therefore isolates the local dense-successor and unbounded-splitting hypotheses used by the general branch construction.

Facts & Assumptions

Given: the explicit finite tree 23; coordinates are numbered 0,1,2 from the root, and binary successors have their usual order.

[F1]

In the Suslin-tree construction, maximal branches are compared at their least differing coordinate after every immediate-successor set is ordered densely and without endpoints. The first-difference order on branches

[F2]

The general construction obtains a ccc dense order without endpoints and uses splitting above arbitrary countable height bounds to prove that no nonempty open interval is separable. The first-difference order on branches

Verification

1.1

A branch maximal under inclusion must end on level 3, so it is determined by one of the eight words 000,001,010,011,100,101,110,111. Conversely each such word is terminal and hence maximal. For distinct terminal words s,t, the comparison rule is s<ts(k)<t(k) for k=min{i<3:s(i)t(i)}. The displayed list in the statement follows by sorting first by coordinate 0, then by coordinate 1, then by coordinate 2.

F1givenconstruct
1.2

The first-difference coordinates of consecutive branches are explicit:

consecutive pairfirst differencecomparison at that coordinate
000,00120<1
001,01010<1
010,01120<1
011,10000<1
100,10120<1
101,11010<1
110,11120<1

For nonconsecutive words the same minimum-coordinate formula applies; for example d(001,111)=0 and d(100,111)=1. Thus the table and formula determine every pairwise comparison, not only the adjacent ones. [step 1.1, construct]

2.1

The adjacent pair 000<001 has no branch strictly between it, so this order is not dense. The elements 000 and 111 are respectively a first and last element, so it is not endpoint-free. Moreover (000,010)={001} is a nonempty open interval with the countable dense subset {001}; hence the nowhere-separable conclusion fails. Since the whole order has only eight elements, every antichain of pairwise disjoint nonempty open intervals is finite, so its ccc assertion holds only trivially.

step 1.1step 1.2construct
3.1

The failed order-density and endpoint conclusions in step 2.1 come from the local successor order {0<1}: it has neither an intermediate successor nor successors on both sides. In [F1], replacing each successor set by a dense order without endpoints supplies exactly those insertions. By contrast, the failure of nowhere separability reflects bounded height: the argument in [F2] first bounds an arbitrary countable candidate dense set by a height δ and then splits above δ, which is impossible in 23. Thus local dense successor orders yield density and no endpoints, while unbounded normal splitting is essential to the later nowhere-separability construction.

F1F2step 2.1
4.1

The root is the empty word, but no empty terminal branch occurs; at height zero it has exactly the two successors 0,1. Every top word has length exactly three, so there is no hidden extension beyond the listed eight. All lists, comparisons, and witnesses are explicit finite data in ZF, and no choice principle is used.

givenstep 1.1step 1.2step 2.1

Remarks

  • Finite ccc is not evidence for the Suslin-tree extraction argument: there is no uncountable family here to test.
  • A finite-height tree with densely ordered infinite successor sets could have a dense endpoint-free branch order. The calculation shows that bounded height and binary local branching are distinct obstructions.

Depends on

Used by

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Sources