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A long chain with club continuity below aleph omega
Statement
Assume AC. Put , , and . Every family in of cardinality less than has a strict eventual upper bound in .
There is a strictly increasing sequence in with for every uncountable regular . More precisely, whenever and for such a , there is a club for which , with the supremum taken pointwise. The strong-increase assertion means every unbounded set of indices contains an order-type- subsequence that is pointwise strictly increasing outside the union of two individual finite exceptional sets.
Facts & Assumptions
Given: AC and the cardinals and product just defined; eventual comparison means comparison at all but finitely many .
Club continuity at cofinality gives when is uncountable regular, and the sequence has regular length (Club continuity produces strongly increasing subsequences).
Subsets smaller than a regular cardinal are bounded in it, and each limit ordinal has a cofinal subset of size its cofinality (; and ; for a limit ordinal the value is an infinite cardinal with , so it is regular; and every cofinal subset of has cardinality at least , a value that is attained, (c)–(d)).
Under AC all successor alephs are regular; has countable cofinality and every smaller infinite cardinal is a finite-index aleph ( is regular in ZF; assuming the Axiom of Choice every successor aleph is regular; , so is singular, and under choice it is the least singular infinite cardinal).
Specified transfinite rules recurse on well-orders (Transfinite recursion).
AC supplies choice functions on nonempty sets of witnesses (The Axiom of Choice).
Proof
If has size , set when , and otherwise. At a large coordinate, each successor is below the infinite cardinal , and F2–F3 place the supremum below . Only finitely many coordinates have , since . Thus and for each . The empty family gives and imposes no comparisons. If instead , take a bijection and let , so and each . The preceding formula gives a bound for each subfamily and a bound for the countable family . For , compose outside the union of the two finite failure sets. AC ensures these cardinalities and the indicated enumeration; thus every family of size less than has a strict bound.
For each nonzero limit , F2 gives a cofinal sequence of length . Turn it into a continuous increasing sequence in : at successor stages choose a value above both the preceding value and the next cofinal-sequence term, and at nonzero limits take the supremum of earlier values. Each intermediate supremum is below because fewer than terms were used, by F2. Successor choices are possible because is a limit. The range is unbounded and closed in , hence is a club of order type ; closure follows because every nonzero limit point below occurs at a limit index of the continuous sequence. F4 supplies the recursion, and A1 chooses these clubs simultaneously from their nonempty witness sets. Also fix by A1 a choice function on the nonempty subsets of the set for use in choosing bounds.
Recursively set . At every , the previous functions form a family of size at most , so step 1.1 and the choice function of step 1.2 specify a strict eventual bound . If for an uncountable regular , put at coordinates with , and put at the others; then set . The large-coordinate supremum is below because by F2–F3. Its successor is still below that infinite cardinal. At all other stages put . Distinct cardinals have distinct double successors, so the special rule is unambiguous. Every value lies in , and F4 implements this specified recursion through . For every we have , proving strict increase.
Fix any uncountable regular . By F3 it is for some positive finite , so , and is regular by F3. At each index of that cofinality, step 2.1 used the special rule, giving whenever . The other coordinates form a finite set. This is the required club-continuity premise with bounding index exactly . The countably infinite coordinate set and strict chain from step 2.1 meet F1's remaining hypotheses. F1 gives for this same chain. As was arbitrary, all stated properties hold simultaneously. QED.
Depends on
- Club continuity produces strongly increasing subsequences
- $\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ and $\operatorname{cf}(\alpha + 1) = 1$; for a limit ordinal $\lambda$ the value $\operatorname{cf}(\lambda)$ is an infinite cardinal with $\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda)$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\operatorname{cf}(\lambda)$, a value that is attained
- $\aleph_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph $\aleph_{\alpha+1}$ is regular; $\operatorname{cf}(\aleph_\omega) = \aleph_0$, so $\aleph_\omega$ is singular, and under choice it is the least singular infinite cardinal
- Transfinite recursion
- The Axiom of Choice
Used by
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