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I am trying to make sense of what this theorem from C.I. Steinhorn, Borel Structures and Measure and Category Logics, says.

Theorem 1.3.3. A Borel linear order cannot have an uncountable increasing or decreasing chain.

I didn't find definitions in the paper, but I am quite sure that a pair $(A,\leq)$ is a Borel linear order iff $A$ is a Borel set in a standard Borel space $X$ and $\leq$ is Borel as a subset of $X^2$. I presume that an increasing chain is just a linear order with no greatest element and similarly for decreasing.

Here is the problem. Intuitively, I would guess that the lexicographic order on Baire space is a Borel linear order. But then Baire space itself should be an uncountable increasing chain under this order.

What am I misunderstanding?

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  • $\begingroup$ This is true if chain is replaced by wellordered chain. In other words, a Borel linear order cannot contain a copy of $\omega_1$ or its reverse. I would guess that's what is meant, otherwise "increasing or decreasing" is not very meaningful. Where is this from? $\endgroup$
    – François G. Dorais
    Aug 19, 2012 at 6:04
  • $\begingroup$ @FrançoisG.Dorais Does the well-ordered chain have to be Borel? $\endgroup$
    – William
    Aug 19, 2012 at 6:08
  • $\begingroup$ @FrançoisG.Dorais The paper is linked to in the question. projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – Rachel
    Aug 19, 2012 at 6:10
  • $\begingroup$ @Rachel Mildly related to this result : A consequence of the Kunen-Martin Theorem asserts that all strict boldface $\Sigma_1^1$ (analytic) wellfounded relations have countable length. Boldface $\Sigma_1^1$ include the Borel sets. $\endgroup$
    – William
    Aug 19, 2012 at 6:14
  • $\begingroup$ I had missed the link! The author cites Harrington and Shelah, who did prove the result I recalled earlier. I guess it would be best to check that reference. @William: No. There are no uncountable wellordered Borel chains at all so that weaker variant is vacuously true. $\endgroup$
    – François G. Dorais
    Aug 19, 2012 at 6:28

1 Answer 1

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The reference is to a result of Harrington and Shelah, who showed that a Borel linear order does not contain an $\omega_1$-chain. Since every uncountable wellordering contains an initial copy of $\omega_1$, it follows that a Borel linear order cannot contain an uncountable increasing wellordered chain. By reversing the linear order and applying the theorem again, we see that there are no uncountable decreasing wellordered chains either.

I conclude from this that the author probably meant for "chain" to be "wellordered chain" (which also explains the otherwise unusual phrase "increasing or decreasing").

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  • $\begingroup$ Thanks for the answer. Elsevier's site is not letting me access the original article (though it should). I emailed them, so now I am just waiting to double-check it. I suppose you are right about "increasing", but what would you say instead, for a linearly-ordered set without a greatest element? I would think "unbounded" might be confusing here. $\endgroup$
    – Rachel
    Aug 19, 2012 at 13:56

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