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I'm told there is no Borel well-order of the reals (in ZFC). I'm told, in fact, that this is because of Borel determinacy. However, this is usually a vague handwave of the form (a) take the usual proof which well-orders the reals and makes an unsolveable game, and then (b) if the well-ordering is Borel, so is this game, which contradicts determinacy.

But when I actually check the details on this, (b) doesn't actually follow. It probably depends on your version of the "usual proof." Can anyone give a reasonably precise proof that a Borel well-order of the reals contradicts Borel determinacy?

Note: I do have a proof of this fact. But I'm not happy with it; it seems to use more machinery than it really needs to. Not that Borel determinacy is anything to sneeze at, I guess...

Asaf Karagila
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Richard Rast
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  • Related: Is there a known well-ordering of the reals? – hardmath Apr 17 '15 at 19:59
  • Right, the accepted answer to that one inspired this question. They make a precise statement but offer no proof. I wanted a precise argument of their claim (specifically, the first bullet point). – Richard Rast Apr 17 '15 at 20:01
  • Seems like an essential duplicate of http://math.stackexchange.com/questions/88757/nice-well-orderings-of-the-reals – Asaf Karagila Apr 17 '15 at 20:44
  • It's not intended to be! I'm looking for a direct proof that Borel well-orderings contradict determinacy. That question is more general and the answer involves different tools. So it doesn't really answer this question. – Richard Rast Apr 17 '15 at 20:48
  • @RichardRast: I do not know which proof you do have but the following should be simpler than proving Borel determinacy: If there existed a Borel well-ordering of $S^1$ (which is "essentially" the reals), then there would exist a Borel set intersecting every $E$-class at exactly one point, where $E$ is the orbit equivalence relation of the $\mathbb{Z}$-action on $S^1$ given by $k \cdot e^{i \theta}=e^{i (\theta+k)}$. But this cannot happen since there exists a rotation invariant Borel probability measure on $S^1$. – Burak Jan 22 '16 at 02:25

1 Answers1

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This argument is due to Sierpinski. Towards a contradiction, suppose $<_1$ is a Borel well ordering of reals. Let $r$ be the $<_1$-least real such that $\{x: x <_1 r\}$ is not null. The argument is essentially the same if no such $r$ exists. It follows that the set $W = \{(x, y): x <_1 r \wedge y <_1 r\} = A^2$ where $A = \{a: a <_1 r\}$ is Borel too. Now for each $y \in A$, $W^y = \{x: x <_1 y\}$ is null and for each $x \in A$, $W_x = \{y: x <_1 y\}$ is non null. But this contradicts Funibi's theorem.

Guesta
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  • On other words, even if you don't know about Borel determinacy, you can solve this. – GEdgar Apr 18 '15 at 18:56
  • I agree that this is a proof, and I like the proof. I believe it is essentially the same, though more concise, than the answer Asaf linked. However, I was looking for the argument wherein this directly contradicts determimacy. – Richard Rast Apr 19 '15 at 01:17