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...
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.