While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then wondered whether this was unique. When I got back to my computer, I looked on google and came up with this, which indicated that is not. However, the proof given there makes heavy use of Choice, and based on similar things, I am guessing that this cannot be avoided.

Define DC($\omega_1$) as:

For all trees $T$, $\quad$ $T\:$ has a branch $\ $ or $\ $ $T\:$ has a chain of length $\omega_1$ $\quad$.

which, if I haven't messed up the simplification, ZF proves is equivalent to the definition given at shelah.logic.at/files/446.ps .

Does $\ ZF+DC(\omega_1)\ $ prove that there is a unique subfield of $\mathbb{C}$ which is both isomorphic to $\mathbb{R}$ and has Baire property?

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    $\begingroup$ +1 just for blowing my mind with the fact that there are subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ other than $\mathbb{R}$! $\endgroup$ – Zev Chonoles Jan 4 '11 at 4:54
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    $\begingroup$ @Zev: The fact that $\mathbb C$ has uncountably many field automorphisms does not seem so surprising to me. After all the only obstruction that keeps $\mathbb R$ from having many automorphisms is the fact that its order can be described algebraically, hence is preserved. Since there is no such obstruction on $\mathbb C$, there should be plenty of automorphisms, and in fact there are. $\endgroup$ – Tobias Hartnick Jan 4 '11 at 21:51
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    $\begingroup$ There seems to be something wrong with the formatting? $\endgroup$ – Harry Altman Jan 6 '11 at 4:12
  • $\begingroup$ Ah, thanks for the explanation Tobias. I found the result very counterintuitive until you pointed out that there was no reason to have that intuition to begin with :) $\endgroup$ – Zev Chonoles Jan 6 '11 at 4:35
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    $\begingroup$ @Ricky: Your translation of DC_omega1 to trees is not correct. The above statement is obviously false, for instance since clearly the one element tree has no chain of length omega1. $\endgroup$ – Justin Moore Apr 3 '11 at 0:04

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