2

In this question, we have proved that $\langle \Bbb R,+ \rangle \cong \langle \Bbb C,+ \rangle$:

  1. Pick your favourite Hamel basis $H=\{U_\alpha \mid \alpha \in I\}$ where $I$ is an indexing set.
  2. Then, $H \cup iH$ is a basis of $\Bbb C$ as a vector space over $\Bbb Q$.
  3. Pick your favourite bijection $\varphi:H \mapsto H \cup iH$.
  4. This gives an isomorphism between the two groups concerned.

Choice is used in the 1st step and, I believe, in the 3rd step.


Let $P$ be the proposition $\langle \Bbb R,+ \rangle \cong \langle \Bbb C,+ \rangle$ under $ZF$. Which of the following statements are true?

  1. $\vdash P$
  2. $\vdash \neg P$
  3. $\operatorname{Con}(ZF \cup \{P\})$
  4. $\operatorname{Con}(ZF \cup \{\neg P\})$
Kenny Lau
  • 25,049
  • Out of the gate, if it is provable from ZFC, then it is consistent with ZF. So option 2 is immediately false; and option 3 is immediate true. Now, 4 holds if and only if 1 fails. And as luck would have it, that is indeed the case. – Asaf Karagila Sep 16 '17 at 07:10
  • Also, c'mon, https://math.stackexchange.com/questions/2424295/do-we-need-the-axiom-of-choice-to-show-that-mathbbr-and-mathbbc was asked very recently. Not even a week ago! :) – Asaf Karagila Sep 16 '17 at 07:12
  • @AsafKaragila a gold badge in the axiom of choice... – Kenny Lau Sep 16 '17 at 07:16
  • I've spent the last seven years working on choice related research. So that kind of makes sense, why not? – Asaf Karagila Sep 16 '17 at 07:24
  • @AsafKaragila that's awesome – Kenny Lau Sep 16 '17 at 07:26

0 Answers0