TITLE: Are $C$ and $\bar{Q_p}$ isomorphic?
QUESTION [16 upvotes]: There is a famous passage on the third page of Deligne's second paper on the Weil conjectures where he expresses his dislike of the axiom of choice, as manifested in the isomorphism between $C$ and $\bar{Q_p}$. The proof of said isomorphism runs as follows. Both $C$ and $\overline{Q_p}$ have transcendence bases, $S$ and $T$. Then $C\simeq \overline{Q(S)}$ and $\overline{Q_p}\simeq \overline{Q(T)}$. But $C$ and $\overline{Q_p}$ have the same cardinality, and hence, so do $S$ and $T$. Therefore, $Q(S)\simeq Q(T)$ and, from there, $C\simeq \overline{Q_p}$. 
For myself, this proof is quite convincing. Recently, Torsten Ekedahl expressed his opinion to the contrary, and this led to
the following exchange:
Why worry about the axiom of choice?
So I wondered about other expert opinions on this matter. Do you find the isomorphism unbelievable and, if so, why?

REPLY [11 votes]: I don't think one should try to determine whether to accept or reject the axiom of choice or any other independent axiom by appealing to "believability" of some consequences of it. With infinite sets, our intuition is just too often misleading. We get used to certain "paradoxa" like Hilbert's hotel because we see them very early in our mathematical life, but nobody should ever claim that he has a complete intuition for set theory.
As for the example, $\bar{\mathbf{Q}}_p$ and $\mathbf{C}$ are isomorphic if the axiom of choice is true, and that's that. Both are constructed using a completion, which makes them topological fields, and they are not homeomorphic or normed isomorphic, that's probably why it feels a bit wrong to some of us.