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If $|\mathbb{C}|=|\mathbb{R}|$ and thus $\mathbb{C}$ is isomorphic to $\mathbb{R}$, why can't $\mathbb{C}$ be totally ordered?

Cam White
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    there is a lexicographic order for $\mathbb C$, but it's not compatible with multiplication; cf. this answer – J. W. Tanner May 12 '20 at 00:47
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    It can be (assuming AC). But there's no way to do it in a way that is compatible with the field operations. For instance. Is $i$ positive, or negative? Either way you'll reach a contradiction by multiplying by $i$ repeatedly – HallaSurvivor May 12 '20 at 00:47
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    @HallaSurvivor: you don't need AC if you don't insist on a well order. – Ross Millikan May 12 '20 at 00:53
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    Note that the word "isomorphic" requires context. It means there is a bijection preserving some structure, but what structure? They are isomorphic as sets, but not as fields. There exist bijections between them, but those bijections cannot be field isomorphisms; they cannot preserve the algebraic structure. – Nate Eldredge May 12 '20 at 01:48
  • @NateEldredge They are isomorphic as vector spaces over $\mathbb Q$. :) – Nothing special Mar 03 '24 at 08:28

1 Answers1

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$\mathbb{C}$ can indeed be totally ordered as a set, since you can find a set bijection to the reals. But for a field the definition of a total order requires that all squares be positive. Since $1$ is positive and $-1$ is negative the fact that $i^2 = -1$ implies that no total order exists.

Ethan Bolker
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