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    $\begingroup$ Wait, the one that's above the real line is $i$, and the one below is $-i$, right? ;) $\endgroup$ Commented Aug 16, 2013 at 1:26
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    $\begingroup$ If you construct $\mathbb{C}$ as $\mathbb{R}^2$ with product $(a,b)\cdot (c,d):=(ac-bd,bc+ad)$, then $i:=(0,1)$ is a standard definition. Also if you construct $\mathbb{C}$ as $\mathbb{R}[x]/(x^2+1)$ you have a standard choice: $i:=x\mathrm{mod}(x^2+1)$. $\endgroup$ Commented Oct 8, 2013 at 23:13
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    $\begingroup$ Another fine equivalence to AC: every set has a unique cardinality. This and your first example are the main things that convince me AC is not so strange. $\endgroup$ Commented Oct 10, 2013 at 14:59
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    $\begingroup$ @NeilToronto: “every set has a unique cardinality” can be phrased in a lot of different ways, plenty of which are provable without choice (e.g.: there is a class $\mathbf{Card}$ and a “cardinality” map $V \to \mathbf{Card}$, such that sets have the same cardinality iff they are isomorphic). The only versions of the statement I know that are equivalent to AC are ones which insist that each cardinality should be represented by some ordinal — but this is a very thinly veiled version of the well-ordering principle, and I think not at all so intuitively obvious. $\endgroup$ Commented Oct 10, 2013 at 17:21
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    $\begingroup$ It isn't natural to believe that a product of nonempty sets is nonempty once you generalize a bit: an $I$-indexed family of sets $\{ J_i \}_{i\in I}$ is the same as an epimorphism $J \to I$, $J = \sum_{i\in I} J_i$. An element of a product of this family is the same as a section of this epimorphism --- and of course epimorphisms in categories can have no sections! This is true even in categories that are a model of (extensional) set theory, i.e. in toposes like a category of sheaves of sets on a space. E.g. for any nontrivial manifold $X$ its $Sh(X)$ will not satisfy AC. $\endgroup$ Commented Jan 23, 2018 at 18:22