User blog comment:BlauesWasser/Why Zero Shouldn't be considered a cardinal/@comment-30754445-20180501091716/@comment-27513631-20180504195555

That's an odd way of thinking about it.

Yes, you can define 1 to be {0}, and 2 to be {0, 1}, but you can also define 3 to be {4} and 4 to be {1, 2} (or {1, 3} with AFA).

Personal opinion, but I think it's the opposite of helpful to think of 1 to be {0} or 3 to be either {2} or {0, 1, 2}; the opposite of helpful to consider (a, b) to be {{a}, {a, b}}. There are two reasons for this:


 * It breaks abstraction. If you think of these as actually true, then it makes sense to ask whether \(\cdot/V_{43}\) is a manifold, where \(\cdot\) is the multiplication operation on the semiring of hyperreals.
 * It's unnecessary. As in, you literally don't need to do anything like this, at all. If you work in type theory, and have the \(\cong\to =\) fragment of the Univalence Axiom (which can only be formulated in type theory), you not only can't ask the above question, due to it not making sense structurally, but can't ask any question which depends on the encoding. All properties are isomorphism-invariant. And yes, you can still encode "normal" sets.

if you want me to do a blog post explaining this in more detail, I'd be happy to. I'm tempted to make one for inductive types (which really really should be well-known here, really) anyway.

Oh, and to keep on the topic of the question, that's actually true in a lot of places, not just cardinal arithmetic. Take any (commutative) semiring (which roughly contains exactly the rules of finite cardinal arithmetic) - you have \(a+0 = 0 = 0+a\) and \(a0 = 0 = 0a\). You can even have everything equal \(0\) - this is aptly named the zero semiring.