# Definition talk:Regular Cardinal

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From what I understand, each of $0, 1, 2$ are also regular. I believe it is not necessary to specify that a regular cardinal is infinite. --prime mover (talk) 08:19, 9 September 2012 (UTC)

- For the sake of working with inaccessible cardinals, it is good to have a definition of regular cardinals that only work with infinite cardinals. That way, we can prove that the von Neumann hierarchy associated with any weakly inaccessible cardinal is a model of ZFC. This will all be worked out and justified as new theorems are added.

- Moreover, $2$ would not actually be regular, even if we dropped the infinite cardinal proviso, since its cofinality is $1$. But you are correct in saying that $\operatorname{cf} (0) = 0$ and $\operatorname{cf}(1) = 1$. --Andrew Salmon (talk) 08:59, 9 September 2012 (UTC)

- Unless your definition of regularity differs from that in 1982: Rudy Rucker:
*Infinity and the Mind*, 2 is definitely regular: "you can't make 2 by aggregating no more than one 1's." --prime mover (talk) 12:16, 9 September 2012 (UTC)

- Unless your definition of regularity differs from that in 1982: Rudy Rucker:

- It must differ, then. See here: "the cofinality of any successor ordinal is $1$" --Andrew Salmon (talk) 17:10, 9 September 2012 (UTC)