User blog comment:Vel!/Ultracardinals/@comment-5150073-20140619060630

''Ultracardinals are larger than Reinhardt cardinals, so they are inconsistent with the axiom of choice. However, not only are they inconsistent with ZFC, but their existence is also inconsistent with ZF's axiom of extensionality! However, unlike the Reinhardt cardinals, it is easy to confirm the existence of ultracardinals. We're unsure about whether Reinhardt cardinals exist, but we can be certain with ultracardinals.''

I see no proof for this.

Ultracardinals are strong enough that the Rathjen psi function fails to return any value with an ultracardinal as its input.

Psi function fails to return value from U, because it is just undefined for it. If I define U as $$\psi_U(0) = K_{K_{K_{\cdots}}}$$, I see no problems.

''But we can go further than U by devising higher-order ultracardinals. If we dive into the definition and redefine C(α)=the order-2 cofinality of α, we get the order-2 ultracardinals, the smallest of which is U 2. The μ function must be 2-argument in order for us to collapse these cardinals. Extensions are fairly intuitive:''

If the way in which your cardinal is really so strong, these naive extensions are no better than adding 1. By the way, we can do the same thing letting $$\psi(\Omega_\alpha) = \psi(\Omega,\Omega)$$ and extending through $$\psi(\Omega,\Omega,\Omega), \psi(\Omega,\omega (1) 2), \psi(\Omega,\omega (\Omega) 2)$$, etc. and we'd come nowhere close to U.