cardinals implied by Replacement

Question

I just want to make sure I have my head around something. Replacement implies the existence of a cardinal number, k, larger than all f(i), where f(0) = Aleph_0 and f(i + 1) = Aleph_f(i). So, just to confirm, f(0) is omega, and while f(1) is Aleph_omega. Is that right? Man that function grows fast!

Answer 1

Yes, your argument is right: Replacement implies that such cardinals exist.

(Incidentally, Replacement is necessary for this - without Replacement, we can't even prove that the ordinal $\omega+\omega$ exists!)

More generally, Replacement implies:

For any class function $F$ on ordinals (= formula $\varphi$ in two free variables with parameters such that for each ordinal $x$ there is exactly one ordinal $y$ satisfying $\varphi(x, y)$) satisfying $F(x)>x$ for all $x$ and $F(\sup(A))=\sup(F(A))$ for every set of ordinal $A$ (this is equivalent to $F$ being continuous in the order topology on the class of ordinal), there are arbitrarily large fixed points of $F$.

(Note that continuity is necessary: e.g. there is of course no fixed point of the successor function $\lambda\mapsto\lambda^+$.)

The proof is the following: supposing $F$ is an increasing continuous class function and $\lambda$ is an arbitrary ordinal, consider the sequence $$\lambda, F(\lambda), F(F(\lambda)), ...$$ which exists by Replacement. Let $\kappa$ be the supremum of the sequence. Since $F$ is increasing, $\kappa$ is a limit ordinal $>\lambda$; since $F$ is continuous, we have $$F(\kappa)=\sup\{F(\eta): \eta<\kappa\}=\sup\{F^n(\lambda): n\in\omega\}=\kappa$$ (with the second equality holding since $\{F^n(\lambda): n\in\omega\}$ is cofinal in $\kappa$ by definition of $\kappa$).

---

In your case, you're looking at the function $f:\alpha\mapsto \aleph_\alpha$. It's a good exercise to check that in fact $f$ satisfies the continuity condition above.