Here are two pretty standard results about elementary embeddings that I don't understand.

(1) Let $\pi:V \to M$ be a non-trivial elementary embedding with $M$ a transitive class. Let $crit(\pi)=\kappa$. Let $C \subset \kappa$ be a club in $\kappa$. Then, we know $\pi(C)$ is a club in $\pi(\kappa)$ by elementarity of $\pi$. Also, since $\pi(\alpha)=\alpha$ for all $\alpha < \kappa$, $\pi(C) \cap \kappa = C$. So, $\pi(C) \cap \kappa$ is unbounded in $\kappa$, so $\kappa \in \pi (C)$ since it is closed.

My problem is understanding what's going on when $\kappa \in \pi (C)$. If this is the case doesn't there have to exist some $\alpha \in C$ such that $\pi(\alpha)=\kappa$, which there isn't? Something similar happens in (2):

(2) The following is an initial part of a proof of Kunen's theorem that there is no non-trivial elementary embedding $\pi:V \to V$. Let $\kappa_0 = crit(\pi)$. Recursively define $\kappa_{n+1}=\pi(\kappa_{n})$. Set $\lambda = \sup_{n<\omega}\kappa_n$. Let $S=\{\alpha < \lambda^+: cf(\alpha)=\omega\}$. S is stationary, so take it apart into $\kappa_0$ disjoint stationary subsets $S_i , i<\kappa_0$. Then, the proof goes, set $(T_i: i<\kappa_1)=\pi((S_i: i<\kappa_0))$. I don't see how this last part is possible. Where does, for example, $T_{\kappa_0}$ come from if it's not in the range of $\pi$? I'm not even sure what $T_{\kappa_0}$ is.


1 Answer 1


For the first question, the elementary embedding is "jumping" at $\kappa$. Namely, the universe of set theory is the same all the way below $\kappa$, and then it get stretched up, somehow. This is the meaning that $\pi(\kappa)>\kappa$.

But at the same time, $M$ is a transitive class. So if $\pi(\kappa)>\kappa$, then $\kappa\in M$.

The key, I think, for understanding this (and it took me several years to grasp that fine point), is that the elements of the ultrapowers are really functions from $\kappa$ to $V$. The embedding just maps $x$ to a constant function $c_x$. To say that $\kappa$ is not in $\pi(\alpha)$ is to say that it is not represented by a constant function.

So, what does that have to do with anything? Well, $\kappa$ is a club in $\kappa$. So the question why $\kappa\in\pi(C)$ is a generalization for why $\kappa<\pi(\kappa)$. Since $\pi''\kappa\neq\pi(\kappa)$, it is natural that $\pi''C\neq\pi(C)$. $C$ gets "stretched" up.

For the second question, Think about $\vec S$ as a function from $\kappa_0$ to stationary subsets of $\lambda$, then $\pi(\vec S)$ is a function from $\pi(\kappa_0)=\kappa_1$ to stationary subsets of $\pi(\lambda)=\lambda$.

This is again the same issue, the function does not operate pointwise, it moves things around, and if need be, it gives us new information. It's just something to get used to, I suppose.

  • $\begingroup$ Is there a problem with the answer that is merits a downvote? $\endgroup$
    – Asaf Karagila
    Jan 19, 2015 at 20:39
  • $\begingroup$ I upvoted, the answer helped. It is just so strange to me that functions don't operate pointwise... I guess it's something to get used to. Can anything concrete be said of $T_{\kappa_0}$, or in general $T_\alpha$ where $\alpha$ is not in the range of $\pi$? I don't even know what they are. If I wanted to know what elements in M are elements of $T_\alpha$, what would I do? $\endgroup$
    – Taro
    Jan 19, 2015 at 20:44
  • $\begingroup$ Mmm... that's a great question. I'm not sure what to tell you. $\endgroup$
    – Asaf Karagila
    Jan 19, 2015 at 20:48

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