# I Should Probably Explain the Title

The url for this blog is actually a bad math joke. So, I’m using this space to explain it.

We say that $\kappa$ is a Reinhardt cardinal if there exists an elementary embedding $j: V \rightarrow V$ with $crit (j)=\kappa$. The following result is due to Kunen, but the proof we will be  giving of the result is due to Zapletal. We will work in Kelly-Morse set theory.

Theorem (Kunen): If $j: V\rightarrow M$ is a nontrivial elementary embedding, then $V\neq M$.

Another way of saying this is that the theory “ZFC + There exists a Reinhardt Cardinal” is inconsistent. So, that’s the joke (also explains the tagline). Of course, there was some reason to even discuss the existence of such a cardinal, else no one would have even considered it, and it wouldn’t be a big deal that the existence of such a thing is inconsistent with ZFC.

We say that a cardinal $\kappa$ is measurable if there exists a nontrivial elementary embedding $j: V \rightarrow M$ such that $crit(j)=\kappa$.

For an ordinal $\alpha$, a cardinal $\kappa$ is said to be $\alpha$strong if there exists a nontrivial elementary embedding $j: V\to M$ such that $crit(j)=\kappa$, $j(\kappa)>\alpha$, and $V_{\kappa +\alpha}\subseteq M$. We say that $\kappa$ is strong if $\kappa$ is $\alpha$-strong for every ordinal $\alpha$.

We say that a cardinal $\kappa$ is $\lambda$supercompact (for a cardinal $\lambda\geq\kappa$) if there exists a nontrivial elementary embedding $j: V \rightarrow M$ such that $crit(j)=\kappa$, $j(\kappa)>\lambda)$, and ${}^\lambda M\subseteq M$. We say that $\kappa$ is supercompact if it is $\lambda$-supercompact for every cardinal $\lambda\geq \kappa$.

We see here that we can ask for stronger and stronger large cardinal axioms by asking for elementary embeddings that agree more and more with $V$. So, the natural step was to ask for the embedding to just go straight to $V$. We will now prove Kunen’s inconsistency theorem above using the following characterization of successors of singular cardinals of countable cofinality due to Shelah.

Theorem (ZFC; Shelah): Let $\lambda$ be a singular cardinal of countable cofinality. Then, there exists a countable set $C\subseteq \lambda$ of regular cardinals such that $C$ is cofinal in $\lambda$, and, letting $J$ denote the ideal of bounded subsets of $C$,

$\lambda^+=tcf(\prod C/J)$.

That is, there exists a sequence of functions $\langle f_\xi : \xi<\lambda^+\rangle$ in $\prod C$ that is increasing and cofinal in $\prod C$ modulo $J$.

The above theorem can be found as an exercise in Abraham and Magidor’s chapter in the Handbook of Set Theory. I’ll probably type out the proof in full at some point just for the sake of completeness, but the focus here is not on pcf theory. I will now give Zapletal’s proof of Kunen’s theorem.

Suppose the theorem is false. That is, there exists a nontrivial elementary embedding $j:V\to V$, and let $\kappa=crit(j)$, the critical point of $j$. Now define a sequence $\langle \kappa_i : i\in\omega\rangle$ by $\kappa_0=\kappa$, and $\kappa_{i+1}=j(\kappa_i)$. Of course, by elementarity, this sequence is strictly increasing, and each $\kappa_n$ is a cardinal in $V$. Let $\lambda=\sup_{i\in\omega}\kappa_i$, and note that $\lambda$ is a singular cardinal of countable cofinality. Since $j$ is continuous at ordinals of cofinality $<\kappa$, we see that $j(\lambda)=\lambda$, and by elementarily $j(\lambda^+)=\lambda^+$. In fact, it is straightforward to check that $\lambda$ is the first fixed point of $j$ above $\kappa$.

Using the above theorem, fix a sequence $C=\{\lambda_i : i\in\omega\}$ of regular cardinals cofinal in $\lambda$ such that

$\lambda^+=tcf\Big(\prod C/J\Big).$

Again, here $J$ denotes the ideal of bounded subsets of $C$. We may assume, without loss of generality, that $\kappa<\lambda_i<\lambda$ for each $i\in\omega$. Let $f=\langle f_\xi : \xi<\lambda^+\rangle$ be $<_J$-increasing and cofinal in $\prod C/J$. By elementarity, $j(f)$ is cofinal and increasing in $j(\prod_{i\in\omega} \lambda_i)$ modulo the ideal of bounded subsets. Define $g\in\prod C$ by $g(i)=\sup j''\lambda_i$. Since $j(\lambda_i)$ is regular for each $i\in\omega$, and $\kappa<\lambda_i<\lambda$, it follows that $g(i) (here we use that $\lambda$ is the first fixed point of $j$ above $\kappa$).

Now, let $h\in \prod C$, and note that $g$ dominates $h$ everywhere by definition. We also note that by elementarity, $j''\lambda^+$ is cofinal in $j(\lambda^+)=\lambda^+$. Thus, we also see that $j'' f$ is cofinal in $j(f)$, and therefore cofinal in $\prod_{i\in\omega} j(\lambda_i)$ modulo the ideal of bounded subsets of $\{ j(\lambda_i) : i\in\omega\}$. But, $j(f_\xi)=j\circ f_\xi$, and thus it follows that $g$ dominates each $f_\xi$ everywhere, contradicting that $j(f)$ is cofinal in $j(\prod C/ J)$.