Let $\kappa$ be a measurable cardinal (or other large cardinal) and let $j:V\longrightarrow M$ be a witness. We know that $j(\kappa)$ has large cardinal properties in $M$, but what about $j(\kappa)$ in $V$?

Let me give a couple tentative nonstandard definitions:

  • Call a measurable cardinal $\kappa$ weakly compact preserving, if there is an elementary embedding $j$ witnessing measurability of $\kappa$, such that $j(\kappa)$ is weakly compact in $V$.

  • Call a measurable cardinal $\kappa$ reflecting preserving if $V_\kappa\prec V_{j(\kappa)}$, again in $V$.

What I am really interested in is a weakly compact preserving–reflecting preserving measurable cardinal $\kappa$ such that there is a weakly compact cardinal $\lambda>\kappa$ such that $V_\lambda\prec V_{j(\lambda)}$.

Question: What is the place of such a cardinal in the large cardinals hierarchy?


2 Answers 2


If $\kappa$ is measurable and there is a weakly compact cardinal $\lambda$ above, then there is an elementary embedding $j:V\to M$ with critical point $\kappa$ and $j(\kappa)=\lambda$. The reason is that one may simply iterate a normal measure, which pushes $j(\kappa)$ higher, until it hits that $\lambda$. The same argument works with any kind of large cardinal $\lambda$. But the embedding $j$ is not an ultrapower embedding. Of course, with an ultrapower embedding by an ultrafilter on $\kappa$, the value of $j(\kappa)$ is never a cardinal in $V$, since it is strictly between $2^\kappa$ and its successor.

For your second question, if $V_\kappa\prec V_\lambda$ for a class club of $\lambda$, then again we can find $j:V\to M$ with $j(\kappa)$ being one of those $\lambda$, by iterating. The consistency strength of that situation is strictly higher than a proper class of measurable cardinals, since $V_\kappa$ itself must be a model of that theory. But it is less than a stationary proper class of measurable cardinals, since from that assumption one can find a model with a measurable cardinal $\kappa$ with $V_\kappa\prec V$ and indeed a stationary class of measurable $\lambda$ above which form an elementary chain $V_\kappa\prec V_\lambda$.

  • $\begingroup$ Thank you, it is nice, I need time to take a careful look at. $\endgroup$
    – Rahman. M
    Mar 22, 2017 at 11:27
  • $\begingroup$ What about the situation one wants to control cardinals above critical point. I mean $\kappa$ is measurable, $\lambda>\kappa$ is weakly compact and $V_\kappa\prec V_{j(\kappa)}$ and $V_\lambda\prec V_{j(\lambda)}$ such that $j(\lambda)$ is inaccessible or at least wordly. $\endgroup$
    – Rahman. M
    Mar 24, 2017 at 9:29
  • 1
    $\begingroup$ Note that if we relax the phrasing of the first question a bit, we can get positive results: Let $U$ be a 'sufficienlty nice' measure on $\kappa$ (e.g. be derived from an ultrapower embedding) and let $i_U \colon L \to L$ be the restriction of the ultrapower embedding to $L$. This itself is an ultrapower embedding of the structure $(L; \in, U \cap L)$ (i.e. definable in this structure by the usual construction) and in there $i_U(\kappa)$ is weakly compact. $\endgroup$ Mar 30, 2017 at 15:43

If $j$ is a superstrongness embedding, $j(\kappa)$ is a large cardinal:

  • If $j : V \to M$ witnesses that $\kappa$ is superstrong (that is, $V_{j(\kappa)} \subset M$) then $j(\kappa)$ is worldly since $V_{\kappa} \prec V_{j(\kappa)}$ but not generally inaccessible, for by theorem 3.3 of Perlmutter 2013, a cardinal that is high-jump for strongness, which is implied by superstrongness, is superstrong with target equal to the clearance of the high-jump for strongness embedding and by lemma 3.1 of the same paper the clearance is singular.
  • If $j : V \to M$ witnesses that $\kappa$ is almost huge (that is, $\phantom{M}^{\lt j(\kappa)}M \subset M$), then $j(\kappa)$ is inaccessible since any short cofinal sequence would be a subset of $M$ of cardinality less than $\kappa$, and even $\Sigma_\omega$-Mahlo (defined by Bosch 2006, which seems to have become paywalled after I last read it) since $V_{\kappa} \prec V_{j(\kappa)}$ and $V_{\kappa} \vDash \text{"Ord is Mahlo"}$, but not generally Mahlo as the characterization in lemma 5.4 of Sato 2007, minus inaccessibility, is satisfied by a club of cardinals $\lambda \lt j(\kappa)$ in place of $j(\kappa)$ so if $j(\kappa)$ is Mahlo, there are stationarily many $\lambda \lt j(\kappa)$ that are almost-hugeness targets of critical point $\kappa$.
  • If $j : V_{\kappa+1} \to V_{j(\kappa)+1}$ witnesses that $\kappa$ is 1-extendible (by the coding argument of this Mathoverflow answer by Gabe Goldberg, 1-extendibility can equivalently be characterized by an elementary embedding $j : H_{\kappa^+} \to H_{j(\kappa)^+}$) then $\kappa$ is Woodin, locally measurable (defined by Holy and Lücke 2020) and thus strongly Ramsey (defined by Gitman 2008), and also $\Pi^1_n$-indescribable for every $n \lt \omega$ since those are properties of $H_{\kappa^+}$ so elementarity of $j$ applies. However, $j(\kappa)$ is not generally super-Ramsey (defined by the same paper by Gitman) or even super weakly Ramsey (defined by Holy and Schlicht 2018) since for any $k : M \to N$ as in the definition of super weakly Ramsey such that $j \in M$, $N \vDash \text{"$\kappa$ is 1-extendible with target $j(\kappa)$"}$ as $H_{j(\kappa)^+}^N=M \prec H_{j(\kappa)^+}$. They are probably not generally $\Pi^2_1$-indescribable as 1-extendibility targets are probably $\Delta^2_1$-definable since the satisfaction predicate for $V_{j(\kappa)+1}$ is $\Delta^2_1$-definable and being a 1-extendibility target involves no other third-order quantifiers, and probably not generally completely ineffable as that property has a characterization involving transitive models $M \prec H_{\beth_{j(\kappa)+1}^+}$ (Nielsen and Welch 2018, theorem 3.12).
  • If $\kappa$ is subcompact (that is, for every $A \subset H_{\kappa^+}$ there is a 1-extendibility embedding $j : (H_{\lambda^+}, B) \to (H_{\kappa^+}, A)$) then, in addition to the properties of 1-extendibility targets, $\kappa$ is fully Ramsey (defined by Holy and Schlicht 2018), for if there was a winning strategy $\sigma$ for player I in the Holy-Schlicht game (and whether a strategy is winning is definable over $H_{\kappa^+}$ since any play of the game is an element of $H_{\kappa^+}$), then, by subcompactness, there is an elementary embedding $j : (H_{\lambda^+}, \tau) \to (H_{\kappa^+}, \sigma)$ so that $\tau$ is a strategy witnessing that $\lambda$ isn't fully Ramsey, which is impossible since $\lambda$ is 1-extendible. However, $\kappa$ is not generally measurable for if $k : V \to M$ witnesses that $\kappa$ is measurable, $M \vDash \text{"$\kappa$ is subcompact"}$ because $M$ contains the true $H_{\kappa^+}$ and every $A \subset H_{\kappa^+}$ in M is in V, and not generally $\Pi^2_1$-indescribable as subcompactness is $\Pi^2_1$-definable.
  • If $\kappa$ is quasicompact (that is, for every $A \subset H_{\kappa^+}$ there is a 1-extendibility embedding $j : (H_{\kappa^+}, A) \to (H_{j(\kappa)^+}, j(A))$) then some of the targets of the witnessing embeddings are measurable, as can be seen by taking $A$ to be a measure witnessing that $\kappa$ is measurable.
  • If $j : V \to M$ witnesses that $\kappa$ is 2-fold 1-strong (that is, $V_{j(\kappa)+1} \subset M$; Sato 2007, definition 9.1) then $j(\kappa)$ is measurable since, if $U$ is a measure witnessing that $\kappa$ is measurable, $j(U)$ witnesses that $j(\kappa)$ is measurable as it measures all subsets of $j(\kappa)$.
  • If $j : V \to M$ witnesses that $\kappa$ is huge (that is, $\phantom{M}^{j(\kappa)}M \subset M$), then $j$ also witnesses that $\kappa$ is 2-fold 1-strong, so $j(\kappa)$ is measurable.

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