3
$\begingroup$

So the large cardinal axioms are for the most part considered to be linearly ordered by consistency strength. For the large cardinals between extendibility and rank-into-rank (i.e. the $n$-huge cardinals), by using a little help from algebra, I have formulated large cardinal axioms where I have no idea about the consistency strength of these axioms.

Suppose that $\gamma$ is a limit ordinal. If $f$ is a function, then define $f\upharpoonright_{\gamma+1}:V_{\gamma+1}\rightarrow V_{\gamma+1}$ be the function defined by $f\upharpoonright_{\gamma+1}(x)=f(x)\cap V_{\gamma}$ for each $x\subseteq V_{\gamma}$. Define $\mathbf{EE}_{\gamma}$ to be the collection of all functions $f:V_{\gamma+1}\rightarrow V_{\gamma+1}$ such that for each $\alpha>\gamma+1$ there is a $\beta$ and an elementary embedding $j:V_{\alpha}\rightarrow V_{\beta}$ with $j\upharpoonright_{\gamma+1}=f$.

Then $\mathbf{EE}_{\gamma}$ can be endowed with a self-distributive operation $*$ defined by $f*g=j(k)\upharpoonright_{\gamma+1}$ where there are ordinals $\gamma\leq \delta<\alpha\leq\beta$ with $j:V_{\alpha}\rightarrow V_{\beta},k:V_{\gamma}\rightarrow V_{\delta}$ and where $j\upharpoonright_{\gamma+1}=f,k\upharpoonright_{\gamma+1}=g$.

If $X$ is a finite self-distributive algebra, then let $\mathrm{LC}(X)$ be the axiom that posits that there is some limit ordinal $\gamma$ such that $\mathbf{EE}_{\gamma}$ has a subalgebra isomorphic to $X$.

For example, if $(X,*)$ is the algebra where $X=\{0,\dots,n\}$ and $x*y=1$ whenever $x\leq y$ and $x*y=y$ whenever $y<x$, then $\mathrm{LC}(X)$ is equivalent to the existence of $n$ extendible cardinals.

The axioms of the form $\mathrm{LC}(X)$ are large cardinal axioms, but there does not seem to be any natural linear pre-ordering on the class of all finite self-distributive algebras $\leq$ where if $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are both consistent, then $X\leq Y$ if and only if $\mathrm{Con}(\mathrm{LC}(X))\rightarrow\mathrm{Con}(\mathrm{LC}(Y)).$ If large cardinal axioms are supposed to be linearly ordered by consistency strength, then why does there seem to be no linear ordering on the axioms of the form $\mathrm{LC}(X)$?

  1. Are the consistent axioms of the form $\mathrm{LC}(X)$ linearly ordered by consistency strength?

  2. Is the collection of consistent axioms of the form $\mathrm{LC}(X)$ well-founded when ordered by consistency strength?

  3. Is the problem of determining whether $\mathrm{LC}(X)$ has greater consistency strength than $\mathrm{LC}(Y)$ computable in polynomial time?

  4. Is the problem of determining whether $\mathrm{LC}(X)$ has consistency strength equal to $\mathrm{LC}(Y)$ computable in polynomial time?

  5. Does there exist a consistent large cardinal axiom of the form $\mathrm{LC}(X)$ which does not follow from the existence of a non-trivial elementary embedding $j:V_{\lambda}\rightarrow V_{\lambda}$?

  6. If $\mathrm{LC}(X)$ is consistent, then is it also consistent that there exists some cardinal $\lambda$ and a limit ordinal $\gamma$ where $X$ is isomorphic to some subalgebra of $\mathcal{E}_{\lambda}/\equiv^{\gamma}$?

  7. If $\mathrm{LC}(X)$ is consistent, then is $\mathrm{Con}(\mathrm{LC}(X))$ a consequence of the existence of an $n$-huge cardinal for some $n$?

  8. If $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are consistent, then is the conjunction of $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ also consistent? If $\mathrm{LC}(X)$ and $\mathrm{LC}(Y)$ are consistent, then does there exist an algebra $Z$ that contains both $X,Y$ as subalgebras and where $\mathrm{LC}(Z)$ is consistent?

  9. If the axioms of the form $\mathrm{LC}(X)$ are well-ordered by consistency strength and if the problem $\mathrm{Con}(\mathrm{LC}(X))\rightarrow\mathrm{Con}(\mathrm{LC}(Y))$ is computable , then what is the order type of this well-ordered set?

The axioms $\mathrm{LC}(X)$ are strengthenings of the notion of extendibility. I would also like to eventually see other large cardinal notions generalized to their $\mathrm{LC}(X)$ versions and their consistency strength levels compared so that we can have a much finer picture of the large cardinals between extendibility and the rank-into-rank cardinals.

$\endgroup$
  • $\begingroup$ I see that this question has downvotes. Is there anything technically incorrect about this post that I need to be aware of? Or are the downvotes simply from people who do not like algebra or something like that? $\endgroup$ – Joseph Van Name Feb 24 at 13:43
  • 1
    $\begingroup$ There can be a problem with asking so many questions in a single post; a general rule of thumb is that this should be avoided, and that asking questions one at a time is usually better. It's true that this rule of thumb is often flouted, but one problem is that no one answer is likely to address all the questions (cf. the comments under this post: mathoverflow.net/a/15857/2926), and another is that such questions lend themselves to long discussions which SE software is meant to discourage. (No doubt there are differing views on the seriousness of these problems.) $\endgroup$ – Todd Trimble Feb 24 at 18:36
  • $\begingroup$ @ToddTrimble. I see. So perhaps I should have put a main but open ended question "Can someone provide a description or possibly an algorithm of the function $X\mapsto\mathrm{Con}(\mathrm{LC}(X))$?" and have a few of the nine questions as specific facts I would like to know about the function and which should be deduced from a description of the function? $\endgroup$ – Joseph Van Name Feb 24 at 19:49
  • $\begingroup$ Joseph, I'm not sure what the best solution is. Your proposal might work, but my gut feeling (and I say this as a non-set-theorist) is that probably any one of these questions, say 1. to begin with, is going to be hard and would probably make a good MO question on its own. I'd actually be pleasantly surprised if someone at MO could answer such a question, as I would bet that among active MO members, you are currently the number 1 expert on these sorts of issues. $\endgroup$ – Todd Trimble Feb 25 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.