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        Questions tagged [large-cardinals]

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        Why does $p_{n}(i,1)=1$ so often where the polynomials $p_{n}$ are obtained from the classical Laver tables

        So I was doing some computer calculations with the classical Laver tables and I found polynomials $p_{n}(x,y)$ such that $p_{n}(i,1)=1$ for many $n$. The $n$-th classical Laver table is the unique ...
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        Infinite products of complex numbers or matrices arising from rank-into-rank embeddings

        I wonder what kinds of closed form infinite products of matrices, elements of Banach algebras, and complex numbers arise from the rank-into-rank embeddings. Suppose that $\lambda$ is a cardinal and $...
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        Growth rate of the critical points of the Fibonacci terms $t_{n}(x,y)$ vs $t_{n}(1,1)$ in the classical Laver tables

        The classical Laver table $A_{n}$ is the unique algebra $(\{1,\dots,2^{n}\},*_{n})$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ and $x*_{n}1=x+1\mod 2^{n}$ for all $x,y,z\in A_{n}$. Define the ...
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        Attraction in Laver tables

        If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x*x^{[n]}$. The motivation for this question comes from the following fact about ...
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        Amalgamation via elementary embeddings

        Can there exist three transitive models of ZFC with the same ordinals, $M_0,M_1,N$, such that there are elementary embeddings $j_i : M_i \to N$ for $i<2$, but there is no elementary embedding from $...
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        Multiple roots in the classical Laver tables

        The classical Laver table $A_{n}$ is the unique algebraic structure $$(\{1,\dots,2^{n}\},*_{n})$$ such that $x*_{n}1=x+1\mod 2^{n}$ and $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ for all $x,y,z\in\{1,...
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        Can we have $\sup\{\alpha\mid(x*x)^{\sharp}(\alpha)>x^{\sharp}(\alpha)\}=\infty$ in an algebra resembling the algebras of elementary embeddings?

        A finite algebra $(X,*,1)$ is a reduced Laver-like algebra if it satisfies the identities $x*(y*z)=(x*y)*(x*z)$ and if there is a surjective function $\mathrm{crit}:X\rightarrow n+1$ where $\mathrm{...
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        In the classical Laver tables, do we have $o_{n}(1)<o_{n}(2)$ for any $n>8$?

        The classical Laver table $A_{n}$ is the unique algebraic structure $(\{1,\dots,2^{n}\},*_{n})$ where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and where $$x*_{n}1=x+1\mod 2^{n}$$ for $x,y,z\in\{1,\...
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        66 views

        For each $n$ is it possible to have $\mathrm{crit}(x^{[n]}*y)>\mathrm{crit}(x^{[n-1]}*y)>\dots>\mathrm{crit}(x*y)$?

        Suppose that $(X,*,1)$ satisfies the following identities: $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$. Define the Fibonacci terms $t_{n}(x,y)$ for $n\geq 1$ by letting $$t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=...
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        26 views

        Vastness of inverse systems of Laver-like algebras

        Suppose that $(X,*,1)$ satisfies the identities $x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$. Then we say that $(X,*,1)$ is a reduced Laver-like algebra if whenever $x_{n}\in X$ for all $n\in\omega$, there is ...
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        Can we always extend a finitely generated reduced Laver-like algebra to a vast inverse system of Laver-like algebras?

        An $(X,*,1)$ that satisfies the identities $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$ is said to be a reduced Laver-like algebra if whenever $x_{n}\in X$ for $n\in\omega$, there is some $N\in\omega$ where $x_{...
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        653 views

        Higher $\infty$-categories

        Is there a reason we consider $\infty$-categories to be the $\omega^{th}$ step in the 2-internalization inside Cat (or enrichment over Cat if you prefer)* process made invertible above some finite ...
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        1answer
        206 views

        Locally presentable categories, universes, and Vopenka's principle

        Some aspects of the theory of locally presentable categories depends on set-theoretic assumptions. Namely, a large cardinal axiom known as Vopenka's principle implies several nice properties of such ...
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        122 views

        Ordering large cardinal axioms around the level of $n$-huge by consistency strength?

        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 ...
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        41 views

        Calibrating the strength of the quotients of subalgebras of the classical Laver tables

        Define an algebraic structure $A_{n}$ by letting $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ where $*_{n}$ is the unique operation such that $x*_{n}1=x+1\mod 2^{n}$ for $$x\in\{1,\dots,2^{n}-1,2^{n}\}$...

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