Search Results

stronger than $V=L$ (i.e., imply it) and which exclude cardinals at an even smaller size than measurable ones. … (Perhaps the law of the excluded middle can be considered in the line of $V=L$ and Choice?) …

What reasons are there to believe in their truth, rather than merely in their consistency, or at most their truth in some transitive model? … One might be $V = L$, or even the axiom of restriction, but most set theorists dismiss them, as they rule out large large cardinals and inaccessible cardinals, respectively, and instead seek core models …

Working in ZF, it's well-known that for any $n \ge 2,$ the claim that there is a $\Sigma_n$ well-ordering of the universe is equivalent to the axiom $V=HOD.$ It seems natural to believe there should be … It would suffice to check that if there a $\Sigma_1$ well-ordering of $V,$ then all sets of ordinals are in $L.$ This is true for subsets of $\alpha$ for any $\alpha$ countable in $V,$ by applying Mansfield's …

$\vdots$ These are a part of arguments which could be used against large cardinal axioms, but many set theorists not only believe in the existence of large cardinals, but also refute every statement like … $V=L$ which is contradictory to their existence. …

Let $L=L(x,y)$ be the free Lie algebra generated by letters $x,y.$ For a vector subspace $V\leq L$ we denote by $[V,L]$ the vector space spanned by brackets $[v,l],v\in V,l\in L.$ A vector subspace $V\ … leq L$ is an ideal of $L$ if and only if $$V=V+[V,L].$$ Consider the following increasing sequence of vector spaces which starts from the ${\rm span}(x):$ $$V_0={\rm span}(x),$$ $$V_1=V_0+[V_0,L],$$ $$ …

I'm not sure if this is the kind of thing you're looking for, but the assumption $V=L$ settles most "interesting" questions in set theory. … However, most (though not all) set theorists don't believe that $V$ is "really" equal to $L$, so the "immunity to independence" enjoyed by $V=L$ doesn't excite people as much as you might naively expect …

retains its large cardinal property in L, so we get consistency with V=L. … These are all independent of ZFC+V=L, since they are independent of ZFC, and their truth is the same in V as in L. …

\end{cases} $$ It is easily seen that the variational formulation of the above equation is $$(\nabla u,\nabla v)_{L^2}+(u,v)_{L^2}=(f,v)_{L^2},\;\;\;\;\text{ for all } v\in H^1_{per}(\Omega):=\{v\in H^ … (I believe it's some eigenspace). How to represent solution in this case? …

I believe the main problem with $V=L$ is how it restricts reflection. It has been discovered that $V$ obeys several reflection properties, the most famous theorem. … And here we get to the root of the problem; $V=L$ postulates, in a sense, the finitude of mathematics; that every object has an elemntary characterization in terms of parameters and formulas, which just …

It is interesting to note that Sacks' proof of Theorem 3.1 from his paper gives an indication of what must happen if $V\neq L$ (most set theorists do not believe that $V$=$L$ is an 'acceptable' axiom): … Suppose $\forall x(x\in L\leftrightarrow\{e\}(x)\downarrow)$ and $V\neq L$. Then for some $b\notin L$, $\{e\}(b)\uparrow$. …

V=L isn't alone in this ...). … V=L, by contrast, is genuinely complicated. …

carelessness) I realized that I do not understand how the "enriched" composition morphism is constructed in $_AV$: $$ \circ_{L,M,N}:V(M,N)\otimes V(L,M)\to V(L,N). $$ I think that this must be the "lift … in $V$). …

and so you will be always able to define two consistent theories, one in which V=L and one in which $V\not =L$). … have models where V=L is true (when $\kappa$=lowest R), but that is not what I mean). …

I will assume $V \in L^\infty(\Omega)$ is smooth, $V>0$ and $g \in C^1(\partial \Omega)$. … {L^\infty}}{\lambda_1(\Omega)} \int_\Omega |\nabla u|^2 + 2\|V\|_{L^\infty} \int_\Omega w^2 \,, $$ hence $$ J(u) \ge \left(1-\frac{2\|V\|_{L^\infty}}{\lambda_1(\Omega)} \right) \int_\Omega |\nabla u|^2 …

Assume $V=L$ and let $\kappa$ be a Mahlo cardinal. … Is there any $\aleph_2$-Souslin tree in $L[G]?$ I assume the believe is that there are, but I don't know how to prove it. …