complex numbers – Is $mathbb{C}$ still algebraically closed in ZF-models without the Axiom of Choice?

Is $mathbb{C}$ still algebraically closed in ZF-models without the Axiom of Choice (AC) ?

I know several results about algebraic closures of fields in models with AC become really delicate in ZF-models without AC.

So this must be one of the basic questions, I guess (but I cannot find a clear reference or definite proof online).

Side question: do finite fields have “unique” algebraic closures (i. e. up to isomorphism) in ZF-models without AC?

Why isn’t this an axiom of choice?

The standard axiom of choice can be expressed in the first order language of ZF by $$forall Xbig((Xneqemptysetwedgeemptysetnotin X)impliesexists f:Xtobigcup X forall Yin X(f(Y)in Y)big).$$ Naively, a simpler version would seem to be $$forall X(Xneqemptysetimpliesexists f:{X}to X)$$ since we then have that $forall Ain{X}(f(A)in A)$, as specified explicitly in the choice axiom. This version is provable in ZF according to the discussion on this answer, however, and is thusly not an axiom of choice. How exactly do we lose choice moving from the first formulation to the second?

lo.logic – Would the axiom of determinacy imply RH?

I read years ago in the great French book “l’aventure des nombres” (Gilles Godefroy, Odile Jacob, 1997) that the axiom of determinacy was incompatible with the axiom of choice. The latter seems to be needed to allow the existence of discontinuous field automorphisms of $C$. As some of you may know, I have been working for some time on an approach of RH based on automorphisms of rigs (rings without negative) whose maximal one I conjecture to correspond to “the” algebraic closure of $Q$.

In other words, denoting by $mathcal{M}$ the maximal “L-rig”, i.e. rig whose elements are L-functions, I expect $operatorname{Aut}(mathcal{M})$ to be isomorphic to $operatorname{Gal}(bar{mathbb{Q}}/mathbb{Q})$. The key idea is to define the symmetry group $operatorname {Sym}(F)$ of a given element $F$ of $mathcal{M}$ as made of elements of $operatorname {Aut}(mathcal{M})$ preserving $F$. Then we want to establish the isomorphy of $operatorname{Sym}(F)$ and the isometry group of the multiset of non trivial zeroes of $F$. The latter group is to be seen as the automorphism group of the Euclidean domain where this multiset lies, i.e. the critical strip. That way, one has to restrict the potential candidates in $operatorname{Aut}(mathcal{M})$ to elements of finite order or to continuous elements of this absolute Galois group. The elements of finite order are the identity and the conjugates of the complex conjugation. The continuous elements are the identity and the complex conjugation. Such a restriction seems rather artificial, and is not quite satisfying. So, my question is:

Would working with the axiom of determinacy instead of the axiom of choice provide a more natural framework for RH? Has this conjecture been proven to be provable in ZF without the requirement of the axiom of choice?

logic – Do FOL theorem provers accept axiom schemata?

Axiom schemata (such as ZFC) are, in a sense, infinite sets of axioms. Do the ATPs designed to work with FOL (such as Vampire) accept axiom schemata?

I looked in the Vampire “manual” briefly, but couldn’t answer this question. Is this possible? How would one do it in TPTP syntax? Are problems involving axiom schemata part of the CASC FOF division?

Axiom collapsed? 公理崩塌?

Pass a point that is not on a straight line, and there is only one straight line that does not intersect the straight line.

The concept of limit

Left limit and right limit

Isn’t there three lines parallel?

Did i get something wrong

Shouldn’t you use limit definitions to challenge axioms

Still logically wrong

If and only if thinking wrong

Or is the limit not understood like this?

Is there a mistake in my thinking?











dg.differential geometry – Geometric intuition for $R[x,y]/ (x^2,y^2)$, kinematic second tangent bundle, and Wraith axiom

This is a sort of continuation of this question.

In synthetic differential geometry (SDG), we have $Dsubset R$ comprised of the second order nilpotents. The Kock-Lawvere axiom (KL axiom) implies that a function $Dtimes Dto R^n$ is of the form $a_0+a_1d_1+a_2d_2+a_3d_1d_2$. This is like a 2-jet without square terms $f(a)+partial_xf |_ad_1+partial_yf|_ad_2+partial_{xy}f|_ad_1d_2$.

In SDG, the infinitesimal rectangle $Dtimes D$ represents the second tangent bundle. In light of the KL-axioms I expect the classical second tangent bundle $mathrm T^2X=mathrm {TT}X$ of a $C^infty$ manifold admits the following kinematic description: elements are equivalence classes of germs of $C^infty$ maps $I^2to X$ where $I$ is an interval about zero, and we identify such germs if upon composing with any germ in $C_{X,x}^infty$ the partials and mixed partials coincide. Let us call such things “microsquares”. They formalize the “2-jets without square terms” above.

If correct, this kinematic description is very geometric. For instance, it allows to define the flip on $mathrm T^2X$ by flipping the $x,y$ coordinates of $I^2$. The two maps $mathrm T^2Xrightrightarrows mathrm TX$ given by $mathrm Tpi_X,pi_{mathrm TX}$ are respectively given by restricting a microsquare to the $x$-axis and the $y$-axis. These fiber $mathrm T^2X$ in two different ways: the fiber of $mathrm Tpi_X$ over a kinematic tangent $dot gamma$ consists microsquare which restricts to $gamma$ on the $x$-axis, and analogously for $pi_{mathrm TX}$.

The vertical lift applied to the tangent bundle gives a bundle isomorphism $mathrm T(mathrm TX/X)cong mathrm TXtimes_Xmathrm TX$ over $mathrm TX$, where the LHS is the vertical bundle of the tangent bundle, i.e the kernel of $mathrm Tpi_X$. For all vector bundles this acts by taking a kinematic tangent (to a fiber of the bundle) to its derivative (which is a vector in the fiber).

Question 1. How to geometrically interpret the vertical lift for a “vertical microsquare”? A microsquare lies in the vertical bundle if its restriction to the $x$-axis is “constant”, i.e the derivative of the restriction is zero. This is like saying the associated “2-jet without square terms” has $partial_xf|_a=0$. What is the vertical lift doing with a microsquare that only makes sense if its restriction to the $x$-axis is zero?

My question is motivated by another one about a seeming discrepancy between SDG and the classical $C^infty$ world:

  • In the $C^infty$ world, the vertical lift $
    mathrm T(mathrm TX/X)cong mathrm TXtimes_Xmathrm TX$
    is defined on any vertical microsquare. There is no further requirement for also being in the kernel of $pi _{mathrm TX}$ (restriction of a microsquare to its $y$-axis), and I see no reason for these kernels to coincide.

  • In SDG, the Wraith axiom says that a function $Dtimes Dto R^n$ which is constant on the axes uniquely factors through the multiplication map $Dtimes Dto D$. This factorization takes such a function to a tangent vector, and this is the analog of the vertical lift. The $C^infty$ version of being constant on the axes is having the $partial_x,partial_y$ coefficients of the ‘2-jet without square terms’ vanish $partial_xf|_a=0=partial_yf|_a$. The remaining mixed partial term indeed factors through the multiplication map because that’s how Taylor series are. The point is that the Wraith axiom asks for both partials to vanish, as opposed to the vertical bundle which involves only vanishing $partial_x$.

Question 2. What is going on here, geometrically? Why does SDG want both partials to vanish while the $C^infty$ world only cares about one of the partials?

Lastly and perhaps most fundamentally: I don’t understand the geometric meaning of a microsquare. I understand 2-jets since we retain the information of the Hessian, but retaining only the mixed partials – I don’t get it.

Question 3. What is the geometric content of a microsquare / an element in the second tangent bundle?

lo.logic – Is axiom of constructibility $V = L$ consistent with Tarski–Grothendieck set theory?

I wonder what is the relationship between ZF + $V = L$ and Tarski–Grothendieck set theory, because I haven’t found any bibliographic references.
If they are compatible, it is possible to introduce $V = L$ using Tarski–Grothendieck as basis instead of ZF.

database design – What axiom is for this functional dependency?

I have recently come across this example that is not in my textbook. I can find the type of axiom to infer in most of the functional dependency except this. Can anyone point me what is the axiom to infer this functional dependency?



I am thinking if union rule is applicable here where AB->C? However, my textbook example wrote for union rule, X -> Y and X -> Z then X->Y,Z. Does union apply to the LHS too?

set theory – Wiki for consequences of axiom of choice?

I raised the following question as part of another MO question, but I am following the suggestion of Nate Eldredge to make it a question in its own right.

For many years, there has a been a valuable web resource, hosted by Purdue, on the
Consequences of the Axiom of Choice. Unfortunately, the page is no longer functioning, as you will quickly discover if you try submitting a form number. The URLs have changed. I suspect that Purdue redesigned its website at some point, changing the URLs, and that since Herman Rubin died a couple of years ago, there is now nobody responsible for maintaining the Axiom of Choice page. I tried emailing a couple of random people in the Purdue mathematics department to find out if something could be done to revive the page, but have received no response.

I am wondering if there is a way to revive this resource, ideally in a way that will prevent it from suffering a similar extinction risk a few years down the line. Perhaps some people can turn the page into a wiki, much in the way the OEIS evolved from a person project of Neil Sloane’s into a wiki? Also, maybe someone reading this knows more than I do about the situation at Purdue and can comment on what would be involved in making the data publicly available again.

set theory – Why if ZF is consistent, neither the axiom of choice nor the generalized continuum hypothesis cannot be disproved?

I think it’s because since by godel incompleteness theorem ZF cannot be proved to be consistent (and therefore it can’t also be disproved to be consistent the consequent of the implication “if ZF is consistent, then so is ZFC and the generalized continuum hypothesis” cannot hold (since the antecedent cannot be proved to be true) or equivalently contrapositive of the implication “if ZFC and the generalized continuum hypothesis are not consistent, then ZF is not consistent” in this case since the consequent cannot be proved or disproved, one cannot say anything about the whole implication, correct?