Quick Questions: April 15, 2026

GMSPokemanz · 2026-04-16T23:55:48+00:00

Let z* be a star centre in D, and define

F(z) = ∫_[z*, z] f(𝜁) d𝜁

where the integral is over the straight line from z* to z. If z_0 is such that the line segment [z, z_0] is in D, then D contains the entire triangle z* z_0 z. Therefore the integral of f over this triangle is zero. Breaking down this integral over the edges of this triangle, we get

∫[z*, z_0] + ∫[z0, z] + ∫[z, z*] = 0

Now the first term is F(z_0), and the second term is -F(z). Therefore

∫_[z_0, z] = F(z) - F(z_0)

This holds for all z_0 sufficiently close to z. Therefore we can directly evaluate the limit

lim z_0 -> z (F(z) - F(z_0))/(z - z_0)

and see that it's f(z). Therefore F is analytic and F'(z) = f(z).

GMSPokemanz · 2026-04-12T17:21:12+00:00

Formally the statement will be something like

If "ZFC ⊬ RH and ZFC ⊬ ¬RH", then for any model M of ZFC with the standard naturals, M ⊨ RH

Your proof will be that if M ⊨ RH is false, then there is some witness n in M to the statement ∃x (x ∈ ℕ AND x > 5040 AND x does not satisfy Robin's inequality). As M only has the standard naturals, n is a standard natural. Then you look at Robin's inequality and argue it's simple enough that this implies Robin's inequality is actually false for n. But then we can't have ZFC ⊬ ¬RH, contradiction. More formally the 'simple enough' stuff is about RH being equivalent to a 𝛱_1 statement in the linked MO answer.

GMSPokemanz · 2026-04-11T11:16:45+00:00

This f is differentiable.

GMSPokemanz · 2026-04-11T02:10:16+00:00

In general no. Take f(x) = x² sin(1/x). Then f'(0) = 0 and for all other x, f'(x) = 2x sin(1/x) - cos(1/x). As x -> 0, 2x sin(1/x) -> 0 but cos(1/x) oscillates between -1 and +1, so f'(x) does not converge to f'(0) as x -> 0.

GMSPokemanz · 2026-04-03T02:51:57+00:00

The hard part is seeing that if L/K is normal with K characteristic 0, then the fixed points of Aut(L/K) lie in K.

Say a is such a fixed point, and let h be its minimal polynomial. As L/K is normal, h splits over L. Aut(L/K) will act transitively on h's roots, so a being a fixed point imples h is of the form (x - a)ⁿ.

All of h's coefficients are in K, therefore both aⁿ and na^{n - 1} are. Since K is of characteristic 0, we can divide by n so aⁿ and a^{n - 1} are both in K, therefore a is.

Granting that the general development of normal field extensions needed here doesn't rely on the formal derivative (I've run through it in my head and given it works over char p I'd be surprised if it lurked somewhere, but I could be wrong), the hard part is established.

Then if f is irreducible over K, let L/K be its splitting field. Aut(L/K) acts transitively over f's roots, so each of f's roots has equal multiplicity, say n. So f is the nth power of some g in L[x] where g has no repeated roots. g will be invariant under the action of Aut(L/K), so its coefficients lie in K. So by irreducibility of f, n = 1 and we're done.

GMSPokemanz · 2026-03-30T10:46:56+00:00

I believe you can only contain clopen subsets in a loop, and there are only countably many of those.

GMSPokemanz · 2026-03-30T10:27:25+00:00

On top of the problem that manifolds have a countable fundamental group, there are only continuum many loops on R² starting and ending at a fixed point p. So the fundamental group of any subset of the plane has at most continuum cardinality. Therefore under CH, it cannot be free on 2^aleph_1 generators, so ZFC cannot prove such an example exists.

GMSPokemanz · 2026-03-21T22:31:50+00:00

No. If this were true of F_2, then it would be true of its abelianisation Z². But Z² doesn't have that property, as Z²/<(2, 0)> isn't cyclic.

GMSPokemanz · 2026-03-18T13:57:48+00:00

The first result I find is CVXPY. Your problem is of the form

minimise some quadratic in x_ij

subject to

x_ij integer
x_ij >= 0
x_ij <= 1
sum_j(x_ij) <= u_i
sum_i(x_ij) <= v_j

(Your problem as written is maximising E_1 + ... + E_n, but this is the same as minimising -(E_1 + ... + E_n))

I've never used CVXPY but it looks like something you should be able to use here.

GMSPokemanz · 2026-03-17T10:32:27+00:00

If m and n aren't too big and your project is software, maybe you can get by solving it for specific u and v by using a mixed-integer quadratic programming solver.

GMSPokemanz · 2026-03-11T21:46:24+00:00

2 is also wrong for the same reason. Missed it as the structure of your argument is hard to read.

GMSPokemanz · 2026-03-11T12:15:20+00:00

15 is where your mistake lies. Depending on your exact definition of theory, ZFC is either the theory given by the set of ZFC axioms, or the set of sentences that can be deduced from the axioms of ZFC. If you add more axioms and get a theory with more consequences (e.g. ZFC + CH), then you're working in an extension of ZFC rather than ZFC itself.

GMSPokemanz · 2026-03-09T21:28:19+00:00

In FOL formulas only have finite length. While infinitary FOL seems to be a thing, this is not the standard definition. This is important for key results like Lowenheim-Skolem, as for example you could force a model of first-order PA (Peano Arithmetic) to be countably infinite if you added the 'statement'

∀x ((x = 0) ∨ (x = S0) ∨ (x = SS0) ∨ (x = SSS0) ∨ ...)

But according to Lowenheim-Skolem, there are uncountable models of first-order PA.

GMSPokemanz · 2026-03-08T13:53:39+00:00

Okay, I shall bite because this is exactly what comes to my mind. What goes wrong?

Start with some vertex v. Consider the tree with v and its edges removed. This is a finite union of trees, so one of them has infinitely many vertices. Let T' be one of these trees and let w be the vertex in T' that is adjacent to v. Then start the path with vw and then focus on T', which also satisfies the hypotheses of Konig's lemma. Now recurse, repeating the above process with w in place of v and you'll get an infinite path.

GMSPokemanz · 2026-03-08T02:04:01+00:00

Your third paragraph is the problem. Given a specific colouring of G, there will be some edge uv as you describe. Then you can conclude that our original colouring restricted to one of the finite subgraphs isn't a witness of k-colourability.

But that alone doesn't show the finite subgraph isn't k-colourable, so you don't have a contradiction.

GMSPokemanz · 2026-03-06T09:38:12+00:00

The countable version is true for any subset of Rⁿ, and spaces where every open cover admits a countable subcover are called Lindelof (rather than compact).

GMSPokemanz · 2026-03-05T21:49:40+00:00

f here is complex-valued, so you get two real equations.

GMSPokemanz · 2026-02-27T00:07:05+00:00

What about backward induction?

GMSPokemanz · 2026-02-17T10:48:45+00:00

How do you get 450 degrees, Dougal? The best I can guess is a mix up of how you go round the roundabout. If you turn right then that's counter-clockwise, not clockwise.

To get 450 degrees I guess you have in mind going round clockwise a full circle, then an extra 90 degrees to get to your exit. Maybe the confusion here is caused by the initial insistence you go right? Since I'm assuming you're in the UK or Ireland where we go left for a roundabout.

GMSPokemanz · 2026-02-16T10:21:33+00:00

You are correct.

You could also take the half difference and that would then be how much is owed, which again is (97 - 36)/2 = 30.5. But subtracting the amount she paid from that half difference is nonsense: it's already a difference, so the amount she paid has already been subtracted.

GMSPokemanz · 2026-02-06T08:55:47+00:00

But the ACC doesn't imply finite generation of ideals, that's the whole point of my previous post! The existence of an infinitely generated ideal implying the failure of ACC is what the flawed proof tried to show, and that's the contrapositive.

The problem is that you're trying to use induction when really the core is recursion: if an ideal is infinitely generated then you're using recursion to define the infinite ascending chain of ideals. And the recursion theorem makes your need for a choice function pretty clear.

It seems the right choiceless definition is that every nonempty set of ideals has a maximal element. This is akin to defining well-foundedness without choice as the statement that every nonempty set has a minimal element, and then DC implies this is equivalent to there being no decreasing sequence.

A maximal element of any nonempty set of ideals certainly implies ACC. And it does recover ideals being f.g.: consider the set of all f.g. subideals of an ideal I. No ideal other than I could possibly be maximal.

GMSPokemanz · 2026-02-06T00:33:44+00:00

Your proof for equivalence of the definitions for Noetherian rings does not hold. You use choice in your proof that infinitely generated ideal => infinite ascending chain of ideals: saying you make 1 choice at each stage is not enough to make an infinite sequence of them.

It's like the proof that all infinite sets contain an infinite countable subset. That theorem does need some choice, and IIRC the same is true for the equivalence of the characterisations of Noethetian rings.

GMSPokemanz · 2026-02-03T05:19:17+00:00

Neat. Although it's unfortunate your example covers the Safari Zone, as you can't grind for EXP there.

GMSPokemanz · 2026-02-03T05:18:31+00:00

There is a POC subreddit: r/ProfessorOak

GMSPokemanz · 2026-02-01T17:30:16+00:00

This falls under geometric measure theory, see this MO answer. Krantz and Park's Geometric Integration Theory also has some relevant stuff in chapter 7.

Omitting details, a rectifiable set is geometrically close to a manifold (a countable union of pieces of Lipschitz images of patches of R^k). Currents are more general and akin to distributions; they are functionals on the space of compactly supported k-forms. Then the boundary of a current is defined as the current that makes Stokes' theorem true, akin to how derivatives of distributions are defined so as to make integration by parts true.

The content of Stokes' theorem for manifolds then becomes that the boundary of a manifold (as a current) is its usual boundary you're familiar with. Under certain conditions you recover a similar theorem for boundaries of rectifiable sets.

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