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Chatgpt (free version) is NOT solving this: Q: If the countable totally ordered set (X<) has no upper bound, must there exist a strictly increasing unbounded sequence x1<x2<x3...? The answer is yes, but chatgpt is NOT offering up a correct proof.

submitted 3 days ago • 0 comments

I wrote about our foundational supremacist culture, its inevitable outcomes of genocide and slavery, and suggest 3 principles for building counterculture. Imagining the best so we will know it when we move toward it. www.the-reframe.com/building-cou...

submitted 6 days ago • 7 comments

Nice article by David Bessis! davidbessis.substack.com/p/the-real-m...

submitted 10 days ago • 0 comments

All my life I’ve been unable to read a mathematical text, however trivial or simple it may be, unless I’m able to give this text a ‘meaning’ in terms of my experience of mathematical things, that is unless the text arouses in me mental images, intuitions that will give it life. (Grothendieck).

submitted 10 days ago • 0 comments

Just realized the are only COUNTABLY many continuous functions f:C-->{0,1} if C is Cantor Space. AI does NOT seem to know this. Pf. Any clopen is a unit of finitely many basic clopens, and C has a countable basis of clopen sets.

submitted 10 days ago • 0 comments

Sorting out the maps between a coupla discrete groups, and THIS emerges. Hom( lim__> A_n, B) = lim_< Hom (A_n, B). It was topology all along, the 2-adic integers are compact. Here B = lim --> Z_(2^n), the countable discrete Prufer 2 group. It was you category theory. It was you topology.

submitted 11 days ago • 0 comments

Great writing, including `...a man who seems like he would have a hard time solving the word jumble on the children's menu at Elias Brothers Big Boy'.

submitted 12 days ago • 0 comments

Shall we count the endormorphisms of the 2-Prufer group? (additive dyadic rationals mod 1 with the discrete topology). We've got 2 distinct choices for where to map 1/2: 0 or 1/2. Then 2 choices for where to map 1/4. And so on. We get the inverse limit of Z_2 <Z_4<Z_8... The 2-adic integers!

submitted 13 days ago • 0 comments

Mucking about with solenoids, and solenoid accessories, is a compact connected ring emerging? Nope! See Tom Leinster's thread from 2014, with weigh ins from Terence Tao, Todd Trimble, Qiaochu Yuan, and others. golem.ph.utexas.edu/category/201...

submitted 13 days ago • 0 comments

Chatgpt, on the other hand, correctly serves up the Prufer groups Z(p^(infinity)). Such a group is infinite and abelian, but each proper subgroup is finite and cyclic.

submitted 13 days ago • 0 comments

Hey Deepseek! Does there exist an infinite abelian group such that each proper subgroup is finite? Deepseek. No...(nonsense follows) Our jobs are safe for now. For an example consider a colimit indexed by the natural numbers, of cyclic groups of order 2^n, under monomorphic bonding maps.

submitted 13 days ago • 0 comments

There exists a discrete infinite abelian group whose proper subgroups are finite and cyclic, one subgroup for each nonnegative power of 2. This group is unique up to isomorphism. It's the colimit under monomorphisms C_2 --> C_4 --> C_8...

submitted 13 days ago • 0 comments

In which the time to think is now.https://www.nytimes.com/2025/02/15/opinion/trump-autocracy-bad-ideas.html?smid=nytcore-android-share

submitted 14 days ago • 4 comments

Is mathematics obsolete? ~ Jeremy Avigad. www.andrew.cmu.edu/user/avigad/... #Math #ITP #LeanProver #AI #LLMs

submitted 16 days ago • 2 comments

To compactify the totally ordered set X: Let Y denote the lower sets of X. Let Z denote the quotient of Y, identifying (,x) and (,x], if and only if x is not a succesor in X. Embed X in Z via the map x--> (*,x].

submitted 16 days ago • 0 comments

Constructing from scratch a model of the reals as ordered field creates a LOT of re-use-able tools. Here's one: In a DIVISIBLE totally ordered abelian semigroup, we can solve a+x=b when a<b.

submitted 17 days ago • 0 comments

great talk on 3blue1brown with Terrence Tao. www.youtube.com/watch?v=YdOX...

submitted 20 days ago • 0 comments

Reloading and trying to give deepseek grace, job is still safe. DeepSeek: ....For example, the discrete topology on Z is not the order topology induced by the usual order ≤.

submitted 22 days ago • 0 comments

Actually DeepSeek did not QUITE have a clear definition of orderable topological space, it should use a sub-basis of open left and open right rays, and NOT this: DeepSeek. The order topology is generated by the basis of open intervals (a,b)={x∈X∣a<x<b}(a,b)={x∈X∣a<x<b}.

submitted 22 days ago • 0 comments

Hey DeepSeek! Is every discrete topological space orderable? (PF: Yes, but I don't see how to prove this in general without axiom of choice). Warning, Bloodbath ahead. DeepSeek: No, Consider the set of integers Z with the discrete topology. (PF: Maybe our jobs are safe for now)

submitted 22 days ago • 0 comments

Hey DeepSeek! Is every finite discrete topological space orderable? DeepSeek nails this correctly, but contradicts a claim it made it in an earlier answer, namely that orderable spaces are connected.

submitted 22 days ago • 0 comments

Hey DeepSeek! With the subspace topology, must every subspace of an orderable topological space be orderable? Bogus! Deepseek reports the correct definition of orderable space, but then THIS nonsense: Deepseek: Orderable spaces are connected in their order topology.

submitted 22 days ago • 0 comments

Hey DeepSeek! If every subgroup H of the finite group G is normal in G, must G be abelian? DeepSeek nails it, finds a counterexample Q(8) right out of the gate.

submitted 22 days ago • 0 comments

With the subspace topology, subspaces of orderable spaces need not be orderable. However, continuous order preserving injections between ordered spaces are always topological embeddings. The domain sub-basic U=(-infty,x) maps to the subbasic h(U) intersect (-infty, h(x)) in the codomain.

submitted 22 days ago • 0 comments

Must a separable, contractible, uniquely arcwise connected, locally path connected space be metrizable? No! Take the quotient of [0,1] x{1,2,3,...}, identifying all pairs (0,n) and (0,m). The quotient is not first countable at the special point.

submitted 22 days ago • 0 comments

Suppose every subgroup H of the finite group G is normal. Must G be abelian? No! The Quaternion group G= Q(8) has 3 subgroups of index 2, the center has order 2, and G and {1} are the only other subgroups. All are normal, yet G is nonabelian.

submitted 22 days ago • 0 comments

With the subspace topology, a subset of a totally orderable space need not be totally orderable, even though we are allowed to rearrange the order of the points in the subset. The real subpace [0,1) U [2,3] is NOT a counterexample, however the subspace (0,1) U [2,3] is NOT totally orderable.

submitted 22 days ago • 0 comments

A 2nd countable totally ordered compact space might NOT be order preserving embedable in the real line. Double cut open each point of [0,1], so that there are uncountably many successors. However this space CAN be topologically embedded in the real line, since it's homeomorphic to Cantor space.

submitted 24 days ago • 0 comments

What does the re election of donald trump portend? DeepSeek: Trump has often expressed admiration for strongman leaders (Vladimir Putin, Kim Jong-un). His re-election could lead to closer ties with authoritarian regimes, potentially undermining efforts to promote democracy and human rights globally

submitted 26 days ago • 0 comments

DeepSeek offers a thoughtful, nuanced, informed answer: Is Miller's Crossing a good film? (Personal opinion: Yes, obviously)

submitted 26 days ago • 0 comments

Does DeepSeek understand basic topology? Hold fast... Must a sequentially compact totally ordered topological space have a minimal element? Great answer, great process. Offers up the long line, but neglects to clarify reverse ordering.

submitted 30 days ago • 0 comments

My collaborator(s?) dropped a good lead. Theorems are for closers but I gotta LOTTA coffee.

submitted 31 days ago • 0 comments

Mathematical research is mostly trying decent ideas that don't quite work. But then you come you back much later, with renewed energy, and rediscover that the same ideas still don't work.

submitted 31 days ago • 0 comments

"Their response was far less confused about the bishop's meaning than it had been about the billionaire's: As far as they were concerned, it was a vicious attack. And it was an attack, because any expression of decency is an attack against indecency." www.the-reframe.com/the-bishop-a...

submitted 33 days ago • 5 comments

Construct a model of the familiar additive reals A as complete ordered abelian group. Then Hom(A,A) is a complete ordered ring (field), and is unique up to unique isomorphism.

submitted 39 days ago • 0 comments

Up to UNIQUE isomorphism, the ring of integers is the ring Hom(A,A) of an infinite cyclic group A. There is an analagous story for the real numbers.

submitted 39 days ago • 0 comments

To construct an interval from a totally ordered Cantor set, identify x with y if y is the successor of x. With the order topology, we now have [0,1], up to isomorphism.

submitted 39 days ago • 0 comments

To accidently construct an ordered Cantor set, start with ANY countable totally ordered set (X,<) with the between-ness property. THEN construct the LEFT SETS. B subset X is a LEFT SET if x is in B, whenever y is in B, and x<y. The left sets form a Cantor set under inclusion, with order topology.

submitted 39 days ago • 0 comments

We have a good mental model of the reals as a straight line, and later as complete ordered field. The puzzle of constructing them from basic set theory, AND proving uniqueness up to unique isomorphism, has LOTS of pieces.

submitted 39 days ago • 0 comments

There is a an easy proof, avoiding Cantor Shroeder Bernstein, that up to isomorphism there is a unique countable, totally ordered set, with no max, no min, and the between-ness property. Construct a model as a colimit from an infinite cyclic group, using x-->2x.

submitted 39 days ago • 0 comments

The based totally ordered set, (Z,0,+) is initial in a category whose 1) Objects are totally ordered sets so that all points both are and have successors, and whose 2) Morphisms preserve basepoints and successors. Two copies of Z attached tip to tail creates an example of a noninitial object.

submitted 39 days ago • 0 comments

Basic counterexample of a category with two dots but only one nontrivial morphism, f:x-->y. The morphism f is both left and right cancelable but not an isomorphism.

submitted 40 days ago • 0 comments

It ain't over. The only thing that's over is the comforting illusion that someone else is going to do the work.

submitted 41 days ago • 7 comments

The group (G,*) is abelian iff the function F:G-->G defined as F(g)=g^2 is a homomorphism.

submitted 41 days ago • 0 comments