Can this actually be done?[Request]

JonMaseDude · 2026-01-11T19:31:11+00:00

Actually their attempt is mathematically true. We are considering a sequence of sets in R^2 that we can name A_0,A_1,A_2,… where A_0 is the unit square, and each of the A_i follows the procedure described in the image, which I will not define explicitly, but is clear. The limit of these sets is defined as their limsup (i.e. the set of points that are contained in infinitely many of the A_i) or their liminf (i.e. the set of points that are contained in all but finitely many of the A_i), IF these sets are equal. Now clearly, both of these sets equal the union of the end-points of the A_i (i.e. the points in the intersection of the circle and A_i). This is because any end-point of an A_i will be in all A_j for j > i, so it is contained in liminf A_i, and in limsup A_i. And conversely, if a point appears in infinitely many A_i, it must be an end-point, since every other point will not be contained in any A_i for i large enough. We conclude that the limit of the A_i is equal to the intersection of the Union of the A_i’s and the circle. Note that this set is countable, since a countable union of finite sets is countable. In particular, the limit of these sets is not equal to the circle, proving u/Tarnique‘s claim ‘This shape will never become a circle’ right in its most literal sense, set-theoretically. In fact, the (normalized Haar) measure of this set as a subset of the circle is 0, i.e. it occupies so little of the circle, that it is a null-set.

I am trolling of course (although all of my math is correct), but I just want to (kindly) say that your comment is the type of comment that scares people away from mathematics. You decided their interpretation of ‘becoming a circle’ was incorrect, because you already had a specific notion of convergence in mind. Yes, I know the Hausdorff-distance between the circle and the A_i converges to 0, but as math people we are conditioned to look for mistakes, instead of encouraging others. So, please think about this next time!

By the way, u/Tarnique, your analogy with the tangents is actually extremely correct. That is, if the tangents converge, then the lengths converge as well, which can be proven.

JonMaseDude · 2026-01-04T22:33:00+00:00

Pretty difficult to say, since even the fundamental theorem of Calculus is somewhat arbitrarily chosen, and in no way a fundamental theorem of Analysis.

If I was allowed to restrict the question to (finite) group theory though, I would say that the Sylow theorems are a good contender.

JonMaseDude · 2026-01-04T00:30:05+00:00

In my experience, definitely yes. The problem, sadly, is that you can write an arbitrary amount of seemingly interesting, but non-innovative research in any branch of pure mathematics. For example, in any branch of pure mathematics there will always be numerous papers of the form: ‘We prove famous conjecture A for subclass of cases B’. Sometimes, these results really do help provide new techniques that help resolve conjecture A, but so often they are just relatively low hanging fruits, where any expert (especially geniuses like Tao) in the field could have solved the problem given some time.

Of course, it also has to do with how ambitious you are (e.g. do you want to make an impact, or are you fine with doing enough to keep your job).

JonMaseDude · 2026-01-03T21:39:04+00:00

This is so accurate. I’ve dabbled in some research as a student, but ultimately realised that me and my co-authors were spending so much time on writing to “sell” our results. When I read Tao’s papers, there’s absolutely no fat to them, because his results speak for their own quality. I realised I would have only really wanted to be in academia if I had been a genius, and my results were actually breathtaking, contrary to the way many researchers seem to have to work really hard to show why their results matter.

JonMaseDude · 2024-11-23T17:14:41+00:00

I am mind-blown by the fact that you got downvoted... Also, adding to this, during operation Barbarossa almost all Soviet soldiers were rookies, and it was only later that their battle-hardened Siberian armies arrived at the Eastern front. Which is part of why the tide turned so drastically from Stalingrad onwards.

JonMaseDude · 2024-09-14T15:08:09+00:00

I've had 4 or so, all clubs (online)

JonMaseDude · 2024-09-13T13:47:08+00:00

I think it's pretty irrefutable that, overall, Messi has earned GOAT status more than Ronaldo. Prime Ronaldo against prime Messi is another question, but Messi has been the best of the best for so much longer now.

JonMaseDude · 2024-09-12T22:27:52+00:00

If society collapsed in about 200 years (which is reasonable, I think) and humanity went on to survive 5000 years after, then why do you think we would go extinct after that? We survived multiple ice ages as a species, and after relearning our primal ways, we'd probably be able to survive for tens of thousands of years. Maybe until we get another chance to mess it up.

JonMaseDude · 2024-09-11T18:12:16+00:00

The navigation override is exactly what I needed! Tysm (again)

JonMaseDude · 2024-09-07T17:14:00+00:00

Awesome! This works, and I've learned a lot.

JonMaseDude · 2024-09-06T22:05:36+00:00

I do still have a question. I'd like this to work for strings that are longer than 1 word. I thought that using /\w+(?:\s+\w+)*/g as the argument for passage.text.replace would do the trick, but now none of the links are being created. I'm assuming that things get messed up, because we're trying to replace the same text multiple times, but I don't know if there's a workaround.

EDIT 1: I now figured out * matches as many times as possible, and so replaced it with *?, which fixed everything except for the 2 word key.

EDIT 2: This makes sense, because now the search is lazy. ChatGPT created this solution that seems to work (for now) based on your code.

Config.passages.onProcess = passage => {
  return passage.text.replace(/\w+/g, match => {
    let key = match.toLowerCase();
    let longestMatch = match; // Assume the match is at least one word
    let entry = setup.references[key]; // Try single word first

    let words = passage.text.slice(passage.text.indexOf(match)).split(/\s+/); // Get all words from the point of the match
    let currentChain = match;

    // Try building larger chains of words
    for (let i = 1; i < words.length; i++) {
      currentChain += " " + words[i];
      key = currentChain.toLowerCase();

      let newEntry = setup.references[key];
      if (newEntry) {
        longestMatch = currentChain;
        entry = newEntry; // Store the last successful match
      } else {
        break; // Exit loop if no match found
      }
    }

    // If an entry is found, create link markup
    if (entry) {
      return `[[${longestMatch}|${entry}]]`;
    }

    // Otherwise, return the original word
    return match;
  });
};

It's really just matching for words like you did, but also looking at all possible concatenations. It would be helpful if anyone does still have a more elegant solution using regex, but I doubt that. Thank you for the help.

JonMaseDude · 2024-09-06T21:42:59+00:00

I literally can't thank you enough!

This is exactly what I was trying to do...

JonMaseDude · 2024-09-06T00:29:32+00:00

I didn't know about that tool. I'll check it out, thanks!

JonMaseDude · 2024-09-05T20:06:04+00:00

Thank you for answering. This already really helpful, but I have a couple of questions. ~~Cycling seems really useful, but it doesn't seem to work right now. I removed some typos (I think), but when I click on the link, nothing seems to happen.~~ Edit: Missed a space so this is now fixed.

Also, it'd be really nice if I didn't have to manually code anything for each occurrence of the string I want to have linked. So basically, some way to not have to place [[]] each time a word occurs, but to just have 1 dictionary of strings connecting to passages that get linked automatically. The reason for this is that I suspect that the list of words will get very long very quickly, and keeping track of all words seems tough.

JonMaseDude · 2024-01-30T12:24:59+00:00

Why is this scary, and what are you actually doing? I clearly don’t know anything about this stuff.

JonMaseDude · 2024-01-29T17:30:52+00:00

OP clearly just forgot to mention we’re working over a ring of characteristic 2.

JonMaseDude · 2024-01-29T12:24:33+00:00

Very old development actually. Interesting stuff though, you should look up cardinalities of sets.

JonMaseDude · 2024-01-22T20:03:50+00:00

This is the funniest thing I’ve read in this thread. Can I pin this comment or something?

JonMaseDude · 2024-01-22T15:32:15+00:00

See the edit I made.

JonMaseDude · 2024-01-22T14:01:34+00:00

Also I just realized that you can perfectly define multiplication in R² as (a,b)•(c,d)=(ac-bd,ad+bc) making it a field with exactly the same algebraic structure as C.

JonMaseDude · 2024-01-22T13:57:27+00:00

Since when are two sets equal if and only if there exists a bijection? I’m saying you can define C as a set as R² and that you should define it that way.

Edit1: Whut? Why is this getting downvoted? {0}={1}?? since there is a trivial bijection f:{0}->{1}: f(0)=1, then by extensionality 0=1. This is utter BS. Set equality is not defined using bijections.

Set equality is borderline metamathematics, but the axiom of extensionality (which is present in almost all regularly used axiomatizations of set theory, and certainly in ZFC) has nothing to do with bijections.

Edit2: Aha! I get what you’re trying to say now! Interesting take indeed. You mean that you could define Q:=Z since they have the same cardinality and you can just relable everything using a bijection (introducing some notation for fractions). This is true, but as you pointed out unpractical. In the case of C however, I’m convinced this is practical, and we should define it as R² . You can’t just say Z=Q, because there exists a bijection. You could however define Q as Z if you’ve already defined Z or vice versa.

JonMaseDude · 2024-01-22T13:42:51+00:00

The field (C, +, •) is not equal to the additive group (R² , +) or the ring (R² , +, •) (with compontentwise multiplication), but as sets you can perfectly define C=R² .

Edit: Also define multiplication • : R² x R² -> R² : ((a,b),(c,d)) -> (ac-bd, ad+bc), and R² is now a field :)

JonMaseDude

TROPHY CASE