Seeker of Carcosa

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  • 21 Comments
Joined 1 year ago
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Cake day: June 28th, 2023

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  • Seeker of Carcosa@feddit.uktomemes@lemmy.worldInfinity
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    8 months ago

    Sorry if you’ve seen this already, as your comment has just come through. The two sets are the same size, this is clear. This is because they’re both countably infinite. There isn’t such a thing as different sizes of countably infinite sets. Logic that works for finite sets (“For any finite a and b, there are twice as many integers between 1 and 100 as there are even integers between a and b, thus the set of integers is twice the set of even integers”) simply does not work for infinite sets (“The set of all integers has the same size as the set of all even integers”).

    So no, it isn’t due to lack of knowledge, as we know logically that the two sets have the exact same size.



  • Seeker of Carcosa@feddit.uktomemes@lemmy.worldInfinity
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    8 months ago

    There’s no problem at all with not understanding something, and I’d go so far as to say it’s virtuous to seek understanding. I’m talking about a certain phenomenon that is overrepresented in STEM discussions, of untrained people (who’ve probably took some internet IQ test) thinking they can hash out the subject as a function of raw brainpower. The ceiling for “natural talent” alone is actually incredibly low in any technical subject.

    There’s nothing wrong with meming on a subject you’re not familiar with, in fact it’s often really funny. It’s the armchair experts in the thread trying to “umm actually…” the memer when their “experience” is a YouTube video at best.


  • Seeker of Carcosa@feddit.uktomemes@lemmy.worldInfinity
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    8 months ago

    I like this comment. It reads like a mathematician making a fun troll based on comparing rates of convergence (well, divergence considering the sets are unbounded). If you’re not a mathematician, it’s actually a really insightful comment.

    So the value of the two sets isn’t some inherent characteristic of the two sets. It is a function which we apply to the sets. Both sets are a collection of bills. To the set of singles we assign one value function: “let the value of this set be $1 times the number of bills in this set.” To the set of hundreds we assign a second value function: “let the value of this set be $100 times the number of bills in this set.”

    Now, if we compare the value restricted to two finite subsets (set within a set) of the same size, the subset of hundreds is valued at 100 times the subset of singles.

    Comparing the infinite set of bills with the infinite set of 100s, there is no such difference in values. Since the two sets have unbounded size (i.e. if we pick any number N no matter how large, the size of these sets is larger) then naturally, any positive value function applied to these sets yields an unbounded number, no mater how large the value function is on the hundreds “I decide by fiat that a hundred dollar bill is worth $1million” and how small the value function is on the singles “I decide by fiat that a single is worth one millionth as a set.”

    In overly simplified (and only slightly wrong) terms, it’s because the sizes of the sets are so incalculably large compared to any positive value function, that these numbers just get absorbed by the larger number without perceivably changing anything.

    The weight question is actually really good. You’ve essentially stumbled upon a comparison tool which is comparing the rates of convergence. As I said previously, comparing the value of two finite subsets of bills of the same size, we see that the value of the subset of hundreds is 100 times that of the subset of singles. This is a repeatable phenomenon no matter what size of finite set we choose. By making a long list of set sizes and values “one single is worth $1, 2 singles are worth $2,…” we can define a series which we can actually use for comparison reasons. Note that the next term in the series of hundreds always increases at a rate of 100 times that of the series of singles. Using analysis techniques, we conclude that the set of hundreds is approaching its (unbounded) limit at 100 times the rate of the singles.

    The reason we cannot make such comparisons for the unbounded sets is that they’re unbounded. What is the weight of an unbounded number of hundreds? What is the weight of an unbounded number of collections of 100x singles?



  • Seeker of Carcosa@feddit.uktomemes@lemmy.worldInfinity
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    8 months ago

    This kind of thread is why I duck out of casual maths discussions as a maths PhD.

    The two sets have the same value, that is the value of both sets is unbounded. The set of 100s approaches that value 100 times quicker than the set of singles. Completely intuitive to someone who’s taken first year undergraduate logic and calculus courses. Completely unintuitive to the lay person, and 100 lay people can come up with 100 different wrong conclusions based on incorrect rationalisations of the statement.

    I’ve made an effort to just not participate since back when people were arguing Rick and Morty infinite universe bollocks. “Infinite universes means there are an infinite number of identical universes” really boils my blood.

    It’s why I just say “algebra” when asked what I do. Even explaining my research (representation theory) to a tangentially related person, like a mathematical physicist, just ends in tedious non-discussion based on an incorrect framing of my work through their lens of understanding.






  • Seeker of Carcosa@feddit.uktoMemes@lemmy.mlCommunism bad
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    1 year ago

    The good old Narcissist’s Tankie’s Prayer:

    That didn’t happen,

    And if it did, it wasn’t that bad,

    And if it was, that’s not a big deal,

    And if it is, that’s not my fault,

    And if it was, that’s Western propaganda,

    And if it isn’t, you deserved it.




  • The Witcher 2, though it’s more of an exploit.

    If you get the sword from the Lady of the Lake in the previous game, you start the prologue of TW2 with two silver swords, one being the Lady of the Lake sword. Unequip the Lady of the Lake sword so you don’t lose it to the dragon.

    You now have a mid-tier silver sword that is good for half the game. You also don’t need to find a new silver sword at the start of the game.