A set is good if, for a set n, the following is true: for numbers k1, k2 both less than an odd number 2n+1, k1 (let's say this is chosen by Alice) is either the same parity as (and smaller than) k2, or the other parity and bigger. Assuming k1 and k2 are different, these are the conditions for Alice to win. The conditions for Bob to win are just swapping k1 for k2. However, this is not determinate: if k1=k2, that's a tie. You might recognize this game- it's Rock Paper Scissors (the case of n=1) and beyond!

Doesn't Bob's set in the power-of-two example also include infinitely many powers of two?

BT doesn't show a problem with AoC but with nonmeasurable sets. Eliminate those from your universe and both AoC and AoD are true. (Yes, this means that you only get AoC in limited form, but that's not a problem.)

The axiom of choice is by far my favorite concept in mathematics.

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There should be a problem grouping all AS sets together by transitivity. If you can have infinitely many "transitions" to populate the group doesn't that break the "finitely many changes" assumption for AS sets?

A set is good when it's bad. Now who will win?

Your explanation of the complexity of Chomp was pretty awful.

Damnit...good content but it's what I do so I hate it when KZbin starts recommending it to me. The point is to listen to something *else* while I work on my research.

I can count to 2047 in binary

I don't get fact 1. If A and C are entirely different to each other, then B being different from both would mean it is similar. It seems to me like A and B can be AS, when B and C are AS, but A and C are not AS.

Thank you so much for making this video! I've tried to understand the axiom of determinacy before but the wikipedia page always loses me haha

Ah, you have to make ONE rule that applies to ANY set. That wasn't clear for any other video. Of course, if sets can contain anything it is obviously (...) impossible. I can't come up with a good rule to choose between ∛π, a tomato and anger. For finite sets of numbers, it becomes easy but let's say we have an infinite set of numbers. Let's say any element in ℝ but not in ℚ. I can come up with a rule, but it is a bit of a cheat, inspired by Rayo's number: describe all numbers by an expression in the language of first-order set theory, using the least number of symbols possible. Then order by the number of symbols. If multiple numbers need the same number of symbols, the smallest number wins. (Note: this of course fails, because there are [infinitely many] numbers that cannot be described by an expression in the language of first-order set theory at all. Side note: did anybody think of giving the set of numbers that cannot be described in the language of first-order set theory its own letter/symbol? Maybe 𝕏 or something?) Of course the non-deterministic rule "pick one element at random" already satisfies the axiom of choice. Not that for an element of my collection 𝕏, you may have trouble describing it...

Sigh. Why are you all jewish

Maximum recursion depth reached :(

6:38 Since the sum of the reciprocals of the natural numbers diverges (harmonic series), it's always possible for Alice to choose enough numbers each turn such that the sum of their reciprocals is greater than any finite number you choose. As long as they are consecutive (because of the way the game works, they must be) and she chooses enough of them (there is no limit to how many she can choose), it doesn't matter how large the numbers get. So no matter how many numbers Bob chooses, Alice can always choose enough numbers to increase her sum of reciprocals by some number greater than 1 for example, guaranteeing her sum diverges and her set is good.

whyd you call me a pervert

So, a topological sort. Reverse postorder it is

"A set is good if Alice goes last."

I am a mathematician and already knew about Determinacy ... and still, I think this is a brilliantly informative video.

I LOVE this channel soo much!

6:59 what if the condition is: "a good number is blue minus red equals 0"? 4:02 if they continue playing forever, how can the game ends?

12:30 okey yeah but then maybe bob has a second book (and an infinite family of them) in case Alice changes the roles once (and many times after)

The game is infinite: therefore you would have to have a pattern in order to say the play continues thus. So what if every turn picks a random amount to color.

I think that what Banach-Tarski actually shows us is that our human intuitions when it comes to measure theory when applied to infinite sets are screwy. There's no paradox at all. It's just that infinities don't work the way that finite numbers work, and our minds are not used to dealing with infinities. I am not actually certain that there are any infinities in the real world. There may be a smallest unit of time, a smallest unit of space, and the universe may be finite. It can be very large but still be finite. Infinities would still be very handy tools for calculation in such a world. But otherwise they may just be mathematical games. Or maybe there are infinities, but we still don't normally deal with those in our everyday lives on the level on which we live.

I think I get the idea of backwards induction to win Chomp. You need to achieve a board state where: 1. There are only two non poisoned blocks left 2. It's your opponent's turn to move Lets call this S1 Then you have to achieve the following: 1. It's your opponent's turn to move 2. All of their moves will either lead to an immediately llosing state or allow you to achieve S1 on the next move. We'll call this S2... And so on

Alice and Bob, the eternal victims of mathematical games.

Another remark and I think the Crux of the problem: The set of normal distrib. Guessed Numbers giving you the right answer is of the same size as the set of Numbers giving you the wrong answer. The Argumentation of the video should be abel to be refuted with the following Let‘s say x is the reveald and y the hidden number and smaler(does Not matter if bigger or smaler) Lets say d is the distance between x and y Yes, there is a chance that the randomly normal dist. Number z is between x and y and, yes, that would mean a 100% of winning. But so would every other number smaler than x because you have fixed the Position of y in relation to x with speaking about the Range of numbers between them. But as Said before: the size of the set with the right Numbers is the same as the Set with the wrong so your chance would still be 50/50 Furthermore, the Argument of the Video could also be mirrowed: the Chance that z is between x and y ist the same as z beeing between x and x-d or x+d (y> or<x) with the second leading to an automatic los

But what would be the difference to this experiment: I choose a random number, then you have to decide if it‘s higher or lower then the next number. After that I choose the second number. You would have decided (with any reasoning you like) that the number is higher or lower. That would have happened before the 2. number is chosen. But the second number is chosen randomly such as it has a50/50 Chance of beeing higher or lower In conclusion, you Must have a 50/50 in guesing correctly

wow, that was a lot of words for no reason. here is a much simple example for "no one can win" the rule is: who ever has more numbers win. no matter what they pick, the next one picks the same number plus 1. no winning strategy

you nailed it ! a nice application of order theory and probability where we can find a order of 2 any element whose probability in partial order sets linear extensions(i.e just all the possible comparison of available partial info. of order or in other words partial order) is between 1/3 and 2/3 and more concretely with proof between 27 % and 73 % so finding this comparison can at least cut down 27 % of linear extension so rapidly decreasing the no. of comparison actually needed to perform just 1 is enough...i read your its use example in a game contest do you have any application in physics point of view ?

Reminds me why I hate set theory arguments

Is it -1/3?

Just use Leibniz's infinitesimal calculus instead of Newton's fluxional calculus. No more contradictions or paradoxes.

Does anyone have any good book recommendations for this topic?

⅓ - ⅔ = -⅓ How do you like THEM apples?

In case this was bothering anyone else: A poset _can_ happen to be totally ordered (a straight line) already, in which case every sorting probability is either 0 or 1 (since every comparison is already determined). In its full form, the conjecture stipulates that the poset isn't totally ordered.

Nerd

This is unironically the only good explanation of the axiom of choice that I have found. I have watched 20 minute videos about it and yet this was more enlightening.

Isn't it easier to include the empty tree in T and write T = 1 + zT²

In the almost-same game, is it possible to come up with a finite description for even one way of categorizing good/bad sets, which fully characterizes which sets are good and bad? I've tried, and can't seem to do it without breaking the game's rules. Is it absolutely required to make an uncountably infinite number of arbitarary decisions (as per the Axiom of Choice) to even come up with one good/bad set categorization?

Can strategy stealing be frustrated (without worrying about AC) by there being no _computable_ winning strategy? For example, what if a good set contains more than half of the "Busy Beaver" numbers? (This satisfies both conditions 1 and 2 of "the plan" provided it's not possible to chose finitely close to half of an infinite set, but that's almost never the result.) At each step each player, to get to a "winning so far" state, merely has to choose a large enough finite number, but there is no way to compute how large is large enough. In fact any computable strategy for choosing will choose too small by nearly infinate margins. (There is also the problem that, despite there (almost?) always being a winner, it's not possible to compute who won.)

This reminds me of how spin-1/2 particles like electrons have to turn 720 degrees to do a full rotation.

"The Banach-Tarski Paradox" is actually a theorem, not a "paradox". It can be proven without error in the ZFC logic system, which almost all math guys use.

Is this reasoning really accurate for the question? When comparing apples to oranges there are only two possible constellations. How does the relation of each to bananas factor into this? The question which one out of the two you like more has nothing to do with bananas. I guess this is a limitation of motivating this problem with fruits the way you did.

hi :3 UwU

This hurts my brain. I am not sure that the collection of sets finitely different from a given set is well behaved. We can restate a lot of this in terms of real numbers between 0 and 1. Define a binary number based on the presence or absense of the corresponding integer in the set so {2,3,5} maps to 0.01101 (that's not quite a 1 to 1 mapping because of the equivalence of .1 and .0111111111...*). So I claim that each collection of sets having finite differences from a given set maps to numbers that are dense on the interval 0 to 1 but have measure 0. I don't think this helps but it does illustrate the pathological nature of these collections. * these awkward numbers correspond to finite sets and their complements, so don't crop up in the game!

I hadn't realised this branch of mathematics existed. It obviously (with hindsight) is a down in the weeds part of sorting. For sorting you generally want each comparison to halve the number of possibilities left but in practice accept much less even results so this conjecture provides a target for the sorting algorithm (in practice simplicity of implementation is also a concern*). * as a programmer debugging an optimal and therefore complex algorithm does not appeal

Such a nice and clear explanation of posets!

Loved this video!