Cantor's Theorem, the Aleph Function, and Beth Numbers

[@musimedmusi8736]

Nice concise basic review

[@alephomega955]

(if I'm wrong, feel free to correct me.) Using the proof on how you cannot list all real numbers, the set of real numbers should be equal to the set of natural numbers. You can list the irrational numbers then make one that doesn't appear on the list, but you could still pair up each number you created with each natural number, you just start with 1 on the new one you created and label them from there on. It's the same thing as adding to an infinite amount, which doesn't change the value. You could describe another infinite amount of numbers using the same method and still not change the value, because you'll never run out of naturals to pair it up with.

[@MikeRosoftJH]

Obviously, there are uncountably many irrational numbers. There are countably many rational numbers, but uncountably many real numbers; and union of two countably infinite sets can't be uncountable. It's also fairly easy to construct an explicit one-to-one mapping between real numbers and irrational numbers.

[@linuxp00]

It's possible to construct an uncontable infinite set from a cartesian product between two countable infinite sets. Though, it's not in any way mappable to a countable set. My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.

[@MikeRosoftJH]

@@linuxp00 "It's possible to construct an uncountable infinite set from a cartesian product between two countable infinite sets." No, that's not possible. Cartesian product of two countably infinite sets is countably infinite; and this result doesn't require axiom of choice. (As opposed to the more general statement that union of a countably infinite collection of countably infinite sets is countably infinite, which *does* require axiom of choice, or a weak variant thereof - axiom of countable choice. For example, it's consistent in absence of axiom choice that the set of real numbers is a union of countably many countable sets; and that doesn't contradict Cantor's result that there are uncountably many real numbers.) There's no such natural number as ...99999. Any natural number is finite in magnitude, and therefore its decimal (or base-n) expansion has finitely many digits. And that's by definition; a set is finite, if its number of elements is equal to some natural number. (It is an empty set, or it has exactly one element, or it has exactly two elements, or it has exactly three elements, or... and so on. Of course, the "and so on" part is a bit involved; you can't have a formula of an infinite length.) The mathematical structure towards which you are heading are the n-adic numbers; in this structure, any base-n expansion (infinite sequence of base-n digits) is an element of this structure, and then you define mathematical operations on these base-n expansions. For example, the number ...999 is equal to -1 - you can verify that indeed ...001+...999=0. (Usually, n is taken to be a prime number, because otherwise you have divisors of zero - a pair of non-zero numbers a and b such that a*b=0.) But the problem is that there are uncountably many n-adic numbers - any infinite sequence of digits corresponds to a unique n-adic number. (From Cantor's diagonal proof it follows that n-adic numbers can't be mapped one-to-one with natural numbers; however, they can be mapped one-to-one with real numbers.)

[@linuxp00]

@@MikeRosoftJH I'm inspired by adics, but I'm not using an actual adic arithmetics. I'm just using naturals as infinite strings of digits in reverse. Naturals are infinitely countable, so are their digits. So, as I see it the whole set of Naturals are contained in the interval [0,1[. Though, it's the same for all the gaps between two consecutive integers, so it is impossible to count the reals. No need for Diagonalization, just Zenon's Paradox in it's essence.

[@MikeRosoftJH]

@@linuxp00Again, every natural number is finite in magnitude, and therefore has finitely many non-zero digits. If you reverse the decimal expansion of a natural number, you get the set of all real numbers in an interval whose decimal expansion has finitely many digits after the decimal point. And that there are countably many such numbers is not a new result; all such numbers are rational, and that there are countably many rational numbers was proven by Cantor himself. And that one specific function from natural numbers to real numbers doesn't cover all reals doesn't by itself prove that real numbers are uncountable. It's a well-known result that an infinite set can be mapped one-to-one with its strict superset or subset. (Precisely, this is true for every set which has a countably infinite subset.) For example, consider the function n->2*n on natural numbers; or, conversely, the same function on an interval from 0 to 1. So in order to prove that set A can't be mapped onto set B, it doesn't suffice that one specific function between the two sets doesn't cover all elements of B; you have to prove that no function does. (Such as by using the diagonal proof, or the earlier Cantor's proof by nested intervals.) In set theory, an infinite sequence is not a process. It's a function whose domain is the set of natural numbers. And a function is itself a set: set of ordered pairs. For example, the function f: x->2*x on the set {1,2,3} is the set {[1,2], [2,4], [3,6]}; if the pair [a,b] is an element of the set f, we by convention say that f(a)=b. Zeno's paradox isn't a theorem of mathematics. In fact, Zeno was heading towards perfectly well-defined concepts of calculus, such as limit of a sequence, infinite sum, or derivative of a function. (Except for that he wasn't able to understand the infinite case, and so he instead concluded that the infinite case is impossible.)

[@ev3rything491]

Dude! Love your videos. I wonder if you could clarify one point about the Beth numbers. I understand how they are defined but I dont follow why the cardinality of the power set of the natural numbers is equal to the cardinality of the real numbers.

[@egheitasean1]

The way that I visualize it, is consider the real numbers in binary (base 2). How many binary numbers can you make with only 2 digits? There are 4, namely 00, 01, 10, 11. For 3 digits there are 8, and generally for k digits there are 2 to the power of k. So, by extension if we consider all digits, we have 2 to the Beth null, because it's a countable infinity. Also, for the Power set of N, consider some natural number n. Every subset of N either contains n or it doesn't. So there are 2 possibilities for every natural number n. Again, we are in a similar situation as before. Multiplying thru a countably infinite number of times gives 2 to the Beth null. I'm not saying this is rigorous, but it's how I visualize it.

[@arkeusalexander9054]

@@egheitasean1 This is the thing that i have a hard time dealing with as a logical person. For example if we worked by combining digits in base 10 1 digit -> 10 possible numbers or codes (0-9) 2 digits -> 100 possible numbers (0-99) 3 digits -> 1000 possible numbers (0-999) 4 digits -> 10000 possible numbers (0-9999) n digits -> 10^n possible numbers ∞ digits (countable infinite digits= Aleph0 digits) -> 10^Aleph0 possible numbers And all of those 10^Aleph0 combinaisons would logically constitute the Cardinal of N if we go by the classical definition of Aleph0. That would mean: |N|= 10^Aleph0= Aleph0 But |R|= 2^Aleph 0 > |N| ??????? 🙉🙉🙉

[@linuxp00]

​@@egheitasean1 that's interesting. I also think we could "visualize" the reals as a tuple of three natural numbers, the first number is the sign restricted to be 1 or -1; the second number, represents the integer part and the third number the decimal part. Disconsidering the sign, it is possible to have infinite pairs of infinities, that is, for each integer part there are infinite decimals. It is where we see that if we have Beth-null naturals for each of the two slots, then we have 2^Beth-null = Beth-one (Cardinality of ℝ).

[@MikeRosoftJH]

@@linuxp00 You misunderstand what cardinal exponentiation means. What you are describing is just the Cartesian product - A×B (i.e. the set of all ordered pairs [a,b], where a∈A and b∈B). And Cartesian product doesn't produce a set with greater cardinality: N×N can be mapped one-to-one with N, and R×R can be mapped one-to-one with R. (Assuming axiom of choice, this is true for all infinite sets.) A^B is the set of all functions from set B to set A. In particular, 2^N is the set of al functions from natural numbers to a two-element set ({0,1}). And it can be seen that this exactly corresponds to the set of all subsets of N - f(n) is either 0 or 1, and we map any function to a set of all elements of N where f(n)=1. N^N (the set of all functions on natural numbers) can be mapped one-to-one with real numbers. R^2 (equivalently, R×R - set of all pairs of real numbers) can be mapped one-to-one with real numbers. R^N (set of all infinite sequences of real numbers) can still be mapped one-to-one with real numbers. But 2^R (equivalently, the set of all subsets of real numbers) can be mapped one-to-one with R^R (the set of all functions on real numbers), and these sets can't be mapped one-to-one with real numbers.

[@tomkerruish2982]

​@arkeusalexander9054 Error creeps in when you consider sequences of infinitely many digits. There are no natural numbers with infinitely many digits, only real number decimal expansions. The number of one-digit, two-digit,... numbers is 10+100+..., which is a countable union of finite sets and, therefore, is also countable.

[@gaylynnblanchard9682]

What comes after Beth numbers,ordinals ?

[@aslpuppy1026]

If the continuum hypothesis can’t be proven false that means that you can’t construct a set with cardinality between aleph_0 and beth_1. This is because constructing a set with cardinality between aleph_0 and beth_1 would disprove the continuum hypothesis. So my question is, if it is impossible to construct a set with a cardinality between aleph_0 and beth_1, wouldn’t that mean that the continuum hypothesis is true.

[@MikeRosoftJH]

You can't prove (without additional axioms) that there is such a set whose cardinality is strictly between Aleph-0 and P(Aleph-0); and you can't prove that there isn't such a set. In other words: there's a model of set theory where the set in question doesn't exist (such as the constructible universe); and there is a model in which the set does exist. (The proof uses the technique of forcing: start with some model of set theory, such as the constructible universe, and then add elements to it so that the new model still satisfies the axioms of set theory, and also satisfies the proposition whose consistency with the rest of set theory axioms we are trying to prove; for example, continuum hypothesis is not true, or real numbers can't be well-ordered, or the like.)

[@linuxp00]

Well, according to my current understanding, any real number can be represented as a triplet: [a sign (-1 or 1), a natural number for the integer part (digits to the left of the decimal point), and a natural number for the decimal part (digits to the right of the decimal point)]. So, between any two consecutive natural or integer (naturals + sign) numbers, there are infinite natural numbers as decimal digits, which hint at a discretization, but that are uncontable because just as integers grow unbound, decimals shrink in the same manner, making their count unfeasible. Because of that I think that if continuum hypothesis is set to false, that also allow us to define infinitesimals conceptually, with their fractal nature (i.e. it's possibly to do magnifications on the real number line and it will self similiar, either by zoom in or zoom outs). P.S: Writing π as 3 + 0.14596... = (+1, 3, ...69541). It is possible to have the infinite digits of π, or any irrational.

[@MikeRosoftJH]

@@linuxp00 It's indeed possible to define infinitesimals, by creating an alternate model of first-order theory of real numbers (i.e. the hyperreals). But continuum hypothesis has nothing to do with it; on real numbers there are no non-zero infinitesimals, regardless of whether or not continuum hypothesis is true. Likewise, fractals have nothing to do with this.

[@linuxp00]

@@MikeRosoftJH yeah, I mistook my words on it. But, surreals have this fractal design with their higher-orders of infinities and infinitesimals. May it help consider them in the CH problem.

[@perpetualrabbit]

I feel stupid but I still don't see the difference between Aleph numbers and Beth numbers.

[@frankfruit]

i have a question about the diagonal argument, why can we use the same argument for the natural number and say that there is always a number not on the list. but is it cause the all naturals number s are list able that why no matter what number we make we can find it on the list thus the argument fails for the natural numbers. if so then wouldn't that apply to real numbers too or there is something i am not able to understand about this argument.

[@ReinLie]

@@bakert11 wonderful explanation.

[@Prasen1729]

Hi, the idea is you assume that the set is listable/countable. Set of natural numbers, N is listable as you can write them down : 1, 2, 3, ..... Given any natural number, I can find it in the above list. I can do the same with Z and Q. But not possible with R. Simple reason for me is, N, Z, Q are all discrete sets whereas R is continuous, so if you suppose it is listable, put them in an order no matter how, I can always construct a new real number which was not in the above list. The same way interval (0, 1) is not listable. Here is a try : Suppose the open interval (0, 1) is countable. So lets list them. Here are the decimal numbers. a_11, a_12 etc ... are the digits. A1 : 0.a_11 a_12 a_13 .... a_1n ... A2 : 0.a_21 a_22 a_23 .... a_2n ... A3 : 0.a_31 a_32 a_33 .... a_3n .... ..... An : 0.a_n1 a_n2 a_n3 .... a_nn .... Now if we can find a decimal number which is outside the above list, then our assumption is wrong. Can we find a new number ? Yes we can. Consider this element A = 0.a_1 a_2 a_3 ... a_n ... with the condition that a_1 ≠ a_11 , a_2 ≠ a_22, a_3 ≠ a_33 .... a_n ≠ a_nn. The 1st inequality above proves A ≠ A_1, 2nd inequality proves A ≠ A_2 etc ... nth inequality proves A ≠ A_n etc ... Hence A is not equal to any of the numbers above. Therefore A was not listed and hence the list was not complete. If (0, 1) is not countable then R can't be either as (0, 1) ⊂ R. Here is video on Numberphile useful. kzbin.info/www/bejne/m53ZgI2jZclnfpI

[@MikeRosoftJH]

The reason is: every natural number is finite, and so it has finitely many digits. (And that's by definition; a set is finite, if its number of elements is equal to some natural number - it's an empty set, or it has exactly one element, or it has exactly two elements, or ... and so on.) If you try to apply the diagonal procedure on the sequence of all natural numbers, you get a sequence which has infinitely many non-zero digits, which doesn't represent any natural number. Likewise, when you apply the diagonal procedure to a sequence of all rational numbers - and such a sequence exists - then the resulting number must be irrational.

[@linuxp00]

My argument in favor of the cardinality: |ℝ| > |ℤ| is that you can imagine the real numbers as the Cartesian product between two natural sets, but with a key twist. While the "integer part" of the reals grows in absolute value: {X ∈ ℕ | 0 ≤ |X| ≤ ...99999} (where we can pair numbers from the interval ]-∞, +∞[ to the interval [0,+∞[ ), the "decimal portion" decreases: {Y ∈ ℕ | 1 > 0.9999.... ≥ Y ≥ 0} (also mappable to the naturals, but read backwards). If we plotted this Cartesian product X × Y, we would have an infinite set of decimals for each of the infinite real integer values, i.e., ∞ ^ ∞ (which is mappable to 2^∞, the power sets of the naturals). However, it is still impossible to list all real numbers, as we can always add a new digit to the left in the integer part or to the right in the decimal part. Moreover, even if you try to "move the decimal point" all the way to the right, attempting to transform reals into integers, it is always possible to add a decimal interval after this integer, turning it back into a real number. This is trying to illustrate, in an intuitive way, that you already "exhaust" the set of naturals/integers just to count between 0 and 1, thus the cardinality of the reals is greater than that of the integers and naturals.

[@wdfusroy8463]

So where is the promised "continuation"?? All of the stuff in this video is very basic.

[@nickosc88]

You can’t measure the size of an infinite set, by definition it has no size. You can’t add or subtract objects that have no size. Again, that is a contradiction in terms. Transfinite arithmetic is at best a fanciful speculation (or rather an example of the limits of mathematics) and at worst meaningless jiberish

[@olbluelips]

Me when I can’t find 2-3 because it’s impossible to subtract 3 from 2 and trying to do so would be meaningless speculation

[@fullfungo]

“[…] by definition it has no size”. Yeah, no, this depends on the definition of “size”, or rather “cardinality”. If we define it in such a way that it makes sense for infinite sets, then by definition they will have a size.

[@markosskace514]

Most mathematics is "fanciful speculation" and also "meaningless jiberish". For instance, there isn't any infinities in the physical universe. Mathematics IS NOT about physical universe, it is about relationships between abstract entities - it doesn't need to have connection to anything real.

[@tomkerruish2982]

There are more things in Heaven and Earth (and Mathematics), @nickosc88, than are dreamt of in your philosophy.