A set is countable when you can pick any number in side of the set and eventually enumerate it. For example the set of all positive numbers. {1,2,3,4,...,n} if u pick 1 billion you will eventually reach it if you working off the equation x+1 where x is all the natural numbers N={0,1,2,3...n}. Some infinities are bigger then others. Such as the Set of all Real numbers. Real numbers include irrational numbers. So how what equation would let you count to pie? Impossible hence we cant count it
@@oliverbeck6839 No, that statement is true. Between each subinterval of two numbers in (0,1) there is an infinity of numbers. E.g between (0, 0.01) there is an inifinty of numbers. You can start with zero and then set the next number to be 0.0...1, a number that is just a tiny bit larger than 0, by a fraction that is infinitely small. Essentially because values in R can be infinetly small you can fit infinities of numbers in each intervall conceivable interval. Even in an interval like (0, 0.0000000001) you can fit an infinite amount of real numbers because you can always increment the serie with something that is infintely smaller than the diameter of a hairstrain. So the intervall (0, 1) is actually an infinity of infinities when you think about it.
@@taggebagge The statement that the countability of the reals is greater than the integers is most likely true. But using the "proof" described in the video is a bit controversial.
I just learned this and class and watched to understand but instead I am more confused than ever. Thanks anyway solid video I'm just too stupid for this shit
I think that all would find this interesting. I would greatly appreciate comments…………The means which Cantor employed in his proposition of diagnalization, i.e., about the infinite string of real numbers being larger than that of natural numbers is discussed and considered in a context in which it is ignored that there is no such thing as infinity in material reality for it defies the means and manner of existence which is that anything that does exist must be distinct, delineable and quantifiable. This understanding includes the products of the realm of the abstract as well in that there is none which is not ultimately the product of the material, contextual referents in reality, that context from which they arise. For example, the abstraction of a pink flying elephant is one formed of the fusion of the material colour pink, the material phenomenon flying and the material entity, elephant. What mathematicians such as Cantor have done is employ the most general understanding of infinity as a concept but ignore the inevitable contradictions which arise, muddying the waters of the context in which their propositions are formulated and presented. 1. Consider that the infinite string of natural numbers is a progression, that which extends outward (forever). Each unit member is a value, the progression advancing by that value plus 1 each time. However, that to which it is being compared, i.e., the infinite set of real numbers is structurally the opposite within the boundaries of the proposition. • In the infinite string or natural numbers, the span between any two unit members is ignored and the line proceeds from each value to the next, extending out forever. • In the proposed infinite string of real numbers, the list of unit members from the first unit member designated to the next, any other which might be identified (e.g., 1 to 2 or perhaps 1 to 1.00000001, etc.) is itself infinite. For this reason, the string cannot exist beyond its consideration as a line segment which is still problematic for reasons I point out below, its overall length a value arbitrarily assigned but finite. So, in the case of the real numbers, the infinite line of unit members would be contained within two designated units with infinite points between and could extend beyond that. The string of numbers does NOT extend outward but rather within itself. This is comparing apples to oranges. - There could be no list of real numbers for the designation of the very first in the list would never be completed or would just be impossible for it would have infinite digits. None of the real numbers could be designated and thus, nor could the list. This is not unlike the problems that arise with line segments in which it is claimed that they are composed of infinite points, yet they cannot be because if of finite length, each end would have to be designated by a point beyond which there was no other which by definition would mean that those points would have to have scope and dimension which would mean that there could not be infinite points composing the line segment. However, if these end points had scope and dimension, what would that be? If 10x, why not 5x then why not 1x, ad infinitum. Thus, the line segment could NOT be composed of infinite points but at the same time would have to be, demonstrating that infinity cannot be paired with material concepts due to such inevitable contradictions. • What then would be the measure by which the string of real numbers was determined to be larger than that of natural numbers? Here we see that the string of natural numbers was being considered by Cantor as such per its unending length, that length forced by the denial of consideration of the span between specified unit members, e.g., 1 to 2 to 3, etc. However the string of real numbers could not be judged in its size in comparison to the string of natural numbers because it would have infinite members within the span of the first two unit members specified. What this means is that both must be considered in terms of unit members only and not by the abstractions of their lengths, as if each at any specified length would have different quantities of unit members. Instead, the string of natural numbers could not be considered in terms of the span between two designated members and the string of real numbers must be considered in and only in that manner. Since length of the string then is not a consideration, we are left to consider only and compare the number of unit members in which case they are equal, their “quantity” being infinite. • I would venture that because of the above, we can only conclude that the list of natural numbers which is infinite, “stretched” along an infinite line could be “aligned” with the real numbers which are infinite while the “quantity” of them is contained within a finite distance, i.e., the length of a line segment arbitrarily defined. So, the comparison of the one quantity with the other is apart from the means of containment of each. This proposition of Cantor’s seems to be a bad analogy to make a mathematical point and is very sloppy in its disregard for the true nature of these concepts of infinity he employs.
I dont think you have shown what you were trying to prove tho. At most, it shows that assuming a countable set can be contained by a square matrix is problematic, if not impossible. If the setup you propose can be shown to be a property of all countable sets when applying this method; then when you assume the interval is countable, and the results are the same for all countable sets of this format ... then what is the contradiction?
Jared Merlo I think you're thinking about countable infinity. Some people use countable to represent countable infinity, though in this case I think he just means you can count the set.
No; natural numbers are by definition countably infinite, because the set of all natural numbers can be mapped one-to-one with itself. Are you trying to apply the diagonal procedure to the sequence of all natural numbers? Then what you get is an infinite sequence of digits, and that doesn't represent any natural number - every natural number is finite. That's by definition: a set is finite, if its number of elements is equal to some natural numbers. But on the other hand, you have successfully proven that the set of all infinite sequences of digits is uncountably infinite.
@@MikeRosoftJH The entire problem is that numbers didn't exist in sets before they were placed in N. Something is "countable" if it can be bijected to N, but that's crazy, because you can count a set in real life before you "bijected it to N." Numbers themselves aren't "countable" - they are "that which is used for counting to begin with." It's insane to define N as isomorphic to itself. N is actually an illusion to begin with, because N is part of R. It makes no sense to compare R to N.
@@PeteRoyJackson That's not the definition of "countable" which is used in mathematics. "Countable" is not the same thing as "finite". As I have said, a set is countably infinite, if it can be mapped one-to-one with natural numbers; in other words, if the elements can be arranged in a sequence such that every element appears at some finite position. For example, integers are countably infinite, because they can be enumerated by the following infinite sequence: 0, -1, 1, -2, 2, -3, 3, ... - any integer n appears at a position no greater than 2*|n|+1. The same can be proven for rational numbers. But real numbers are uncountable; given any function from natural numbers to real numbers (in other words, any infinite sequence of real numbers), there is some real number which the function (sequence) does not cover. And why should natural numbers be an "illusion" solely by virtue of being a subset of real numbers? To the contrary, it is possible to construct real numbers starting from natural numbers.
@MikeRosoftJH "As I have said, a set is countably infinite, if it can be mapped one-to-one with natural numbers" - Right, I actually said that initially above, no? Something is mathematically "countable" if it can be bijected to N. I'm saying that the *set* (or really, "class") called "N" is an illusion, not the integers themselves which stand apart from it. The countability or uncountability does not make it "less" or "more" infinite, by the mathematical definition of equality. Consider: Let A and B be sets. If A ⊆ B and B ⊆ A then we have A = B. Let f and g be functions from set X to set Y say, such that f(x) = g(x) for all x∈X, we have f = g; the two function are equal. If there is a 1-1 bijective function map set P to another set Q we say the number of elements (cardinality of two sets are equal, and denoted by |P| = |Q|). With two finite sets of the same number of elements, there is always a bijective mapping from one to another. Therefore, if we have a bijective mapping from one to another infinite set, the two sets are of equal number of elements by the mathematical definition of equality: One cannot rationally compare the cardinality of two infinite sets, and say "they are more or less infinite," just like we can't say a set of cardinality "5" is somehow "less" than another set with cardinality "5." The attribute of countable vs. uncountable does not make infinite |N| < infinite |R|. Furthermore, it could be said R is not a "set of numbers," but a continuum of 2 dimensional numeric *processes* that are rendered from some kind of algorithm. This is why they can't "rationally" be compared. Sure, there can be more "processes" than integers, but that again doesn't make the infinity of "R" greater than "N." 2 exists in N, but √2 exists in R. 2 is a number, √2 is a numeric process UPON a number (result). Numbers AND numeric processes combined into an entity doesn't make R a "bigger infinity" - it makes a "non-rational thing" that's not to be compared to a "rational thing." E.g., "What's bigger, a piece of fruit or an apple?" All of N is contained "within" R. R is the fruit, N is the apple. The elements of R have interminability, and therefore cannot be technically delimited in a set or class to begin with, because to delimit them is to limit the limitless and to rationalize the irrational. R is therefore not even a set - it is an unbroken continuum of points which are not "elementized," and where each point should not be considered as discretely knowable elements of a set.
I don't get it, and I don't think it's right. If you think of the binary stings with length 2, the set is 00, 01, 10, 11. Every number on the diagonal in tabular form would also be a number in the set. For example the diagonal of the first two would form 01, which is the second item. What am I missing?
The diagonal argument would require your third string to have at least 3 digits, your fourth string to have at least 4 digits, and so on. Personally, I don't like it, because as you pointed out.. geometrically, listing all possible combinations of digits creates a rectangle, not a square; and you can't make a (one down, one over) diagonal between two corners of a rectangle, but it's built in to the argument. The argument allows, perhaps wrongly, the infinity of digits to overcome the shape that would be created by instantiating progressively larger, but finite, strings that account for all possible permutations of digits.
I'm not sure what the heart of your question is. But I can, at least, tell you that what you suggest is impossible. Every infinite set has a proper subset that is the same size as itself. With just a step more of logic, this implies that every infinite set has infinitely many proper subsets that are the same size as itself. So it's impossible for a set to have only one subset that is uncountable (that uncountable subset would have infinitely many uncountable subsets, which would then be uncountable subsets of the original set).
If I'm understanding you correctly, it only takes one uncountable subset to prove the point. No countable subset could disprove the point. It's like saying "No ravens are black." You only need to find one black raven (the metaphor for the uncountable set) to show the contradiction.
+Veritas There are different definitions of N. One is that it is {0,1,2...} and the other one is {1,2,3...}. So according to the first definition N =/= Z+
Instead of having a row for each integer with a real, could you not have a row for each real and a corresponding integer. The number of digits in an integer is boundless because they are also infinite, so can you not perform the same diagonalisation to produce a new integer that is different to all others?
That doesn't even make sense. Digits in a decimal representation are indexed by natural numbers, not by real numbers. And if you were to try to prove that the set of natural numbers can't be mapped one-to-one with itself by applying a diagonal procedure to a sequence of natural numbers, that doesn't work, either, for a different reason: every natural number is by definition finite (a set is finite, if its number of elements is equal to some natural number); so it has finitely many digits. If you apply the diagonal procedure to a sequence of natural numbers, you get a string with infinitely many non-zero digits, which doesn't represent any natural number. You can use a diagonal proof to show that real numbers can't be mapped one-to-one with the set of all functions on reals.
@@MikeRosoftJH I'll have to watch it again to be sure, but I think what I was thinking is that a diagonal proof could be used to prove the integers, or at least natural numbers are infinite, i.e, boundless. Not that there are infinitely many naturals between two naturals, because there is not, there are a finite number of naturals between two naturals.
@@earthstick Well, of course the set of all natural numbers is infinite (for any natural number n there exists n+1). That doesn't need diagonal proof. But every natural number is finite in magnitude; and that doesn't contradict that there are infinitely many natural numbers. That all ultimately depends on how natural numbers are constructed in set theory. We define any natural number as a set of numbers less than itself: 0 is the empty set, 1 is {0}, 2 is {0,1}, 3 is {0,1,2}, and so on. Observe that 1) any natural number n is a set with exactly n elements, and 2) the number n+1 is obtained from n by adding n itself to it: n+1=n∪{n}. Now we want there to exist a set of all natural numbers; but we can't have a formula of an infinite length. So the construction is a bit involved. Set theory has an axiom of infinity, which essentially says that there exists a set which contains all natural numbers. Formally, it says: There exists a set Ω with the following properties: 1) Ω contains 0 (the empty set); and 1) given any set n, if Ω contains n, then it also contains n∪{n} (that's how we have defined n+1). Now we define ω - the set of all natural numbers - as the intersection of all sets which satisfy the axiom of infinity. This set exists by the axiom of separation: Let Ω be any set which satisfies the axiom of infinity; then ω={x: x∈Ω, and ∀Ω': if Ω' satisfies axiom of infinity, then x∈Ω'}. It can be seen that this set ω is non-empty - for example, it contains 0 - and that it itself satisfies the axiom of infinity. Finally, it can be shown that the set has the property of induction: let P(x) be any proposition, then suppose P(0) is true, and that given any n: if n∈ω, then n+1∈ω; then P(n) is true for all elements of ω. We define: a set is finite, if it can be mapped one-to-one with any element of ω (that means that it is an empty set, or it has exactly one element, or it has exactly two elements, or... and so on). However, no two elements of ω can be mapped one-to-one with each other, and no element of ω can be mapped one-to-one with ω itself (in other words, ω is an infinite set).
@@MikeRosoftJH I'll read this, but first I want to say that I am aware of Peano's axioms. It just occurred to me that a diagonal proof could be used to prove the natural numbers are infinite, not that it should be. But I don't think my original comment said it should be used.
@@earthstick I think that it sort of makes sense; let f be any function from a finite set to natural numbers; then construct a diagonal number of this function which is different number than any value of the function. And that's a number with finitely many digits. I am not sure if this is really a valid proof - if it doesn't indirectly assume what it is trying to prove. That every natural number has a unique decimal expansion of a finite length (and conversely, every decimal expansion defines a natural number) is a rather advanced property of natural numbers. At minimum, it requires the + and * operations to be defined; and these don't really make sense without the set of natural numbers being infinite. The proof that the set of natural numbers is infinite only needs the successor operation - S(n)=n∪{n}. (Writing the successor as n+1 is a bit of an abuse of notation, because we haven't defined the + operator yet.)
think about space. If I have two empty containers how much "space is there?" if a is a bigger container than b, a and b both have infinite amount of space, but a has more space. Real world sometimes relies on abstractions
This does not work because there are "decimal numbers" which have infinitely many decimal places, but there are no natural numbers that have infinitely many (nonzero) digits. So there are plenty of decimal numbers which are not mapped to anything under your proposed mapping.
MuffinsAPlenty.. Then what is the maximum number of non-zero digits for a natural number? If it's not infinite, it seems to me it would have to be a finite number.
Just because no individual natural number has infinitely many digits does not imply that there must be a maximum number of digits. Natural numbers can have arbitrarily large numbers of digits (meaning that for any natural number n, there is a natural number with n digits). But this does not imply that there is some global cap on the number of digits natural numbers, in general, could have. Here's a way to turn your question to me around. We know that there are certainly natural numbers with only finitely many nonzero digits. We both agree, no doubt, that such numbers exist. So what's the largest natural number with only finitely many nonzero digits? If your question were based on valid reasoning, then this question should have an answer, based on the same reasoning.
Perhaps I missed some subtlety in your point, but my point was that there is no maximum natural number, and there is no maximum number of digits. There is no upper limit. It is unbounded. Unless I'm missing something, that's what makes it infinite. If there can be an infinite number of natural numbers, each representing the number of digits of some set of natural numbers, I don't see why there wouldn't be infinitely many natural numbers with an infinite number of digits.
I agree that there is no maximum natural number and that there is no maximum number of digits. But this doesn't mean that there is a single natural number with infinitely many digits. You can have infinitely many things that are all finite in value. This is actually pretty common in formal mathematics: the number of sizes of something is bigger than any of those sizes itself. So the number of natural numbers is bigger than any natural number. The number of types of infinity is bigger than any of those types of infinity. etc. My counter to your point was as follows: Your argument seemed to be along the lines of saying that if every natural number had finitely many digits, then there would have to be a largest number of digits. But this logic is faulty. Let's say that you were entirely right about there being natural numbers with infinitely many digits and that your point above was correct. Let's now just focus on the natural numbers with finitely many digits. By your point above, there would have to be a maximum number of digits. As such, there would have to be a largest natural number with finitely many digits. (Because that's what your point was: if every natural number has only finitely many digits, then there must be a maximum number of digits.) So if we restrict our attention to only the natural numbers that have finitely many digits, by that reasoning, there must be a largest one. So my question was: What is that? The point was to demonstrate that your logic was flawed. The argument you were making would imply the existence of a largest finite-valued number. This would also imply the existence of a finite-valued number n so that n+1 was infinite-valued, which is also not possible. It's possible to have infinitely many finite-valued things.
Yo dude. Just FYI, I'm not sure that finite sets are considered "countable." I think that countable means that it has the same cardinality as the natural numbers ... For example, the following theorem says: If A is a subset of B, then A is either countable, finite, or empty....
How are finite sets not countable? You just count the number of elements in the set. Ex: {1, 2, 3, 4, 5} In this set, the cardinality is 5 b/c there are five elements.
@@OhItzAutumn It's a matter of terminology; some mathematicians use the word "countable" to mean "countably infinite" (and, for lack of ambiguity, "at most countable" to mean "finite or countably infinite").
Read Wilfrid Hodges article "An Editor Recalls Some Hopeless Papers". People object ALL THE TIME. Maths journals are flooded every month with (awful) arguments against Cantor's diagonal argument. I've run logic conferences and we get philosophy postgrads + amateurs objecting ALL THE TIME.
It is never used up but it still can't be considered the same infinity You can always count the next natural number, but you can't count the next decimal since you would have to add another 0 first. This means there can't be a one-to-one relationship between the two, even if neither are used up
@@newone2870 I think what's important here is that everyone is willing to question math. Since it's impossible to look at things scientifically by making measurements, we must rely on each other to find flaws
thanks the trollface addition really helped to drive home the argument
I mean this video did literally come out 12 years ago. This video is straight outta 2012
Bruh if this was really a group of students they explained this better than my prof who's earning 6 figures. I'm totally getting my money's worth
Sir Jection is my favorite knight of the round table.
You had me at Trollface.
A set is countable when you can pick any number in side of the set and eventually enumerate it. For example the set of all positive numbers. {1,2,3,4,...,n} if u pick 1 billion you will eventually reach it if you working off the equation x+1 where x is all the natural numbers N={0,1,2,3...n}. Some infinities are bigger then others. Such as the Set of all Real numbers. Real numbers include irrational numbers. So how what equation would let you count to pie? Impossible hence we cant count it
just gonna say this really helped for my calc project and you are great!
The crucial observation is that for each integer n, a differs in the nth decimal
position from the nth number on the list.
It's pretty amazing that there's more real numbers between 0 and 1 than there are integers
Is this false? There are infinite integers aswell as infinite number between 0 and 1? How can you say which infinite is larger?
@@oliverbeck6839
No, that statement is true. Between each subinterval of two numbers in (0,1) there is an infinity of numbers. E.g between (0, 0.01) there is an inifinty of numbers. You can start with zero and then set the next number to be 0.0...1, a number that is just a tiny bit larger than 0, by a fraction that is infinitely small. Essentially because values in R can be infinetly small you can fit infinities of numbers in each intervall conceivable interval.
Even in an interval like (0, 0.0000000001) you can fit an infinite amount of real numbers because you can always increment the serie with something that is infintely smaller than the diameter of a hairstrain. So the intervall (0, 1) is actually an infinity of infinities when you think about it.
@@taggebagge The statement that the countability of the reals is greater than the integers is most likely true. But using the "proof" described in the video is a bit controversial.
10/10 troll face gave me unexpected flashbacks to the 2010s
Way better explaination than my goddamn professor
very detailed explanation, thank you so much!
Very clear explanation. The proof from Axiomatic Set Theory had my head spinning :/
Very helpful, and great job explaining as well.
All infinities are big. But some infinities are more big that others, lol!
nice video. the idea is very clear.
Anyone else notice his voice change at 6:20?
+MasterTallness These videos were part of a group project for teaching discreet math. Different team members did different sections.
+Malkey Ah, very goo.d I just found it somewhat humorous. Sounded like the first guy got a cold all of a sudden.
I just learned this and class and watched to understand but instead I am more confused than ever. Thanks anyway solid video I'm just too stupid for this shit
I think that all would find this interesting. I would greatly appreciate comments…………The means which Cantor employed in his proposition of diagnalization, i.e., about the infinite string of real numbers being larger than that of natural numbers is discussed and considered in a context in which it is ignored that there is no such thing as infinity in material reality for it defies the means and manner of existence which is that anything that does exist must be distinct, delineable and quantifiable. This understanding includes the products of the realm of the abstract as well in that there is none which is not ultimately the product of the material, contextual referents in reality, that context from which they arise. For example, the abstraction of a pink flying elephant is one formed of the fusion of the material colour pink, the material phenomenon flying and the material entity, elephant. What mathematicians such as Cantor have done is employ the most general understanding of infinity as a concept but ignore the inevitable contradictions which arise, muddying the waters of the context in which their propositions are formulated and presented.
1. Consider that the infinite string of natural numbers is a progression, that which extends outward (forever). Each unit member is a value, the progression advancing by that value plus 1 each time. However, that to which it is being compared, i.e., the infinite set of real numbers is structurally the opposite within the boundaries of the proposition.
• In the infinite string or natural numbers, the span between any two unit members is ignored and the line proceeds from each value to the next, extending out forever.
• In the proposed infinite string of real numbers, the list of unit members from the first unit member designated to the next, any other which might be identified (e.g., 1 to 2 or perhaps 1 to 1.00000001, etc.) is itself infinite. For this reason, the string cannot exist beyond its consideration as a line segment which is still problematic for reasons I point out below, its overall length a value arbitrarily assigned but finite. So, in the case of the real numbers, the infinite line of unit members would be contained within two designated units with infinite points between and could extend beyond that. The string of numbers does NOT extend outward but rather within itself. This is comparing apples to oranges.
- There could be no list of real numbers for the designation of the very first in the list would never be completed or would just be impossible for it would have infinite digits. None of the real numbers could be designated and thus, nor could the list. This is not unlike the problems that arise with line segments in which it is claimed that they are composed of infinite points, yet they cannot be because if of finite length, each end would have to be designated by a point beyond which there was no other which by definition would mean that those points would have to have scope and dimension which would mean that there could not be infinite points composing the line segment. However, if these end points had scope and dimension, what would that be? If 10x, why not 5x then why not 1x, ad infinitum. Thus, the line segment could NOT be composed of infinite points but at the same time would have to be, demonstrating that infinity cannot be paired with material concepts due to such inevitable contradictions.
• What then would be the measure by which the string of real numbers was determined to be larger than that of natural numbers? Here we see that the string of natural numbers was being considered by Cantor as such per its unending length, that length forced by the denial of consideration of the span between specified unit members, e.g., 1 to 2 to 3, etc. However the string of real numbers could not be judged in its size in comparison to the string of natural numbers because it would have infinite members within the span of the first two unit members specified. What this means is that both must be considered in terms of unit members only and not by the abstractions of their lengths, as if each at any specified length would have different quantities of unit members. Instead, the string of natural numbers could not be considered in terms of the span between two designated members and the string of real numbers must be considered in and only in that manner. Since length of the string then is not a consideration, we are left to consider only and compare the number of unit members in which case they are equal, their “quantity” being infinite.
• I would venture that because of the above, we can only conclude that the list of natural numbers which is infinite, “stretched” along an infinite line could be “aligned” with the real numbers which are infinite while the “quantity” of them is contained within a finite distance, i.e., the length of a line segment arbitrarily defined. So, the comparison of the one quantity with the other is apart from the means of containment of each.
This proposition of Cantor’s seems to be a bad analogy to make a mathematical point and is very sloppy in its disregard for the true nature of these concepts of infinity he employs.
Is countability predicable? # philosophical question 😕
I dont think you have shown what you were trying to prove tho. At most, it shows that assuming a countable set can be contained by a square matrix is problematic, if not impossible. If the setup you propose can be shown to be a property of all countable sets when applying this method; then when you assume the interval is countable, and the results are the same for all countable sets of this format ... then what is the contradiction?
Thank you so much :) Super helpful
Super helpful thanks dude
Jared Merlo I think you're thinking about countable infinity. Some people use countable to represent countable infinity, though in this case I think he just means you can count the set.
Thank you!
Countable is a set infinite, Natural Number are countable e Infinite.
Then are naturals also uncountable?
No; natural numbers are by definition countably infinite, because the set of all natural numbers can be mapped one-to-one with itself.
Are you trying to apply the diagonal procedure to the sequence of all natural numbers? Then what you get is an infinite sequence of digits, and that doesn't represent any natural number - every natural number is finite. That's by definition: a set is finite, if its number of elements is equal to some natural numbers. But on the other hand, you have successfully proven that the set of all infinite sequences of digits is uncountably infinite.
i can't tell iff your eyes are open or not
@@MikeRosoftJH The entire problem is that numbers didn't exist in sets before they were placed in N. Something is "countable" if it can be bijected to N, but that's crazy, because you can count a set in real life before you "bijected it to N." Numbers themselves aren't "countable" - they are "that which is used for counting to begin with." It's insane to define N as isomorphic to itself. N is actually an illusion to begin with, because N is part of R. It makes no sense to compare R to N.
@@PeteRoyJackson That's not the definition of "countable" which is used in mathematics. "Countable" is not the same thing as "finite". As I have said, a set is countably infinite, if it can be mapped one-to-one with natural numbers; in other words, if the elements can be arranged in a sequence such that every element appears at some finite position. For example, integers are countably infinite, because they can be enumerated by the following infinite sequence: 0, -1, 1, -2, 2, -3, 3, ... - any integer n appears at a position no greater than 2*|n|+1. The same can be proven for rational numbers. But real numbers are uncountable; given any function from natural numbers to real numbers (in other words, any infinite sequence of real numbers), there is some real number which the function (sequence) does not cover.
And why should natural numbers be an "illusion" solely by virtue of being a subset of real numbers? To the contrary, it is possible to construct real numbers starting from natural numbers.
@MikeRosoftJH "As I have said, a set is countably infinite, if it can be mapped one-to-one with natural numbers" - Right, I actually said that initially above, no? Something is mathematically "countable" if it can be bijected to N. I'm saying that the *set* (or really, "class") called "N" is an illusion, not the integers themselves which stand apart from it. The countability or uncountability does not make it "less" or "more" infinite, by the mathematical definition of equality.
Consider:
Let A and B be sets. If A ⊆ B and B ⊆ A then we have A = B.
Let f and g be functions from set X to set Y say, such that f(x) = g(x) for all x∈X, we have f = g; the two function are equal.
If there is a 1-1 bijective function map set P to another set Q we say the number of elements (cardinality of two sets are equal, and denoted by |P| = |Q|).
With two finite sets of the same number of elements, there is always a bijective mapping from one to another.
Therefore, if we have a bijective mapping from one to another infinite set, the two sets are of equal number of elements by the mathematical definition of equality:
One cannot rationally compare the cardinality of two infinite sets, and say "they are more or less infinite," just like we can't say a set of cardinality "5" is somehow "less" than another set with cardinality "5." The attribute of countable vs. uncountable does not make infinite |N| < infinite |R|.
Furthermore, it could be said R is not a "set of numbers," but a continuum of 2 dimensional numeric *processes* that are rendered from some kind of algorithm. This is why they can't "rationally" be compared. Sure, there can be more "processes" than integers, but that again doesn't make the infinity of "R" greater than "N." 2 exists in N, but √2 exists in R. 2 is a number, √2 is a numeric process UPON a number (result).
Numbers AND numeric processes combined into an entity doesn't make R a "bigger infinity" - it makes a "non-rational thing" that's not to be compared to a "rational thing." E.g., "What's bigger, a piece of fruit or an apple?" All of N is contained "within" R. R is the fruit, N is the apple. The elements of R have interminability, and therefore cannot be technically delimited in a set or class to begin with, because to delimit them is to limit the limitless and to rationalize the irrational. R is therefore not even a set - it is an unbroken continuum of points which are not "elementized," and where each point should not be considered as discretely knowable elements of a set.
You didn’t define real number.
I don't get it, and I don't think it's right. If you think of the binary stings with length 2, the set is 00, 01, 10, 11. Every number on the diagonal in tabular form would also be a number in the set. For example the diagonal of the first two would form 01, which is the second item. What am I missing?
You have to be able to draw a diagonal that intersects with the entire list of strings
The diagonal argument would require your third string to have at least 3 digits, your fourth string to have at least 4 digits, and so on. Personally, I don't like it, because as you pointed out.. geometrically, listing all possible combinations of digits creates a rectangle, not a square; and you can't make a (one down, one over) diagonal between two corners of a rectangle, but it's built in to the argument.
The argument allows, perhaps wrongly, the infinity of digits to overcome the shape that would be created by instantiating progressively larger, but finite, strings that account for all possible permutations of digits.
thank you so much
So what if only one subset of a set is uncountable but the others are not? Does the proof not break then?
I'm not sure what the heart of your question is. But I can, at least, tell you that what you suggest is impossible. Every infinite set has a proper subset that is the same size as itself. With just a step more of logic, this implies that every infinite set has infinitely many proper subsets that are the same size as itself. So it's impossible for a set to have only one subset that is uncountable (that uncountable subset would have infinitely many uncountable subsets, which would then be uncountable subsets of the original set).
If I'm understanding you correctly, it only takes one uncountable subset to prove the point. No countable subset could disprove the point. It's like saying "No ravens are black." You only need to find one black raven (the metaphor for the uncountable set) to show the contradiction.
Doug George: That sounds correct to me.
perfect!
Something is terribly wrong with this! 🤔
Strange how you use Z+ instead of N.
+Veritas There are different definitions of N. One is that it is {0,1,2...} and the other one is {1,2,3...}. So according to the first definition N =/= Z+
+Veritas N includes 0, Z+ doesn't
+Maarten Vorwerk Like the guy above you said, there are two definitions for N, one includes 0 and one does not.
Instead of having a row for each integer with a real, could you not have a row for each real and a corresponding integer. The number of digits in an integer is boundless because they are also infinite, so can you not perform the same diagonalisation to produce a new integer that is different to all others?
That doesn't even make sense. Digits in a decimal representation are indexed by natural numbers, not by real numbers. And if you were to try to prove that the set of natural numbers can't be mapped one-to-one with itself by applying a diagonal procedure to a sequence of natural numbers, that doesn't work, either, for a different reason: every natural number is by definition finite (a set is finite, if its number of elements is equal to some natural number); so it has finitely many digits. If you apply the diagonal procedure to a sequence of natural numbers, you get a string with infinitely many non-zero digits, which doesn't represent any natural number.
You can use a diagonal proof to show that real numbers can't be mapped one-to-one with the set of all functions on reals.
@@MikeRosoftJH I'll have to watch it again to be sure, but I think what I was thinking is that a diagonal proof could be used to prove the integers, or at least natural numbers are infinite, i.e, boundless. Not that there are infinitely many naturals between two naturals, because there is not, there are a finite number of naturals between two naturals.
@@earthstick Well, of course the set of all natural numbers is infinite (for any natural number n there exists n+1). That doesn't need diagonal proof. But every natural number is finite in magnitude; and that doesn't contradict that there are infinitely many natural numbers.
That all ultimately depends on how natural numbers are constructed in set theory. We define any natural number as a set of numbers less than itself: 0 is the empty set, 1 is {0}, 2 is {0,1}, 3 is {0,1,2}, and so on. Observe that 1) any natural number n is a set with exactly n elements, and 2) the number n+1 is obtained from n by adding n itself to it: n+1=n∪{n}. Now we want there to exist a set of all natural numbers; but we can't have a formula of an infinite length. So the construction is a bit involved. Set theory has an axiom of infinity, which essentially says that there exists a set which contains all natural numbers. Formally, it says: There exists a set Ω with the following properties: 1) Ω contains 0 (the empty set); and 1) given any set n, if Ω contains n, then it also contains n∪{n} (that's how we have defined n+1). Now we define ω - the set of all natural numbers - as the intersection of all sets which satisfy the axiom of infinity. This set exists by the axiom of separation: Let Ω be any set which satisfies the axiom of infinity; then ω={x: x∈Ω, and ∀Ω': if Ω' satisfies axiom of infinity, then x∈Ω'}.
It can be seen that this set ω is non-empty - for example, it contains 0 - and that it itself satisfies the axiom of infinity. Finally, it can be shown that the set has the property of induction: let P(x) be any proposition, then suppose P(0) is true, and that given any n: if n∈ω, then n+1∈ω; then P(n) is true for all elements of ω. We define: a set is finite, if it can be mapped one-to-one with any element of ω (that means that it is an empty set, or it has exactly one element, or it has exactly two elements, or... and so on). However, no two elements of ω can be mapped one-to-one with each other, and no element of ω can be mapped one-to-one with ω itself (in other words, ω is an infinite set).
@@MikeRosoftJH I'll read this, but first I want to say that I am aware of Peano's axioms. It just occurred to me that a diagonal proof could be used to prove the natural numbers are infinite, not that it should be. But I don't think my original comment said it should be used.
@@earthstick I think that it sort of makes sense; let f be any function from a finite set to natural numbers; then construct a diagonal number of this function which is different number than any value of the function. And that's a number with finitely many digits. I am not sure if this is really a valid proof - if it doesn't indirectly assume what it is trying to prove. That every natural number has a unique decimal expansion of a finite length (and conversely, every decimal expansion defines a natural number) is a rather advanced property of natural numbers. At minimum, it requires the + and * operations to be defined; and these don't really make sense without the set of natural numbers being infinite.
The proof that the set of natural numbers is infinite only needs the successor operation - S(n)=n∪{n}. (Writing the successor as n+1 is a bit of an abuse of notation, because we haven't defined the + operator yet.)
Ayyy I'm taking 203 at Umich right now!
think about space. If I have two empty containers how much "space is there?" if a is a bigger container than b, a and b both have infinite amount of space, but a has more space. Real world sometimes relies on abstractions
My f(x) is to mirror the set about the decimal point.
This does not work because there are "decimal numbers" which have infinitely many decimal places, but there are no natural numbers that have infinitely many (nonzero) digits. So there are plenty of decimal numbers which are not mapped to anything under your proposed mapping.
MuffinsAPlenty.. Then what is the maximum number of non-zero digits for a natural number? If it's not infinite, it seems to me it would have to be a finite number.
Just because no individual natural number has infinitely many digits does not imply that there must be a maximum number of digits. Natural numbers can have arbitrarily large numbers of digits (meaning that for any natural number n, there is a natural number with n digits). But this does not imply that there is some global cap on the number of digits natural numbers, in general, could have.
Here's a way to turn your question to me around. We know that there are certainly natural numbers with only finitely many nonzero digits. We both agree, no doubt, that such numbers exist. So what's the largest natural number with only finitely many nonzero digits? If your question were based on valid reasoning, then this question should have an answer, based on the same reasoning.
Perhaps I missed some subtlety in your point, but my point was that there is no maximum natural number, and there is no maximum number of digits. There is no upper limit. It is unbounded. Unless I'm missing something, that's what makes it infinite.
If there can be an infinite number of natural numbers, each representing the number of digits of some set of natural numbers, I don't see why there wouldn't be infinitely many natural numbers with an infinite number of digits.
I agree that there is no maximum natural number and that there is no maximum number of digits. But this doesn't mean that there is a single natural number with infinitely many digits. You can have infinitely many things that are all finite in value. This is actually pretty common in formal mathematics: the number of sizes of something is bigger than any of those sizes itself. So the number of natural numbers is bigger than any natural number. The number of types of infinity is bigger than any of those types of infinity. etc.
My counter to your point was as follows: Your argument seemed to be along the lines of saying that if every natural number had finitely many digits, then there would have to be a largest number of digits. But this logic is faulty. Let's say that you were entirely right about there being natural numbers with infinitely many digits and that your point above was correct. Let's now just focus on the natural numbers with finitely many digits. By your point above, there would have to be a maximum number of digits. As such, there would have to be a largest natural number with finitely many digits. (Because that's what your point was: if every natural number has only finitely many digits, then there must be a maximum number of digits.) So if we restrict our attention to only the natural numbers that have finitely many digits, by that reasoning, there must be a largest one. So my question was: What is that?
The point was to demonstrate that your logic was flawed. The argument you were making would imply the existence of a largest finite-valued number. This would also imply the existence of a finite-valued number n so that n+1 was infinite-valued, which is also not possible.
It's possible to have infinitely many finite-valued things.
Yo dude. Just FYI, I'm not sure that finite sets are considered "countable." I think that countable means that it has the same cardinality as the natural numbers ...
For example, the following theorem says:
If A is a subset of B, then A is either countable, finite, or empty....
How are finite sets not countable? You just count the number of elements in the set. Ex: {1, 2, 3, 4, 5} In this set, the cardinality is 5 b/c there are five elements.
I was also taught that countable and finite are different categories.
@@OhItzAutumn It's a matter of terminology; some mathematicians use the word "countable" to mean "countably infinite" (and, for lack of ambiguity, "at most countable" to mean "finite or countably infinite").
How come no one studying math ever objects to Cantor's proof based on it's flaw? The countable infinity is never used up, never ended.
Read Wilfrid Hodges article "An Editor Recalls Some Hopeless Papers". People object ALL THE TIME. Maths journals are flooded every month with (awful) arguments against Cantor's diagonal argument. I've run logic conferences and we get philosophy postgrads + amateurs objecting ALL THE TIME.
It is never used up but it still can't be considered the same infinity
You can always count the next natural number, but you can't count the next decimal since you would have to add another 0 first. This means there can't be a one-to-one relationship between the two, even if neither are used up
Why can't you just count 10 times what you've already counted to add a 0?
Because all math seems to assume ad infinitum = 1 based on a leap of faith. Mathematics is based on faith but no one is willing to acknowledge it.
@@newone2870 I think what's important here is that everyone is willing to question math. Since it's impossible to look at things scientifically by making measurements, we must rely on each other to find flaws