Lecture 36/65: Infinity: Countable and Uncountable

[@andrewpersaud4144]

yer a wizard, harry

[@NewCalculus]

Anything but a whiz! At @5:27, I had to laugh because a set need not be finite in order to be countable. You see, mainstream morons don't even understand the theory they propagate. Also, a correspondence with N or a subset of N means a given set is countable. But why N?! The orangutan professors such as Harry have never thought about this. You see, the original definition of a countable set is simply this: 1. A set is countable iff its members can be listed systematically. OR 2. A set is countable iff its members can be given names systematically. OR 3. A set is countable iff it can be indexed. 3 above is the reason Cantor chose N, viz. the "set of natural numbers" can be systematically named in any given radix system. The mythical set of "real numbers" is obviously not countable because most of its members don't exist! There is no such thing as an irrational number. Of course Cantor wouldn't have chosen R as an indexing set in place of N. Chuckle. But mainstream morons would never comprehend these things given that they have no fucking clue what it means to be a "number". Cantor was institutionalised for good reason - he was insane! Set theory is a bunch of crap that has nothing to do with mathematics. Cantor's Diagonal "Argument" is a bowel movement deluxe. Here's the funniest part: if indeed one accepts the bullshit of infinite decimal expansions, then I have proved that R is indeed countable: kzbin.info/www/bejne/np3UhairnbdmY5Y Bullshitology naturally leads to more brain syphilis infections in fools like Prof. Harry because ill-formed concepts always present contradictions and paradoxes. Cantor really fucks up with respect to his theories on bijective cardinality as I explain in the following video: kzbin.info/www/bejne/sGSskJubqcunoZI You are all unbelievably stupid cranks who cannot be turned because you are no better than religious fundamentalists. Priests of the Church of Academia who are incorrigibly stupid, arrogant and ignorant.

[@hectorg362]

@@NewCalculus I think Cantor isn't the only one who's insane on here...

[@Michael19841]

Cantor is the wizard. No one else. This is merely someone explaining Cantor’s work.

[@HNCS2006]

Thank you very much for clearly stating all the assumptions to this proof. I'm a lay person, so i often find other videos explaining this have assumptions or methods that are not made explicit. You have made things clear. For the first thing, I think I'm understanding.

[@anaselmanaa5035]

thank you so much professor i was really struggling with this section and you made it really clear

[@IsaacarWolfsbane]

This point is a bit moot due to how long ago this video was made, but there is a small hole in your argument that there are uncountably many irrational numbers. In fact, there is no reason to believe that the number constructed through the diagonalization cannot be rational (and thus shouldn't be on the list in the first place). This is easily fixed, however, by taking a few extra steps. First, we can use this exact diagonalization argument to show that the interval [0,1] (real numbers between 0 and 1, inclusive) is uncountable. We then have that [0,1] is the union of Q and I, where Q is the set of rationals between 0 and 1, and I denotes the set of irrationals in the same interval. We claim that Q must be countable (as any subset of a countably infinite set is either finite or countably infinite), and therefore we can list it as {q1, q2, q3, ...}. If I were countable, then we could also list its elements as {r1, r2, r3, ...}. From these two correspondences, we can construct a third correspondence {q1, r1, q2, r2, q3, r3, ...} between the natural numbers N and the union of Q and I. Hence Q union I must be countable. This is a contradiction, however, as Q union I = [0,1] which we've shown to be uncountable. Therefore, it cannot be that I is countable and so the set of irrational real numbers must be uncountable as well. Thank you for the videos!

[@hectorramirez626]

Excellent explanation! Great video

[@ghfudrs93uuu]

Geeat video. I was butting heads with this for some days now

[@juditkardos2919]

I would like to know how do you make sure that the number you get is irrational, when you find a number not in the list using diagonalization. The proof presented does not assure this. Without assuring that, you could get a rational as your number not listed, which would not prove that the list of irrationals is uncountable since rationals were not listed to begin with.

[@axel1026]

Great explanation! This helped me a lot!

[@Michael19841]

Perhaps it would be clearer for the average person to understand that anywhere in a sequence of irrationals you can always use Cantor's diagonal argument to prove that it has a greater cardinality than the infinite set of natural numbers.

[@RedShiftedDollar]

A similar diagonal argument can be made for the natural numbers. Take the full list of natural numbers and begin to reveal them one row at a time. After the first number is revealed, to construct a number not in the list you just have to begin the number by starting with a number that differs from the first on the list in at least one place. Then go to the next row. If the value has less digits than the number you have already started to write down or it differs in at least one of the already revealed digits, then the number clearly does not match and proceed to the next row. If the number matches in all digits, then simply write any number in the column to the left of the most significant digit of your number. If the number on the list has more digits than your number then fill in more digits on the left hand side of your number until you at least match the length of the number on the list while making sure it differs in at least one place. When the list is complete, the constructed number will be finished, it will contain a finite number of digits, and it will not be on the list. For example: List Construction of number not in list { 2323 ...3323 change at least 1 digit to start 200 ...3323 200 is guaranteed to be less than the number 3323 ...13323 reveal another digit on most significant side 13322 ...13323 one digit differs. Cannot match. 312313323 ...412313323 match in length and ensure difference in one place 412313323 ...1412313323 ... } When list is complete then stop constructing. As you can see, the constructed number will never be on the list and the final result will have finite length which makes it a natural number. Although this is a bit more complicated of a process than the real number case, the end result is the same. Therefore the set of natural numbers is uncountable.

[@MuffinsAPlenty]

If you work with the list: 1, 2, 3, 4, 5, etc. Then the resulting number will indeed have infinite length. Let's start at 1. We choose to change that number to 2, so our new number is now 2. Next, we get to 2. Our new number is 2, so we change it to be 12. Great. We continue until we reach 12. Our new number is 12, so we change it to be 112. Great. We continue until we reach 112. Our new number is 112, so we change it to be 1112. Great. Repeat ad infinitum. The result will necessarily have infinitely many digits, and thus, not be a natural number.

[@redtree732]

Great video!!! Thank you!

[@afrahnajib1218]

FANTASTIC !

[@rj-nj3uk]

Hi. On what vidoe did you explained about eqyivalence relations and equivalence classes?

[@DanielSColao]

Great video!

[@muthumeena1010]

just love this explanation💗💗 its awesome

[@Nico-xz6th]

Is this like set theory

[@todymaverick]

why are rational numbers a set of N? couldnt they be negative numbers as well?

[@biz_wiz_tech]

excellent job on explaining countable and uncountable

[@syed9576]

Question: Why use Natural numbers for the correspondence to decide whether a set is countable or not countable. Wouldn't it make more sense to use integers? Since integers include negative numbers?

[@asilvap]

a rational number is not that one that can be represented by fraction where the denominator and the numerator are integers? (not only natural number as the video says)

[@kersa96]

Antonio J. Silva P. I agree

[@RJYounglingTricking]

Thanks, Professor. Excellent video!

[@TimberSha]

谢谢

[@NewCalculus]

真的?！这个人是个白痴。

[@shadowslip7125]

Excellent video and certainly cleared a lot of the fog. But, if there is an infinite amount of natural numbers then there will always be enough for an infinite amount of them to correspondence to. I'm using Hillbert's Grand Hotel theory. No matter how many irrational numbers there are there will always be another 'hotel room' available as there are an infinite amount of them. What am I missing?

[@salaharfaj2981]

+Shadowslip 71 There will always be x number not paired to any number in N (So it won't be an onto relation between the two sets). Assume that we reach the last number "infinity" in natural number where it will be mapped to "infinity" number in irrational numbers then we can create x ( x= infinity irrational number that we reached +1) by applying the diagonalization method where that number x isn't mapped to any number in natural numbers. Hope that helps!!

[@shadowslip7125]

Thank you. It took my head a while to get this.

[@divakar6474]

@@salaharfaj2981 continuum says hi to alephnot

[@narendraparmar1631]

Thanks sir god bless you

[@LordBalder]

Why can't I just have the following correspondence between N and R[0..1] ? 1 -> 0.1 | 53 -> 0.53 | 7428583 -> 0.7428583 | 14159265... -> PI - 3 This way, ALL of the real numbers between 0 and 1 are linked to the natural numbers, not just ONTO but CORRESPONDING. I can not find a real number, that is not in that list.

[@flammingodable]

+Lord Balder Hi, I'm not an expert on this, however aren't there more irrational numbers than the set R[0..1] ? Pi ? sqrt(2) ...

[@LordBalder]

Roy Khoury You are right, they are not in this set, but you can show, that there are just as many irrational numbers in R as there are in R[0..1], so the proof of R[0..1] is enough.

[@ahmedelashry8577]

+Lord Balder Good idea, but how about this number 0.0053 ?

[@LordBalder]

Ahmed Elashry I have seen this problem too and came up with a nice solution: put the leading zeros at the end of the number, in the case of 0.0053 -> 53 + 00 -> 5300. This solves 2 problems at once, the missing numbers you mentioned (0.0053 cannot be index 53, because thats 0.53) and double numbers (index 5300 should not be 0.5300, because thats already index 53). It all adds up quite nicely ;-)

[@ahmedelashry8577]

+Lord Balder nice :) it seems good, I couldn't make a counter example

[@jt4401]

you had me up until the very end

[@eugenek951]

rational numbers = m/n such that m, n in Z, n n.e 0

[@christophersedlak1147]

Thanks!!!!!!!!!!!!!!!

[@vaibhav8941]

B is co-domain.

[@kamranl1987]

How is that a proof? we definitely did not list all the irrational numbers, just wrote 6 or 7. how can we proove that based on just 6 numbers? Is there some point I am missing out in all this?

[@NewCalculus]

Nope. Your brain is working perfectly. All the others who claim they understand and are applauding this mainstream crank (prof. harry) are idiots.

[@matthewfrancis5414]

Great explanation Thanks sir :)

[@hardipchander5483]

Ur great

[@henriquembotelho]

How do you film your hands?

[@김지원-m8q]

Similar argument for natural numbers. In the provment of real numbers you showed that you can generate infinitly many new real numbers. I will do the same. First just write the list of odd numbers. And you can generate infinitly many new even numbers. Prove done.

[@codethings271]

where is Hermione harry?

[@Michael19841]

If only you were born before Cantor you’d be a genius 🤔😞😖😣😠

[@lemyul]

good work

[@titotitoburg6298]

I think the diagonal argument is flawed as it doesn't take into account that the vertical plane is many times larger than the horizontal plane(larger ratio of numbers than there are decimal spaces) meaning that a list of all real numbers doesn't form a square, but a long rectangle. Hitting the diagonals on a rectangle you don't touch the bottom the same digit that you touch the right side, meaning that whatever "new" number you form is actually already in the set, lower on the list. additionally if you were to try the same technique(diagonalization) on a set of finite numbers (say one to 999) ------ 1 -818 2 -671 3 -582 4 -173 ... ------ we get 983 Which is already on the list but lower down. see the same proof doesn't hold up for finite sets. Infinite sets are different in that you can't just choose to cut your number off at some point, meaning basically to me, that georg's proof is flawed.

[@MuffinsAPlenty]

"I think the diagonal argument is flawed as it doesn't take into account that the vertical plane is many times larger than the horizontal plane(larger ratio of numbers than there are decimal spaces)" The claim you just made is equivalent to saying that the cardinality of the real numbers is larger than the cardinality of the natural numbers. In other words, you're agreeing with the result of the diagonal argument without realizing it. Let me explain why this is the case. Saying that an infinite set is countable means that there exists a one-to-one correspondence between it and the natural numbers. An infinite "list" is just a metaphor for such a function. The natural number 1 is being mapped to the first number on the list, the natural number 2 is being mapped to the second number on the list, etc. Each real number can be written in decimal notation, and the digits to the right of the decimal point are in one-to-one correspondence with the natural numbers again. In other words, if the set of real numbers between 0 and 1 were countable, then you would necessarily be able to have a list of real numbers which is just as long as it is wide (since the length is in one-to-one correspondence with the natural numbers and so is the width). But you claim the list must necessarily be longer than it is wide. In other words, there are more real numbers than there are decimal places in a real number which (by definition of decimal notation) is the same size as the natural numbers.

[@koskovictor5095]

@@MuffinsAPlenty Unless you assume 2↑Aleph-0=Aleph-0 in the same sense that 2 [ +, × ] Aleph-0=Aleph-0, that is, the power set proof is circular reasoning………

[@MuffinsAPlenty]

Victor Kosko - No, that doesn't contradict what I said. Cardinality is about the _existence_ of a one-to-one correspondence between the two sets in question. It is not a universal statement about all functions between the two sets in question. Even if 2^(Aleph_0) were equal to Aleph_0, the existence of a bijection between the two sets would allow one to create a "square" where the vertical plane is the "same length" as the horizontal plane. Just use the one-to-one correspondence to do this. The fact that the vertical plane is _necessarily_ much longer than the horizontal plane shows that no such bijection is possible.

[@koskovictor5095]

@@MuffinsAPlenty If you want to argue that way nor does 2↑∞=∞ contradict what we said. Well order the reals from 0 inclusive to 1 exclusive to form a set. Since 2↑∞=∞ it's size is w. Pair it's members to a well ordering of the reals from 0 inclusive to ∞=w=Aleph-0 exclusive. The diagonal can't be taken due to the reals thereof are defined as all sequences of 0 and 1 digits binary. TAKING THE DIAGONAL INCREASES THE NUMBER OF DIGITS THUS THE SIZE OF THE SET 2× FOR EACH BIT (due to using the set theory proof techniques so that any time based algorithm is instead instantaneous or outside of time if result can be made by repeated use of powerset axiom. The 'past' then can affect the 'future'.). The number of digits = Aleph-0 and the number of reals = 2↑Aleph-0=Aleph-0 . Same with any permutations of the set of reals from 0 to 1 paired with naturals. It is the diagonal argument we're disproving, not it's conclusion!

[@MuffinsAPlenty]

Victor Kosko - I _think_ you're trying to argue that you can create a "list" of real numbers that is infinite but is not necessarily order isomorphic to the natural numbers, but rather has order type α for some ordinal number α with α > ω. Then, since the list extends past ω, one can create a "new" real number which might not actually be "new". This is because you are only allowed to specify the new number's digits for positive integers (order type ω). As such, the "new" number could actually exist further on the list, in position greater than ω, which would not produce a valid contradiction per Cantor's argument. Is this what you're arguing?

[@serinacat4781]

smart

[@evgeniyan2426]

No correct proof of uncountability.