What if countability is inconsistent? (adopted from Mueckenheim): If the fractions m/n are enumerated by the natural numbers k according to Cantor's function k = (m + n - 1)(m + n - 2)/2 + m then all the fractions of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... are enumerated. But if the natural numbers first are in bijection with the integer fractions of the first column of the matrix 1/1, 1/2, 1/3, 1/4, ... 2/1, 2/2, 2/3, 2/4, ... 3/1, 3/2, 3/3, 3/4, ... 4/1, 4/2, 4/3, 4/4, ... 5/1, 5/2, 5/3, 5/4, ... ... then they must be distributed over the matrix such that no fraction remains without index. That means, there is a permutation such that the X of the first column XOOOO... XOOOO... XOOOO... XOOOO... XOOOO... ... by being exchanged with the O's cover all matrix positions. All O's will vanish. This is obviously impossible because exchanging cannot reduce them. The number of not indexed fractions, represented by O's, will remain constant forever, in infinity. Why do mathematicians believe in Cantor yet?
we just got introduced to discrete maths, but I don't understand why we define ei like ej = djj + 2 (mod 10) and then show a contradiction with that definition. How does that prove that R is unlistable? We had an example on why N and Z are countable and thus why the relation is a bijection by putting a number N; based on if n is even or uneven in a formula and then showing that each n has a specific z and vice versa if I, for example, contradict this 'defined algorithm' is it then proven that there is no bijection between N and Z?
The thing about Cantor's diagonal argument is that it is an _argument._ It isn't a list. It's an argument that can be applied to _any_ list. A set is countable if _there exists_ at least one bijection between that set an N. That set will also have functions between it and N which are not bijections. That's okay. But Cantor's diagonal argument applies to _every_ list of real numbers. For each list of real numbers, there is a real number not on your list! (And you can prove that using Cantor's diagonal argument.) So there can't be a bijection with N. Any attempt to build a bijection with N will fail - because Cantor's diagonal argument will find a real number not hit by your attempted bijection. Make sense?
Anyone noticed that the procedure for determining the digits of the diagonal number has the same form as that for determining the digits of a computable number? That is, given any positive integer n, you have a procedure for determining the nth digit of the number in a finite number of steps. But it has long been known that the cardinality of the set of computable numbers is the same as that of the integers. So therefore the Cantor diagonal construction can only ever produce numbers from a set with the same cardinality as the integers. So how does it prove the existence of a larger set?
I think it would be easier for the student to introduce first the idea that innumerability is not (just) a property of infinite sets. This is easy to show using Cantor's power set theorem. You go through the standard diagonal proof reasonably well. But there is still one point that, in my opinion, needs to be cleared up. The diagonal proof is working on digits, not on the number itself. It does not matter that you have declared the digits to be in a list of partial sums. The proof does not use this information. The point is that the proof is incomplete until you show that new "values" have been created. For example, consider the case where the interval is the natural numbers. Let's assume that the natural numbers are modular at the infinity limit. Then, although an innumerabilty of digit strings is proved, there are no new numbers created. Thus disproving the notion that the Reals are innumerable. It then becomes subject to an axiom, i.e. the Natural numbers at the infinity limit are not modular. I have actually never seen this problem raised, so maybe there is something I have overlooked, but I can't see it myself.
Enjoyed the proof, absolutely correct. But wondering but is there any case where the mod 2 will cause a problem,if not how can we prove it/ understand it?
exactly if we add 2 to all digit then there may be a case like 0.8888.... which will become 0.00000... and if we add 3 the again there may be a case like 0.77777..... which will become 0.00000....ad this will happen with ever digit from 1 to 9. I think this proof is not valid.
This video is pretty confusing. You start by defining a countable set S as a set for which there exists an injection between S and Z+, but then you proceed to *not* use this definition. Are you using an equivalent definition of countability, something to do with surjections?
If the "d" sequence on the diagonal was, say ".23577777777777......", then adding 2 modulo 10, would generate a number ".457999999999999.....". You have proved that this sequence is not on the list, but what about ".4580000000000000", which is, after all, the same number?
But was 0.458 on the original list? It's not the first number, since the first digit of that number is 2 not 4. It is not the second digit, since the second digit in that number is 3, not 5. It is not the third number, because the third digit of that number was 5, not 8. It is not the nth number, n >= 4, because the nth digit of that number is 7, not 0. So 0.458 was not in the list. The proof is not very convincing but it is hard to find a counterexample. If we use binary instead of decimal notation, there is a specific example where the diagonal argument fails.
@@alnitaka Sure - I didn't mean to suggest that 0.458 might, in fact, be on the list, although now I've read back my own comment, it does look like that. I only meant to point out that the justification for using 2 mod 10 as the offset value was incomplete. That aside, I've always felt a little uncomfortable with this proof, without being able to put my finger on exactly why!
@@michellezhuang2483 I'm not a mathematician, but as far as I understand, the definition says "injection" because when proving that N and Q have the same cardinality, you would run into a problem, since you are skipping over 2/2, 3/3, 4/4, etc. in the diagonal enumeration, therefore the correspondence between the two sets is not bijective, only injective. I hope I'm right, but as far as I understand, that is the case.
An injection is sufficient to show that S is countable. An injection from S to N means that the cardinality of S is either the same as N (countably infinite) or finite, which is the definition of countable. To say that there is a bijection between S and N would mean S is definitely countably infinite, and not finite.
Good question! I thought we can not add 1 but now I see we can. I am not sure I get the logic though. Do you mean since 0.99...becomes 1.00..., 0.11... becomes 1.00... as well? I think that actually the number we add to every digit can be 1. Because when we add 1 to each digit of 0.99...under mod 10, 0.99...becomes 0.00..., not 1.00... I looked up Cantor's Diagonal Proof again and watched another video on it so that I understood the logic of the proof better, which is that we can always try to make a comprehensive list of real numbers. The first can be called x_1, the second x_2, and so on. But that the list will never be comprehensive of all the real numbers that there are: We can always make another number besides the list of say, j numbers: Making its first decimal place the same as that of x_1, its second decimal place the same as that of x_2 and so on, then changing each of the decimal digits in this number to another integer that is between 0 and 10. As the first two decimal places came from respectively x_1 and x_2, this number will not be x_1 and not be x_2. By the same argument it will not be any of the other listed numbers, either. Have a good day!
That's a very complicate demonstration to say there is no multiple of a number outside of the integer side. Real numbers are a way to say a perimeter is a infinite divided point.
because he is not listing real numbers, he is listing real line segment lengths, that is what his ellipses points do. each line segment itself contains many points corresponding to real numbers.
The proof shows that beyond every list of real numbers there exists another real number not included in the list. In the case of 0.99..., if adding a 1 to each digit creates 1.00..., then because 1.00... is in fact 0.99...(can be proven by geometric series), we do not successfully prove the list incomplete by finding a number not included by the list: 1.00... is included in the list as 0.99..., a numeral for the same number. 0.99... would become 1.00... if 1 is added to every digit in 0.99 and each digit reduced by a multiple of 10 once it exceeds 10 (calculating in mod 10). But only the decimals are affected -- we are only looking at part of each number, that after its decimal point (the tenth places are all numbered by d_x1, then the hundredth places are the d_x2, the thousandth place d_x3...) -- so when e_i=d_ii+(2 in this case), only the decimals change. 0.999...would become 1.000 making it the same number, but because of the way the digits are numbered, d_ii only refer to the decimals and an increase of 1 on each digit in 0.999... (mod 10) results in 0.000...
The proof is OK for finite sets, but it will not work for infinite sets. By showing using one to add to each digit does not help your argument, you are giving an example that disproves the argument. Try your argument with whole numbers. Starting at zero, adding 2 to each digit will give the following sequence: 0 -> 2 1 -> 22 2 -> 222 3 -> 2222 n -> 2 repeated n+1 times By the diagonal argument, the resulting number is 2 repeating indefinitely which should not be in the set of whole numbers. Obviously a contradiction that disproves the diagonal argument. Let's try a set of real numbers from zero to one. If we create an ordered set by reflecting the whole numbers on the decimal point, our series is .0, .1, .2, .3, .4, .5, .6, .7, .8, .9, .01, .11, .31, ... The real number corresponding to the whole number 59842 would be .24895. Applying the diagonal argument to this series using 2 as the adder results in the new number .22222... Again an obvious contradiction that disproves the diagonal argument. No matter how far down the series of digits you take J, there will be an infinite subset of numbers in your starting set that match your new number up to digit N sub J and an infinite number of digits beyond digit N sub J to still check.
Hello! I'm glad you're thinking about these topics! Allow me to push your reasoning a little further. Why does the Diagonal Argument work on the real numbers, but not on the integers? This, pretty much, is what your first point is about. Let me summarize the main point of the diagonal argument. Using the diagonal argument, you can find a real number which is not in your list of real numbers, regardless of what your list is. So you can conclude that your list is incomplete, since there is definitely something missing from it. The thing I want to highlight about the argument is that the newly constructed number is, in fact, a real number. Because of how real numbers are defined as limits of rational numbers (that's the whole point; they "fill in the gaps" between rational numbers), real numbers can have infinitely many digits to the right of the decimal point. For integers, on the other hand, you cannot have infinitely many (nonzero) digits to the left of the decimal point. Each integer has only finitely many nonzero digits. What is the purpose of an integer? Well, we really only care about positive integers here, and the point of positive integers is, more or less, for counting things. While it is true that no matter what number you can count to, you can always count one number higher (meaning that there are infinitely many positive integers), you are not able to count infinitely far and then keep counting more. If a positive integer had infinitely many digits, it would have indicated that you were able to reach it after already having counted an infinite amount of time. This is all a simplification. All this stuff is defined and axiomatized, and I can use this stuff to _prove_ that each positive integer has only finitely many digits, but I thought I would explain it on an intuitive level. If you want me to go deeper, please ask. Now, this being said, ...222 is _not_ an integer since it has infinitely many nonzero digits going to the left. So you cannot conclude that the list of integers is incomplete because even though the new "number" you found is not on the list, that new number shouldn't be on the list. Therefore, we can see why Cantor's Diagonal Argument works on the real numbers, but not on the integers. The same explanation above applies to your second proposed counter-argument, since it also relies on ...222 being an integer, which is false. Your third argument is a little bit of a different nature, since this gets into philosophy of mathematics. What does it take to prove that something exists? Some mathematicians are finitists, in that they hold the philosophy that you cannot prove something exists mathematically unless you can actually construct it in a finite number of steps. These mathematicians reject any mathematical construction that requires an infinite number of steps. However, this is a fringe philosophy of mathematics. I don't say fringe to undermine it - it is a perfectly valid philosophy; it's just not the mainstream view. And the mainstream view overwhelming eclipses this view. It's not even a contest. Finitism is an extreme minority among mathematicians. So what about the mainstream view? The mainstream view is that, if we can describe something specifically and precisely, then we are able to talk about it, even if that description contains an infinite amount of information. So we don't have to physically write out an infinite number of digits to show that a particular real number exists. We just have to be able to uniquely describe it in order to know it exists. So, yes, if you were trying to physically write out the number "constructed" by Cantor's Diagonal Argument, you would never be finished. You would never be able to physically construct it. But that doesn't matter. We are able to describe it in a rigorous and complete way, as being the real number whose nth decimal place differs from the nth decimal place in the nth number of the list according to some algorithm. Even if we can't physically write it out, we can prove that such a real number actually exists (since it fits the definition of a real number) and that it cannot be on the list. So while finitists would not accept the Diagonal Argument (but then again, they would object to the entire topic as a whole, not just the argument), it is completely valid from a mainstream perspective, and not being able to physically write it down is not an issue.
I understand that an individual integer can not have an infinite number of digits, and I understand your reasoning here. My point is that the reals from zero to one are countable and that the diagonal argument does not work. I'll use binary so the options for each digit are only one and zero. If you only consider the set of real numbers based on their first digit, there are only two possibilities, 0.0000... and 0.1000... I am extending the zeros out to the right to indicate I am referring to only those two specific real numbers. This set is countable. To get the Cantor number that is not in the set, you have to go out to two digits to the number 0.11. Of course there are an infinite amount of numbers that start with 0.11 that are not in our set. Now extend our set to two digits, and we add the numbers 0.01000... and 0.11000... Our set now contains a countable four (2^2) elements, and to get a Cantor number not in our set, we have to go to three digits for 0.111. Going to three digits, we add the numbers 0.001000..., 0.101000..., 0.011000..., and 0.111000... Now we have a countable set of eight (2^3) elements. The Cantor number not in our set (0.1111) needs four digits. You can easily see that going to n digits, we will have a countable set of 2^n elements, and the Cantor number not in the set needs n+1 digits. Each set so far is an exhaustible set up to the number of digits contained , and each set is countable. By extending n to infinity we get the entire countable set of reals from zero to one, and the Cantor number is always one digit farther out than the n we are extending to. Every time I have seen Cantor's diagonal argument explained, the numbers used seem to be random: 0.3748954..., 0.18849375..., 0.94830881... and so on. This seems to hide the fact that the numbers can be put in an order, although not in an order by magnitude. It also seems to be assumed that Cantor's diagonal argument works to generate a number not in the set without any proof. I believe I have given enough evidence here to suggest that it does not work.
You have constructed a number which comes after every number in your list. But the list is endless. You can do the same with a numerical list of the natural numbers to show they are uncountable. .999....... never equals 1. Lim .999...... equals 1 Lim .999....99m equals 1 for any m. Lim is an operation on the number . 999.... As the number of 9’s becomes infinite, .9999..... becomes infinitesimally close to 1. In other words, there are numbers infinitesimally close to 1.
No. On real numbers there are no non-zero infinitesimals. You are conflating two different statements. The number 0.999... *is defined* to be the limit of the infinite sequence 0.9, 0.99, 0.999, ... . No element of the sequence is equal to 1. But the limit *is* equal to 1, and the value of 0.999... is the limit of the sequence, not any its element. Therefore, 0.999... *is* 1.
Thank you for this explanation! I had been looking for an answer to this question for a while and nothing explained it as nicely as this video :)
nice proof, you voice is incredibly nice for some reason
What if countability is inconsistent? (adopted from Mueckenheim):
If the fractions m/n are enumerated by the natural numbers k according to Cantor's function
k = (m + n - 1)(m + n - 2)/2 + m
then all the fractions of the sequence
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ...
are enumerated.
But if the natural numbers first are in bijection with the integer fractions of the first column of the matrix
1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
...
then they must be distributed over the matrix such that no fraction remains without index. That means, there is a permutation such that the X of the first column
XOOOO...
XOOOO...
XOOOO...
XOOOO...
XOOOO...
...
by being exchanged with the O's cover all matrix positions. All O's will vanish. This is obviously impossible because exchanging cannot reduce them. The number of not indexed fractions, represented by O's, will remain constant forever, in infinity. Why do mathematicians believe in Cantor yet?
we just got introduced to discrete maths, but I don't understand why we define ei like ej = djj + 2 (mod 10) and then show a contradiction with that definition. How does that prove that R is unlistable? We had an example on why N and Z are countable and thus why the relation is a bijection by putting a number N; based on if n is even or uneven in a formula and then showing that each n has a specific z and vice versa if I, for example, contradict this 'defined algorithm' is it then proven that there is no bijection between N and Z?
The thing about Cantor's diagonal argument is that it is an _argument._ It isn't a list. It's an argument that can be applied to _any_ list.
A set is countable if _there exists_ at least one bijection between that set an N. That set will also have functions between it and N which are not bijections. That's okay.
But Cantor's diagonal argument applies to _every_ list of real numbers. For each list of real numbers, there is a real number not on your list! (And you can prove that using Cantor's diagonal argument.) So there can't be a bijection with N. Any attempt to build a bijection with N will fail - because Cantor's diagonal argument will find a real number not hit by your attempted bijection.
Make sense?
God bless ya I think I'm going to pass a text bc of this keep up the good work homie
Anyone noticed that the procedure for determining the digits of the diagonal number has the same form as that for determining the digits of a computable number? That is, given any positive integer n, you have a procedure for determining the nth digit of the number in a finite number of steps.
But it has long been known that the cardinality of the set of computable numbers is the same as that of the integers. So therefore the Cantor diagonal construction can only ever produce numbers from a set with the same cardinality as the integers. So how does it prove the existence of a larger set?
What does list mean? Does it mean to explicitly write down every one of the set?
I think it would be easier for the student to introduce first the idea that innumerability is not (just) a property of infinite sets.
This is easy to show using Cantor's power set theorem.
You go through the standard diagonal proof reasonably well.
But there is still one point that, in my opinion, needs to be cleared up.
The diagonal proof is working on digits, not on the number itself. It does not matter that you have declared the digits to be in a list of partial sums. The proof does not use this information. The point is that the proof is incomplete until you show that new "values" have been created. For example, consider the case where the interval is the natural numbers. Let's assume that the natural numbers are modular at the infinity limit. Then, although an innumerabilty of digit strings is proved, there are no new numbers created. Thus disproving the notion that the Reals are innumerable. It then becomes subject to an axiom, i.e. the Natural numbers at the infinity limit are not modular. I have actually never seen this problem raised, so maybe there is something I have overlooked, but I can't see it myself.
Where’s your video smarty pants? 👖
Enjoyed the proof, absolutely correct. But wondering but is there any case where the mod 2 will cause a problem,if not how can we prove it/ understand it?
exactly if we add 2 to all digit then there may be a case like 0.8888.... which will become 0.00000... and if we add 3 the again there may be a case like 0.77777..... which will become 0.00000....ad this will happen with ever digit from 1 to 9. I think this proof is not valid.
Amazing proof but little complicated....
Can you explain more in detail why we didn’t let ei=dii + 1 mod 10?
Because the number constructed 1.000000... is already in the list in the form of 0.9999999.....
So there isn't a contradiction.
Thanks for such amazing explanation
Excellent
This video is pretty confusing. You start by defining a countable set S as a set for which there exists an injection between S and Z+, but then you proceed to *not* use this definition. Are you using an equivalent definition of countability, something to do with surjections?
Can’t that also be used for integers or naturals?
If the "d" sequence on the diagonal was, say ".23577777777777......", then adding 2 modulo 10, would generate a number ".457999999999999.....". You have proved that this sequence is not on the list, but what about ".4580000000000000", which is, after all, the same number?
But was 0.458 on the original list? It's not the first number, since the first digit of that number is 2 not 4. It is not the second digit, since the second digit in that number is 3, not 5. It is not the third number, because the third digit of that number was 5, not 8. It is not the nth number, n >= 4, because the nth digit of that number is 7, not 0. So 0.458 was not in the list. The proof is not very convincing but it is hard to find a counterexample. If we use binary instead of decimal notation, there is a specific example where the diagonal argument fails.
@@alnitaka
Sure - I didn't mean to suggest that 0.458 might, in fact, be on the list, although now I've read back my own comment, it does look like that. I only meant to point out that the justification for using 2 mod 10 as the offset value was incomplete.
That aside, I've always felt a little uncomfortable with this proof, without being able to put my finger on exactly why!
Anyone else here from MACM 101
Isn't it a bijection between the function and the natural numbers ?
i was wondering the same thing!
A bijection in particular is used to show that the two sets are of the same cardinality.
I don't know but he wrote Z^+ which would be the same, as it is the set of all positive integers. :)
@@michellezhuang2483 I'm not a mathematician, but as far as I understand, the definition says "injection" because when proving that N and Q have the same cardinality, you would run into a problem, since you are skipping over 2/2, 3/3, 4/4, etc. in the diagonal enumeration, therefore the correspondence between the two sets is not bijective, only injective. I hope I'm right, but as far as I understand, that is the case.
An injection is sufficient to show that S is countable. An injection from S to N means that the cardinality of S is either the same as N (countably infinite) or finite, which is the definition of countable. To say that there is a bijection between S and N would mean S is definitely countably infinite, and not finite.
Using the same logic (that 0.9999... = 1.000...) can we also say that you cannot define e_i to be d_ii + 9?
Good question! I thought we can not add 1 but now I see we can.
I am not sure I get the logic though. Do you mean since 0.99...becomes 1.00..., 0.11... becomes 1.00... as well?
I think that actually the number we add to every digit can be 1. Because when we add 1 to each digit of 0.99...under mod 10, 0.99...becomes 0.00..., not 1.00... I looked up Cantor's Diagonal Proof again and watched another video on it so that I understood the logic of the proof better, which is that we can always try to make a comprehensive list of real numbers. The first can be called x_1, the second x_2, and so on. But that the list will never be comprehensive of all the real numbers that there are: We can always make another number besides the list of say, j numbers: Making its first decimal place the same as that of x_1, its second decimal place the same as that of x_2 and so on, then changing each of the decimal digits in this number to another integer that is between 0 and 10. As the first two decimal places came from respectively x_1 and x_2, this number will not be x_1 and not be x_2. By the same argument it will not be any of the other listed numbers, either.
Have a good day!
That's a very complicate demonstration to say there is no multiple of a number outside of the integer side.
Real numbers are a way to say a perimeter is a infinite divided point.
Well explained.
I'm disappointed in how lost I got as soon as you started defining your variables
because he is not listing real numbers, he is listing real line segment lengths, that is what his ellipses points do. each line segment itself contains many points corresponding to real numbers.
thanks!!
then it shouldnt also hold for 0.8888888... for ei= 8+2 which is also 10, i didnt get it
In modular arithmetic, this 8+2 would become 0, then there would be a part of the number where the digit would be 0, like 0.88888 ...888808888...
The proof shows that beyond every list of real numbers there exists another real number not included in the list. In the case of 0.99..., if adding a 1 to each digit creates 1.00..., then because 1.00... is in fact 0.99...(can be proven by geometric series), we do not successfully prove the list incomplete by finding a number not included by the list: 1.00... is included in the list as 0.99..., a numeral for the same number.
0.99... would become 1.00... if 1 is added to every digit in 0.99 and each digit reduced by a multiple of 10 once it exceeds 10 (calculating in mod 10). But only the decimals are affected -- we are only looking at part of each number, that after its decimal point (the tenth places are all numbered by d_x1, then the hundredth places are the d_x2, the thousandth place d_x3...) -- so when e_i=d_ii+(2 in this case), only the decimals change.
0.999...would become 1.000 making it the same number, but because of the way the digits are numbered, d_ii only refer to the decimals and an increase of 1 on each digit in 0.999... (mod 10) results in 0.000...
Neat read it in my math book
tnx sir
If dii=9 then eii=11. eii=9+2 (mod10), but eii must have one digit. Is your formula correct?
e_ii = 11 (mod 10) = 1 (mod 10). So it still has one digit
@@MathemaEducation then indeed the formula is (dii+2) ( mod 10)
very well sir....i m from indiaaa....good keep it up
Everything is okay, but still u r not very clear and articulate to justify your definition of ei at the end
Fake.
The proof is OK for finite sets, but it will not work for infinite sets. By showing using one to add to each digit does not help your argument, you are giving an example that disproves the argument. Try your argument with whole numbers. Starting at zero, adding 2 to each digit will give the following sequence:
0 -> 2
1 -> 22
2 -> 222
3 -> 2222
n -> 2 repeated n+1 times
By the diagonal argument, the resulting number is 2 repeating indefinitely which should not be in the set of whole numbers. Obviously a contradiction that disproves the diagonal argument.
Let's try a set of real numbers from zero to one. If we create an ordered set by reflecting the whole numbers on the decimal point, our series is .0, .1, .2, .3, .4, .5, .6, .7, .8, .9, .01, .11, .31, ... The real number corresponding to the whole number 59842 would be .24895. Applying the diagonal argument to this series using 2 as the adder results in the new number .22222... Again an obvious contradiction that disproves the diagonal argument.
No matter how far down the series of digits you take J, there will be an infinite subset of numbers in your starting set that match your new number up to digit N sub J and an infinite number of digits beyond digit N sub J to still check.
Hello! I'm glad you're thinking about these topics! Allow me to push your reasoning a little further.
Why does the Diagonal Argument work on the real numbers, but not on the integers? This, pretty much, is what your first point is about.
Let me summarize the main point of the diagonal argument. Using the diagonal argument, you can find a real number which is not in your list of real numbers, regardless of what your list is. So you can conclude that your list is incomplete, since there is definitely something missing from it.
The thing I want to highlight about the argument is that the newly constructed number is, in fact, a real number. Because of how real numbers are defined as limits of rational numbers (that's the whole point; they "fill in the gaps" between rational numbers), real numbers can have infinitely many digits to the right of the decimal point.
For integers, on the other hand, you cannot have infinitely many (nonzero) digits to the left of the decimal point. Each integer has only finitely many nonzero digits. What is the purpose of an integer? Well, we really only care about positive integers here, and the point of positive integers is, more or less, for counting things. While it is true that no matter what number you can count to, you can always count one number higher (meaning that there are infinitely many positive integers), you are not able to count infinitely far and then keep counting more. If a positive integer had infinitely many digits, it would have indicated that you were able to reach it after already having counted an infinite amount of time.
This is all a simplification. All this stuff is defined and axiomatized, and I can use this stuff to _prove_ that each positive integer has only finitely many digits, but I thought I would explain it on an intuitive level. If you want me to go deeper, please ask.
Now, this being said, ...222 is _not_ an integer since it has infinitely many nonzero digits going to the left. So you cannot conclude that the list of integers is incomplete because even though the new "number" you found is not on the list, that new number shouldn't be on the list.
Therefore, we can see why Cantor's Diagonal Argument works on the real numbers, but not on the integers.
The same explanation above applies to your second proposed counter-argument, since it also relies on ...222 being an integer, which is false.
Your third argument is a little bit of a different nature, since this gets into philosophy of mathematics. What does it take to prove that something exists? Some mathematicians are finitists, in that they hold the philosophy that you cannot prove something exists mathematically unless you can actually construct it in a finite number of steps. These mathematicians reject any mathematical construction that requires an infinite number of steps. However, this is a fringe philosophy of mathematics. I don't say fringe to undermine it - it is a perfectly valid philosophy; it's just not the mainstream view. And the mainstream view overwhelming eclipses this view. It's not even a contest. Finitism is an extreme minority among mathematicians.
So what about the mainstream view? The mainstream view is that, if we can describe something specifically and precisely, then we are able to talk about it, even if that description contains an infinite amount of information. So we don't have to physically write out an infinite number of digits to show that a particular real number exists. We just have to be able to uniquely describe it in order to know it exists. So, yes, if you were trying to physically write out the number "constructed" by Cantor's Diagonal Argument, you would never be finished. You would never be able to physically construct it. But that doesn't matter. We are able to describe it in a rigorous and complete way, as being the real number whose nth decimal place differs from the nth decimal place in the nth number of the list according to some algorithm. Even if we can't physically write it out, we can prove that such a real number actually exists (since it fits the definition of a real number) and that it cannot be on the list.
So while finitists would not accept the Diagonal Argument (but then again, they would object to the entire topic as a whole, not just the argument), it is completely valid from a mainstream perspective, and not being able to physically write it down is not an issue.
I understand that an individual integer can not have an infinite number of digits, and I understand your reasoning here. My point is that the reals from zero to one are countable and that the diagonal argument does not work.
I'll use binary so the options for each digit are only one and zero. If you only consider the set of real numbers based on their first digit, there are only two possibilities, 0.0000... and 0.1000... I am extending the zeros out to the right to indicate I am referring to only those two specific real numbers. This set is countable. To get the Cantor number that is not in the set, you have to go out to two digits to the number 0.11. Of course there are an infinite amount of numbers that start with 0.11 that are not in our set.
Now extend our set to two digits, and we add the numbers 0.01000... and 0.11000... Our set now contains a countable four (2^2) elements, and to get a Cantor number not in our set, we have to go to three digits for 0.111. Going to three digits, we add the numbers 0.001000..., 0.101000..., 0.011000..., and 0.111000... Now we have a countable set of eight (2^3) elements. The Cantor number not in our set (0.1111) needs four digits.
You can easily see that going to n digits, we will have a countable set of 2^n elements, and the Cantor number not in the set needs n+1 digits. Each set so far is an exhaustible set up to the number of digits contained , and each set is countable. By extending n to infinity we get the entire countable set of reals from zero to one, and the Cantor number is always one digit farther out than the n we are extending to.
Every time I have seen Cantor's diagonal argument explained, the numbers used seem to be random: 0.3748954..., 0.18849375..., 0.94830881... and so on. This seems to hide the fact that the numbers can be put in an order, although not in an order by magnitude. It also seems to be assumed that Cantor's diagonal argument works to generate a number not in the set without any proof. I believe I have given enough evidence here to suggest that it does not work.
Contradiction ,by Logic, means Cantor said meaningless words as "proof".
You have constructed a number which comes after every number in your list. But the list is endless. You can do the same with a numerical list of the natural numbers to show they are uncountable.
.999....... never equals 1.
Lim .999...... equals 1
Lim .999....99m equals 1 for any m.
Lim is an operation on the number . 999....
As the number of 9’s becomes infinite, .9999..... becomes infinitesimally close to 1. In other words, there are numbers infinitesimally close to 1.
No. On real numbers there are no non-zero infinitesimals. You are conflating two different statements. The number 0.999... *is defined* to be the limit of the infinite sequence 0.9, 0.99, 0.999, ... . No element of the sequence is equal to 1. But the limit *is* equal to 1, and the value of 0.999... is the limit of the sequence, not any its element. Therefore, 0.999... *is* 1.
Quit smacking