Refuting the Anti-Cantor Cranks

[dcpo]

I find hard to believe there is no a theory already in place that fully explain this kind of numbers which, obviously, not me nor you are aware of.

Well, as it stands these new 'numbers' are not properly defined, so it cannot be said whether or not existing theory fully explains them. My guess is that any way they could be defined would fit esily into existing theory, and likely wouldn't yield anything particularly useful. You could get something similar, for example, by considering the interval [0,1) \subset \mathbb{R}^+ and taking [0,1) \times[0,1) equipped with a kind lexicographic ordering, so (x,y)\leq(x',y')\iff \text{ either } x< x' \text{ or }(x=x' \text{ and } y\leq y'). Here the left element of the pair gives you something you might like to describe as an 'infinite digit whole number', and you can interpret the right element as the decimal part. This will be a lower bounded dense total order, so elementarily equivalent to \mathbb{R}^+ (as an order at least), and it will have the same cardinality. Completeness too will be inherited from the completeness of \mathbb{R}, so this structure will be in many ways similar to \mathbb{R}^+, though I can't say off the top of my head if the two will be order isomorphic. You could likely define arithmetic operations without too much trouble too.

 

[viraltux]

Well, as it stands these new 'numbers' are not properly defined, so it cannot be said whether or not existing theory fully explains them. My guess is that any way they could be defined would fit esily into existing theory, and likely wouldn't yield anything particularly useful. You could get something similar, for example, by considering the interval [0,1) \subset \mathbb{R}^+ and taking [0,1) \times[0,1) equipped with a kind lexicographic ordering, so (x,y)\leq(x',y')\iff \text{ either } x< x' \text{ or }(x=x' \text{ and } y\leq y'). Here the left element of the pair gives you something you might like to describe as an 'infinite digit whole number', and you can interpret the right element as the decimal part. This will be a lower bounded dense total order, so elementarily equivalent to \mathbb{R}^+, and it will have the same cardinality. Completeness too will be inherited from the completeness of \mathbb{R}, so this structure will be in many ways similar to \mathbb{R}^+, though I can't say off the top of my head if the two will be order isomorphic. You could likely define arithmetic operations without too much trouble too.

I think I see what you mean dcpo but I maybe disagree with them having the same cardinality than \mathbb{R}. Think about this, imaging you have the π number without the decimal dot: 314159... now you can decide to place the dot in any position, which means that for every number of this kind you have infinite real numbers. Wouldn't this show they have both different cardinalities?

 

[dcpo]

I think I see what you mean dcpo but I maybe disagree with them having the same cardinality than \mathbb{R}. Think about this, imaging you have the π number without the decimal dot: 314159... now you can decide to place the dot in any position, which means that for every number of this kind you have infinite real numbers. Wouldn't this show they have both different cardinalities?

Well, the way I've defined it the base set is [0,1)\times[0,1), and since [0,1) has the same cardinality as \mathbb{R}, that the cardinality of [0,1)\times[0,1) is the same as that of \mathbb{R} follows from the fact that the cardinality of the product of two infinite cardinals will be the max cardinality of the pair.

ETA: The rule is, when dealing with infinite quantities cardinality gets a bit weird. The first 'weird' result is Cantor's theorem itself, and it only gets worse.

 

[viraltux]

Well, the way I've defined it the base set is [0,1)\times[0,1), and since [0,1) has the same cardinality as \mathbb{R}, that the cardinality of [0,1)\times[0,1) is the same as that of \mathbb{R} follows from the fact that the cardinality of the product of two infinite cardinals will be the max cardinality of the pair.

ETA: The rule is, when dealing with infinite quantities cardinality gets a bit weird. The first 'weird' result is Cantor's theorem itself, and it only gets worse.

That's why I think the definition you give does not quite fit the numbers we're talking about, but anyway, I'm no expert on this so thank you for your explanations and patience!

 

[jgens]

I think I see what you mean dcpo but I maybe disagree with them having the same cardinality than \mathbb{R}. Think about this, imaging you have the π number without the decimal dot: 314159... now you can decide to place the dot in any position, which means that for every number of this kind you have infinite real numbers. Wouldn't this show they have both different cardinalities?

Nope. They will have the same cardinality.

Explanation: It is easy to show that all sequences consisting of just 0s and 1s have the same cardinality as the real numbers. Since these are just a subset of your numbers, it follows that the cardinality of your numbers is at least as great as the cardinality of the reals. On the other hand, clearly each one of your numbers corresponds in a natural way to a unique real number (just put the decimal before the first number). This proves that the cardinality of the reals is equal to the cardinality of your numbers.

 

[viraltux]

Nope. They will have the same cardinality.

Explanation: It is easy to show that all sequences consisting of just 0s and 1s have the same cardinality as the real numbers. Since these are just a subset of your numbers, it follows that the cardinality of your numbers is at least as great as the cardinality of the reals. On the other hand, clearly each one of your numbers corresponds in a natural way to a unique real number (just put the decimal before the first number). This proves that the cardinality of the reals is equal to the cardinality of your numbers.

Well in fact it is not a subset but the same set, the fact that I used base 10 instead base 2 to write the numbers does no make one subset of the other.

Very interesting that all sequences of 0's and 1's have the same cardinality than Real numbers, yet, here we are not talking about all sequences but only about the infinite ones, any finite sequence of 0's and 1's is not in the set. Could this alone change it's cardinality to be smaller than the Reals? After all the correspondence you do with the Real numbers is one among the infinite you can do; you could place the dot after the second number, the third...

 

[micromass]

It does not belong to any conventional number systems and is probably meaningless to discuss. You could try to define specific properties that the infinite sequence of digits might have, but I doubt the theory you could derive would be particularly fruitful.

Numbers which are infinite to the left are certainly useful, but in a surprising way. Check out the p-adic numbers: http://en.wikipedia.org/wiki/P-adic_number

It gives rise to surprising identities such as

...999999999 = -1

 

[jgens]

Numbers which are infinite to the left are certainly useful, but in a surprising way. Check out the p-adic numbers: http://en.wikipedia.org/wiki/P-adic_number

I actually mentioned the p-adics in post 59 of this thread :)

The point of my comment was that if the poster intends to give his infinite sequences of digits some sort of meaning apart from the p-adics or reals, then he/she will have a difficult time doing so. A lot of the meaningful ways of dealing with infinite sequences of digits is captured by the real numbers and by the p-adics, so IMO it would be rather difficult to find an entirely new structure on infinite sequences of digits that proves to be particularly fruitful.

 

[jgens]

Well in fact it is not a subset but the same set, the fact that I used base 10 instead base 2 to write the numbers does no make one subset of the other.

If X is a set, then X is a subset of itself. So my original statement was entirely correct. What you mean is that the sequences of 0s and 1s is not a proper subset. In any case, formally your claim may/may not be true depending on what we mean by "infinite sequences of digits". We can talk about real numbers in base 2 versus base 10 because they can be defined independent of a particular representation. So if you define your numbers in a representation invariant way, then you can do the same thing here. On the other hand, in this case I would tend to define an infinite sequence of digits in terms of its representation in a particular base, and in this case the two are not equal.

Edit: It also worth noting the following. The set of all real numbers in [0,1] whose decimal expansion consists entirely of 0s and 1s is not equal to the set of all real numbers in [0,1]. The set of all real numbers in [0,1] whose binary expansion consists entirely of 0s and 1s is equal to the set of all real numbers in [0,1]. So we can reinterpret strings in a particular way that makes this true, but once we have fixed an interpretation (for example, using the characters 0,1,2,3,4,5,6,7,8,9 as you did in your example indicates at least base 10) we have to stay consistent with that. In my previous post, to stay consistent with the convention of using the symbols 0,1,2,3,4,5,6,7,8,9, the collection of all infinite sequences of 0s and 1s is a proper subset of your numbers.

Very interesting that all sequences of 0's and 1's have the same cardinality than Real numbers, yet, here we are not talking about all sequences but only about the infinite ones, any finite sequence of 0's and 1's is not in the set. Could this alone change it's cardinality to be smaller than the Reals? After all the correspondence you do with the Real numbers is one among the infinite you can do; you could place the dot after the second number, the third...

No. The cardinality is still the same. The infinite sequences of 0s and 1s have the same cardinality as the reals. In fact, if you take all infinite sequences of 0s and 1s that begin with 1, you get a set with the same cardinality as the reals. Or if you take the set of all infinite sequences of 0s and 1s that start with 101110110 this set has the same cardinality as the reals.

 

[viraltux]

aaaaaaand outta here. nice week end everyone :)

 

[mbs]

I'm still here. Been traveling coast to coast.

I'm quite sure I'll never become a crank on this issue. As a non-mathematician though I can see how someone would become one. Maybe if this issue can be zoomed in on, many future cranks-in-waiting might be saved. Please lead me down the reasoning path.

I'll do my best to explain using the language I have available.

Proofs by contradiction make sense. You make an assertion or assumption that may or may not be true, then you follow up with some valid deductions based on the assumption. If your subsequent deductions are valid but you arrive at a contradiction or falsehood, then the original assumption was false. This is proof by contradiction as I understand it.

For example (and I'm making this up on the fly) let's suppose that division by zero were legitimate arithmetic. I can probably form some simple algebraic expessions which would result in a statement like 1=2. Nobody should have a problem with such a proof.

But if you start a proof with 1=2 and then proceed to do valid algebra with it, the contradiction doesn't arise from the proof but is built in at the beginning.

I can't speak for any Anti-Cantor cranks but for me this is an issue.

A few posts back MBS says that the proof of the irrationality of sqrt(2) can begin by assuming the existence of two integers m and n such that n^2/m^2=2. You then perform valid reasoning on this and arrive at absurd conclusions. That's great. I don't have trouble with that because the expression above is legitimate algebra, it just so happens that no two integers will satisfy it.

But I'd bet it's not legitimate logic to start with the absurd result and reason your way backward to the expression above.

What am I missing?

Ok. It seems to me that you are questioning the meaning of an arbitrary "infinite list" that is not defined with a specific function or recursive formula. The existence of arbitrary infinite lists comes down to the existence of arbitrary functions having the natural numbers as their domain. Most mathematician simply accept such things as axiomatic objects. Restricting oneself to finitarily constructable objects severely limits mathematics.

OTOH, even if you don't want to accept an arbitrary list as a well defined object, you can still use the diagonal argument as a second-order logical proof. What I mean is if you did have a logical formula defining a one-to-one function from the natural numbers to the real numbers, such a formula must still lead to a contradiction. We have a second-order logical statement because our statement concerning the countability of the real numbers is no longer a statement about mathematical objects but a statement about logical formulas themselves.

 

[Antiphon]

No, actually I'm ok with that.

I question the validity of reasoning with "illegal" objects.

If one is restricted to the reals, and I begin a proof with sqrt(-1) I can derive all kinds of contradictory results. But clearly this is construct is legal over the complex field.

I need to understand how it is legitimate to postulate the existence of a complete list of reals. Its not an "incorrect statement" like the rational root 2 proof; it's a logical non-sequitur to anyone who know the basic properties of real numbers.

Why is this permissible?

 

[SteveL27]

No, actually I'm ok with that.

I question the validity of reasoning with "illegal" objects.

If one is restricted to the reals, and I begin a proof with sqrt(-1) I can derive all kinds of contradictory results. But clearly this is construct is legal over the complex field.

I need to understand how it is legitimate to postulate the existence of a complete list of reals. Its not an "incorrect statement" like the rational root 2 proof; it's a logical non-sequitur to anyone who know the basic properties of real numbers.

Why is this permissible?

Well, it's not a logical non-sequitur. Before we begin the proof, we have no way of knowing whether the reals are countable or not. In fact at first glance, most people would guess that the reals ARE countable: "The naturals are an infinite set, the reals are an infinite set, infinite is infinite. There's only one infinite, right?"

So we start our proof by assuming we have a bijection from the naturals to the reals; and we derive a contradiction: in fact we did NOT have a bijection, because any map from the naturals to the reals must necessarily fail to hit some real.

In fact we can express Cantor's argument so that it's not a reductio proof. All the proof says is that any map (or function) from the naturals to the reals is not a surjection. Surely there's no contradiction or problem in taking an arbitrary map from the naturals to the reals, and showing that it's not a surjection. Because there ARE lots of maps from the naturals to the reals. And if you take anyone of them, you can prove (via the diagonal argument) that it's not a surjection.

No contradictory or false assumption need be made in the proper formulation of Cantor's proof.

Does that address the question you had in mind?

 

[dcpo]

The logician Wilfred Hodges wrote an article talking about 'refutations' of Cantor's argument. It's worth reading, and available here.

 

[micromass]

No, actually I'm ok with that.

I question the validity of reasoning with "illegal" objects.

If one is restricted to the reals, and I begin a proof with sqrt(-1) I can derive all kinds of contradictory results. But clearly this is construct is legal over the complex field.

I need to understand how it is legitimate to postulate the existence of a complete list of reals. Its not an "incorrect statement" like the rational root 2 proof; it's a logical non-sequitur to anyone who know the basic properties of real numbers.

Why is this permissible?

Why do you think it is not permissible to have a complete list of all real numbers?? What basic properties do you think it violates (without doing a Cantor-like construction)

We CAN make a complete list of all rational numbers, so being able to list all reals doesn't sound far-fetched a priori.

 

[mbs]

No, actually I'm ok with that.

I question the validity of reasoning with "illegal" objects.

If one is restricted to the reals, and I begin a proof with sqrt(-1) I can derive all kinds of contradictory results. But clearly this is construct is legal over the complex field.

I need to understand how it is legitimate to postulate the existence of a complete list of reals. Its not an "incorrect statement" like the rational root 2 proof; it's a logical non-sequitur to anyone who know the basic properties of real numbers.

Why is this permissible?

It is not necessary to postulate the existence of a complete list of reals. It is sufficient to show that any list must be incomplete. Just as it is sufficient to show, in the proof that the square root of two is irrational, that for any pair of relatively prime numbers m,n \in \mathbb{N} it is not the case that m^2 = 2\cdot n^2.

There is absolutely no difference in the logical construction. If you're going to make the argument that the "list" in the Cantor diagonal argument is somehow "illegal", you must also make the same argument that the expression m^2 = 2\cdot n^2 is somehow "illegal".

 

[martix]

Taking a step back, I have a slightly different view on the matter.

First you need to define what "crank" means.
My definition at least makes the OP subject pointless in more ways than one. :)

 

[lugita15]

Taking a step back, I have a slightly different view on the matter.

First you need to define what "crank" means.
My definition at least makes the OP subject pointless in more ways than one. :)

I agree, the term "crank" is perhaps better reserved for the people who have looked into the proof in sufficient detail that they really have no excuse for still attacking Cantor, as opposed to people who are earnestly learning Cantor's proof for the first time and are open to accepting its validity if you can just clear up their objections and misconceptions.

I was hoping to come up with a good answer to the kind of objection raised in the dialogue in my OP, in order to help the second group of people see the light.

 

[SteveL27]

I agree, the term "crank" is perhaps better reserved for the people who have looked into the proof in sufficient detail that they really have no excuse for still attacking Cantor, as opposed to people who are earnestly learning Cantor's proof for the first time and are open to accepting its validity if you can just clear up their objections and misconceptions.

I was hoping to come up with a good answer to the kind of objection raised in the dialogue in my OP, in order to help the second group of people see the light.

You ever see the pamphlets and websites of the circle-squarers and the angle-trisectors? Some of these guys (and they are ALWAYS guys) are very smart ... often retired engineers who learned the practical side of math and built a career, then they start studying a little math, and they just get obsessed with refuting long-established results.

There's a certain willful obtuseness about cranks. It's a strange psychological orientation. They were around long before the Internet and the Web has given them a voice. But it's really not clear why they do it.

I agree that earnest questioning is definitely not the same as outright crankery. Cantor's proof does really ask a lot of us ... imagine an infinite list, imagine each item on the list is an infinitely long string to the right of the decimal point -- but not infinitely long to the LEFT of the decimal point. I can see why a curious and honest skeptic would ask a lot of questions about all this. And we know Cantor received a lot of criticism from contemporaries, particularly Kronecker. A lot of people do have an instinctive aversion to this type of infinitary reasoning.

I was hoping to come up with a good answer to the kind of objection raised in the dialogue in my OP, in order to help the second group of people see the light.

After reading some of these posts I think it would be better to remove the reductio aspect of this proof; and simply show that if you take any function from N to R, that function's not a bijection.

We write down the natural numbers:

1
2
3
...

Nobody objects to that.

Then next to each number we write down where it's mapped by our function f:

1 -> f(1)
2 -> f(2)
3 -> f(3)
...

Now each f(n) is actually a decimal expression; we form the anti-diagonal, and voila: f can not be a surjection.

I think the above is a formulation of the proof that avoids the reductio proof, which troubles people. And instead of using the word "list," with all the preconceived notions people might have about what that word means, we just enumerate the naturals. That's something people are familiar with.

And we're revealing the functional nature of the relationship between the number n, and the n-th item on the list.

I think this is an exposition of the proof that might help people see things more clearly. No reductio, no "list," and we've elucidated the structure of that mysterious list of decimals that people always want to add the anti-diagonal to the end of. Instead, we just take a harmless little function, and prove that it can't possibly be a surjection, because we just constructed a number that function can't hit.

That would be my contribution to the subject of how to make this more clear.

 

[lugita15]

Now each f(n) is actually a decimal expression; we form the anti-diagonal, and voila: f can not be a surjection.

I think the above is a formulation of the proof that avoids the reductio proof, which troubles people.

Unfortunately, I don't think this will help with the kind of people I'm describing in my dialogue. It is the forming of the anti-diagonal that they take issue with. It is true that they object to the reduction ad absurdum version of the proof, because they think that the fact that the anti-diagonal number is not on the complete list of reals just indicates that the anti-diagonal number is not a well-defined real number. But even if you don't make it a proof by contradiction, they still think that in order to prove that the anti-diagonal number is well-defined, you first need to establish by independent means that the anti-diagonal number would not be on the list.

 

[Antiphon]

Unfortunately, I don't think this will help with the kind of people I'm describing in my dialogue. It is the forming of the anti-diagonal that they take issue with. It is true that they object to the reduction ad absurdum version of the proof, because they think that the fact that the anti-diagonal number is not on the complete list of reals just indicates that the anti-diagonal number is not a well-defined real number. But even if you don't make it a proof by contradiction, they still think that in order to prove that the anti-diagonal number is well-defined, you first need to establish by independent means that the anti-diagonal number would not be on the list.

I've had a month to ponder it and I've come over to being a Pro-Cantor crank now.

Not speaking for any other cranks of any particular stripe, let me just say that you almost nailed my objection; the anti-diagonal is a perfectly well-formed real but the moment you form it you have shown that the list that is missing the new anti-diagonal was not properly formed. And you don't need independent means; it's staring you right in the anti-diagonal.

 

[Andromeda12]

I'm not sure any of you have met Cantor-Agnostics yet but, to add to your frustrations, hi.

Suppose there are a countable amount of Rotnac parallel universes, each containing a countable amount of food, money, boredom and time. In one of these universes Rotnac is flipping a countably balanced coin and has nothing better to do than to sit and flip the coin into countability. So every possible Rotnac exists and we have every possible irrational sequence. The countable union of countable sets is countable, so we list all parallel Rotnac coin flips.

Suppose now that a countable amount of parallel Cantor universes exist, in which only impossible things happen. Every Cantor always flips a rational number. This is impossible because the chance of flipping infinite heads (or tails), from any point onward, is zero. Similarly, the chance of randomly flipping an infinitely repeating sequence from any point onward is zero. But the countable union of countable sets is countable, so we union all Cantor and Rotnac universes to get the complete list of reals in [0,1].

The reason we know rationals are countable is because we have a matrix of all rationals staring us in the face. No rational is not in that matrix so snake your way across. But the reals in [0,1] are an infinite binary tree with 2^{\aleph_0} leaves. No real in the interval [0,1] is not in that tree. How do you traverse the tree though? Well I think all Rotnacs and Cantors do so with two countable coin flips.

You can now argue that there are uncountable Rotnac universes. I can't think of any contradictions that will arise from this. If one looks at Cantor's proofs, it sure does smell like undecidable. Assume countable and using nothing but logic, and by breaking nothing else along the way, arrive directly at uncountable. So reals are countable and uncountable?

In other theorems that use proof by contradiction, like the irrationality of \sqrt{2}, or the infinity of primes, some other absurd statement is reached using only logic. Not many cranks will accept the challenge of finding a natural that is both even and odd, or a large prime that divides 1 into whole numbers.

If Godel managed to prove that the continuum hypothesis is undecidable then I'm sure he would have been able to do the same for uncountabillity of reals. So there must be something wrong with my reasoning. On the other hand, Godel believed that humans possesses a 6th sense which can perceive truths in an existent mathematical realm. He sensed the uncountability of reals?

The following theorems make it very hard to believe the reals are uncountable:

1) The countable union of countable sets is countable.

So you are not even bounded by infinity? You can union a countable amount of countable sets to form a countable set, again and again... So, a countable matrix of \aleph_0^{2} sets has \aleph_0^{4} elements, which is countable. A matrix of \aleph_0^{4} sets has \aleph_0^{6} elements, which is countable. So \aleph_0^{n}=\aleph_0 for all n. As long as you don't do that to infinity you still have countable, otherwise it spills over and you get uncountable. And there are that many transcendentals. Okay fine, so the reals are just a slab of transcendentals with a some measly countable sets of algebraics and rationals sprinkled inbetween. The number of reals does have a power of \aleph_0 after all. But,

2) The rationals are dense in the reals, more specifically, between every two transcendentals there exists a rational.

What? The reals are a slab of uncountable transcendentals but the countable rationals are dense in that slab? No wonder there are so many cranks out there.

If maths and physics never yield counter intuitive results they can arguably be abandoned. We could rely on gut feel and intuition. So I hear you, mathematicians and physicists should rightly embrace counter intuitive results. Intuition often fails us, in which case we hand over the reins to reason. However, I have my doubts about this one.

Finally, the argument that no number can differ with itself at position n is wrong. You are not comparing an already generated cantor diagonal with a growing list. You are using any static list of reals to generate a number not in the list. The infinite computer keeps blindly toggling the nth bit.

 

[micromass]

Suppose there are a countable amount of Rotnac parallel universes, each containing a countable amount of food, money, boredom and time. In one of these universes Rotnac is flipping a countably balanced coin and has nothing better to do than to sit and flip the coin into countability. So every possible Rotnac exists and we have every possible irrational sequence. The countable union of countable sets is countable, so we list all parallel Rotnac coin flips.

Suppose now that a countable amount of parallel Cantor universes exist, in which only impossible things happen. Every Cantor always flips a rational number. This is impossible because the chance of flipping infinite heads (or tails), from any point onward, is zero. Similarly, the chance of randomly flipping an infinitely repeating sequence from any point onward is zero. But the countable union of countable sets is countable, so we union all Cantor and Rotnac universes to get the complete list of reals in [0,1].

There are uncountably many Rotnac universes, not countably many.
Furthermore, using terminology like "Rotnac universe" just obfusciates the main point.

The reason we know rationals are countable is because we have a matrix of all rationals staring us in the face. No rational is not in that matrix so snake your way across. But the reals in [0,1] are an infinite binary tree with 2^{\aleph_0} leaves. No real in the interval [0,1] is not in that tree. How do you traverse the tree though? Well I think all Rotnacs and Cantors do so with two countable coin flips.

Try to give a more concrete description. "The tree is being traverses by coin flips" is not mathematically meaningful.

You can now argue that there are uncountable Rotnac universes. I can't think of any contradictions that will arise from this. If one looks at Cantor's proofs, it sure does smell like undecidable. Assume countable and using nothing but logic, and by breaking nothing else along the way, arrive directly at uncountable. So reals are countable and uncountable?

In other theorems that use proof by contradiction, like the irrationality of \sqrt{2}, or the infinity of primes, some other absurd statement is reached using only logic. Not many cranks will accept the challenge of finding a natural that is both even and odd, or a large prime that divides 1 into whole numbers.

The same thing happens with Cantor's proof. An absurd statement leads to another absurd statement.

If Godel managed to prove that the continuum hypothesis is undecidable then I'm sure he would have been able to do the same for uncountabillity of reals. So there must be something wrong with my reasoning. On the other hand, Godel believed that humans possesses a 6th sense which can perceive truths in an existent mathematical realm. He sensed the uncountability of reals?

Godel knew very well that the reals were uncountable. It is not an undecidable statement.

The following theorems make it very hard to believe the reals are uncountable:

1) The countable union of countable sets is countable.

So you are not even bounded by infinity? You can union a countable amount of countable sets to form a countable set, again and again... So, a countable matrix of \aleph_0^{2} sets has \aleph_0^{4} elements, which is countable. A matrix of \aleph_0^{4} sets has \aleph_0^{6} elements, which is countable. So \aleph_0^{n}=\aleph_0 for all n. As long as you don't do that to infinity you still have countable, otherwise it spills over and you get uncountable. And there are that many transcendentals. Okay fine, so the reals are just a slab of transcendentals with a some measly countable sets of algebraics and rationals sprinkled inbetween. The number of reals does have a power of \aleph_0 after all. But,

2) The rationals are dense in the reals, more specifically, between every two transcendentals there exists a rational.

What? The reals are a slab of uncountable transcendentals but the countable rationals are dense in that slab? No wonder there are so many cranks out there.

I don't see how the rationals being dense implies anything about countability of the reals.
Try to use denseness to find an injection of the reals to the naturals. It's a good exercise to see why exactly that fails. This will enhance your intuition on why such a thing is possible.

Finally, the argument that no number can differ with itself at position n is wrong. You are not comparing an already generated cantor diagonal with a growing list. You are using any static list of reals to generate a number not in the list. The infinite computer keeps blindly toggling the nth bit.

If the argument is wrong, then there must be a logical error somewhere. That is: there must be a step in the proof that isn't allowed by using the axioms and inference rules.
Here is the proof broken down to the axioms and inference rules: http://us.metamath.org/mpegif/ruc.html Try to find the mistake in that. Here's a less formalized proof: http://us.metamath.org/mpegif/mmcomplex.html#uncountable
It's easy to say lots of big words, but it's a lot harder to counter a rigorous proof.

 

[viraltux]

I'm not sure any of you have met Cantor-Agnostics yet but, to add to your frustrations, hi.

Haha... I really like this thread

 

[Andromeda12]

There are uncountably many Rotnac universes, not countably many.
Furthermore, using terminology like "Rotnac universe" just obfusciates the main point.




Try to give a more concrete description. "The tree is being traverses by coin flips" is not mathematically meaningful.



The same thing happens with Cantor's proof. An absurd statement leads to another absurd statement.



Godel knew very well that the reals were uncountable. It is not an undecidable statement.



I don't see how the rationals being dense implies anything about countability of the reals.
Try to use denseness to find an injection of the reals to the naturals. It's a good exercise to see why exactly that fails. This will enhance your intuition on why such a thing is possible.



If the argument is wrong, then there must be a logical error somewhere. That is: there must be a step in the proof that isn't allowed by using the axioms and inference rules.
Here is the proof broken down to the axioms and inference rules: http://us.metamath.org/mpegif/ruc.html Try to find the mistake in that. Here's a less formalized proof: http://us.metamath.org/mpegif/mmcomplex.html#uncountable
It's easy to say lots of big words, but it's a lot harder to counter a rigorous proof.

Rotnac is Cantor backwards, I was just thinking of a name for the guy flipping the coin. I will try to figure out why using denseness to find a injection fails.

My last paragraph is an attempt at arguing against cranks that say a number cannot differ at nth position with itself.

 

[Andromeda12]

I don't see how the rationals being dense implies anything about countability of the reals.
Try to use denseness to find an injection of the reals to the naturals. It's a good exercise to see why exactly that fails. This will enhance your intuition on why such a thing is possible.

If the argument is wrong, then there must be a logical error somewhere. That is: there must be a step in the proof that isn't allowed by using the axioms and inference rules.
Here is the proof broken down to the axioms and inference rules: http://us.metamath.org/mpegif/ruc.html Try to find the mistake in that. Here's a less formalized proof: http://us.metamath.org/mpegif/mmcomplex.html#uncountable
It's easy to say lots of big words, but it's a lot harder to counter a rigorous proof.

In terms of trying to use density I think I see why it fails. If you try and 'transfer' the link from a rational to an 'irrational neighbour', you end just isolating that irrational because rationals don't have least upper bound.

I read through the informal proof and I think it's finally clicked for me. So when constructing the diagonal, you're actually creating an infinite increasing rational sequence, which is bounded above so it converges, but has a limit outside the list. So it's almost the nested interval proof in disguise.

Thanks for the help.

 

[Antiphon]

Andromeda12, I bow before you! You are the Mark Twain of Cantorian questioners!

I hereby renounce my allegiance to the Pro-Cantor-crank camp and lend all moral support to Andromeda12 to further advance this worthy cause.

 

[Andromeda12]

Andromeda12, I bow before you! You are the Mark Twain of Cantorian questioners!

I hereby renounce my allegiance to the Pro-Cantor-crank camp and lend all moral support to Andromeda12 to further advance this worthy cause.

lol, thanks!

I think I have found a way to generate and order a Rotnac list. It will be undecidable whether or not the diagonal is in the list. That's if my crack pot is not telling lies.

 

[Antiphon]

Time to pile on in my feeble way. There's a post over here: https://www.physicsforums.com/showthread.php?t=507001

It says that 1.000... = 0.999...

Now what is the status of Cantor's proof when there are multiple representations of the *same* real number? And they are not even near one another as regards the diagonal generation procedure.

Surely this spells doom for proponents of Cantor's proof!

 

[micromass]

Time to pile on in my feeble way. There's a post over here: https://www.physicsforums.com/showthread.php?t=507001

It says that 1.000... = 0.999...

Now what is the status of Cantor's proof when there are multiple representations of the *same* real number? And they are not even near one another as regards the diagonal generation procedure.

Surely this spells doom for proponents of Cantor's proof!

It can be proven that the only numbers which have multiple decimal representation are numbers which end in 00000... or 9999...

If we take the diagonal in Cantor's prove and change all numbers not equal to 5 to 5, and furthermore change all 5's to 6, then we get a number not on the list and the problem will not show up. That is: a number like 0.55555555... has a unique decimal representation.

Furthermore, there are versions of Cantor's proof which do not work with decimal representation at all!