# Cantor's Diagonalization Proof of the uncountability of the real numbers

1. Feb 5, 2012

### Leucippus

I have a problem with Cantor's Diagonalization proof of the uncountability of the real numbers. His proof appears to be grossly flawed to me. I don't understand how it proves anything.

Here is a totally abstract pictorial that attempts to show how Cantor's proof works. However, this pictorial is grossly flawed in two ways. First off, it's perfectly square, and secondly, it abstractly uses letters in place of numerals totally missing the point of what this proof is based upon.

In the above picture we're supposed to assume that we've gone down a nice neat square list of numbers and came up with a brand new number that clearly cannot be on the list.

But is that truly the case? I claim that it can't be the case. It's impossible. Why?

Well to realize what the problem is we must first recognize that this list is entirely based upon numerals and numbering systems. It is not based on abstract letters as show in the previous graphic. So let's get a more suitable graphic like the one below so we can illustrate why this can't work.

Ok now we'll be able to see why this can't work. What he's doing here is listing numerals (please understand that he is listing numerals not numbers). Numerals are a system that we use to write out numbers. He is using a graphic system of writing out numerals and then crossing them off diagonally to obtain a brand new number that is supposedly not already on the list.

And if numerical lists were perfectly square he would indeed have a valid proof. But lists of numerals are not square. They are extremely rectangular in the vertical direction. We have no such thing as a square numerical system in terms of being able to write it out graphically as a square. The decimal notational system is extremely rectangular.

What do I mean by this?

Well, take all possible 2-digit decimal numbers (assume non-negative numbers) and list them out. What do you get? You get a very tall narrow rectangular list of numbers like so:

http://users.csonline.net/designer/me/dec_digit.gif [Broken]

This list is not even remotely close to being square. Add another digit to make it a list a mere 3 digits wide, and it will become almost 1000 rows deep. With every additional digit you add, the list innately becomes longer by powers of 10.

Lists of numbers using the decimal numerical system are far from being square lists.

What does this have to do with Cantor's diagonalization proof, you may ask. Well it has everything to do with it. Cantor is claiming to take a process out to infinity. But I claim that his process can't even be made to work in the simplest finite case.

I would like to use the binary number system to try to illustrate this point. It's far more manageable than trying to do this in base 10. Even the Binary numeralic system isn't square, but it's far more manageable so please bear with me for a few quick examples.

Let me start with a simple 2-digit binary list. This will be four rows long:

Please keep in mind that these are binary numbers now.
http://users.csonline.net/designer/me/2_digit.gif [Broken]

So what do we do? Well, we go down the list diagonally replacing every digit we cross out with some other arbitrary digit that is not the same digit. In binary this is extremely simple because we are forced to swap 1's for 0's and vice versa.

So we go down this list and changing the first 0 to a 1 and the second 1 to a zero and come up with our new number 10. Then we ask. Is that number already on our list. And yes of course it is. In fact it's the very next number on the list. We just weren't able to reach it because a diagonal line doesn't go down the list fast enough to cover all the rows.

So the new number that we created was already on our 2-digit wide list.

Let's try it now on a binary list 3-digits wide:

http://users.csonline.net/designer/me/3_digit.gif [Broken]

Well this time it's easy. When we draw our digaonal line we hit three 0's and create the new binary number 111 which happens to be the last number on the list. But it's on the list!

It fact, any number we create this way must always be on the list. It's the very nature of the numerical system. And this is precisely due to the fact that these lists are far taller than they are wide.

Let's to it one last time for a 4-digit binary list. If the list if 4-digits wide it's going to also be 16 rows tall.

http://users.csonline.net/designer/me/4_digit.gif [Broken]

Yep, sure enough, when we create our new number using Cantor's diagonalization method we come up with another number that is already on our list. This must necessarily always be the case simply due to the very nature of these numerical systems of notation.

Now, you might be tempted to say, "But wait a minute, Cantor is taking this process out to infinity!"

So?

If a diagonal line is already this far behind in a mere finite situation taking it out to infinity is only going to make matters that much worse. Look the binary examples I gave above. When the list was 2-digits wide the diagonal line went half way down the list. When the list was 3-digits wide the diagonal line went only about a third of the way down the list. When the list was 4-digits wide the diagonal line only when down the list about a quarter of the way. The wider the lists become the worse the situation gets.

Using decimal notation (which is what Cantor actually used) the situation is far worse.

If he can't even make it to the bottom of a finite list how could he ever hope to cross off every possible number on a list infinity many digits wide?

He's claiming to have created a brand new number that cannot be on the list.

I say baloney. He could never get to the bottom of the list. How can he claim that the new number that he has constructed "cannot be on the list" when he can't possibly be making enough downward progress in the vertical direction to have ever gotten to the bottom of his list?

As far as I'm concern this so-called "proof" doesn't prove anything other than Cantor must have been mistakenly assuming that it makes sense to assume that lists of numerals could be made square (that is a false assumption).

~~~~~

Please note that this does not mean that the results of Cantor's proof are false. It simply means that his proof does not prove the assertion that he's claiming.

~~~~~

Now if I'm in error please advise. But I honestly don't see how Cantor's diagonalization proof proves anything when it's based on a numerical representation of numbers that simply cannot be made to be square which his "proof" apparently requires.

At no point can he legitimately claim to have reached the "bottom" of his list. And claiming to take this out to infinity doesn't help. Because with every digit wider the list just gets that much taller by an increasing magnitude. He could never hope to reach the bottom of his infinite list to make the proclamation that he has created a new number that's not on the list. His diagonal line simply can't have a steep enough slope to accomplish this task.

I seriously don't see how that could be made to work.

See in these pictures in math books (like the one shown below) they make it look like he can easily reach the bottom of the list. He's going down at what appear to be a nice 45 degree angle on a seemingly square list. But that is a mistaken illusion in these graphic examples. Real numerical lists don't behave this way.

So can anyone explain to me how Cantor's Diagonalization Method can be accomplished if taken out to infinity when it clearly can't even work in simple finite examples?

Seriously, I'd appreciate an explanation.

If you can explain to me why his proof makes sense and why my objections are flawed, I would truly be very grateful because I honestly cannot see how his proof makes any sense at all.

Thank you for taking the time to read my concerns.

Last edited by a moderator: May 5, 2017
2. Feb 5, 2012

### pwsnafu

This here is your problem. Just because a mathematical statement is true (or false) for the finite case does not mean it is true (or false) for the infinite case. For example
1/2 < 1
1/2 + 1/4 < 1
1/2 + 1/4 + 1/8 < 1
and so on
1/2 +1/4 + 1/8 + ... 1/(2n) < 1.
For any finite sum, that sum will be strictly less than 1. But when you take the infinite sum it equals 1.

You can't say, it fails for finite length expressions therefore it will fail for infinite length expressions.

More to the point "taken out to infinity" is wrong. The proof starts with an infinite list of infinite length expressions. Not starting with a finite list and expanding it.

Last edited: Feb 5, 2012
3. Feb 6, 2012

### Leucippus

Thank you for your reply pw, but the problem you referred to doesn't have anything at all to do with the problem I posted.

The problem I posted isn't based solely on the idea that just because something doesn't work in a finite case that means it can't work in an infinite case.

The problem I posted shows clearly that the problem gets increasingly worse as it approaches the infinite case.

The problem you posted actually gets increasingly better as it approaches infinity.

There's no comparison at all.

4. Feb 6, 2012

### pwsnafu

"Better" and "Worse" are useless arguments when it comes to proofs. It's true or false. That's it. Seriously, I have no idea what you mean by "[t]he problem posted actually gets increasingly better". What do you mean by "better"? The inequality fails with the infinite sum.

5. Feb 6, 2012

### Parlyne

All you're demonstrating here is that the diagonalization proof fails to show that the integers are uncountable. But, that's no surprise because the integers are, in fact, countable.

What changes in the case of the reals on [0,1) is that any number in this range may be represented by an infinitely long string of digits, even if all the digits trailing beyond a certain point have the value of 0. The proof works because it can act on these 0s just as easily as on any other digit. This means that it can't run out of digits to change before it runs out of numbers in the list.

6. Feb 6, 2012

### Deveno

there's a couple of things worth pointing out here:

"twice" a countably infinite set is the same size as the original set.

the standard argument here is that there is a bijection between the natural numbers, and the even natural numbers:

n↔2n

if the reals between 0 and 1 WERE countable, then the list WOULD be "square", there would be a 1-1 correspondence between "digit places" and "elements of (0,1)". so the diagonal method WOULD apply. we would exhaust the reals and the number of digits at the same rate (bijectively). so it's not the case that the list is "longer" than it is "wide".

this fact is hard for people to swallow, because it is contrary to our finite experiences.

another way to express this idea is that "countable infinity is elastic", an infinite "part" is equinumerous to the "whole".

only infinite sets have this quality, which some people take as the DEFINITION of infinite (the technical term is Dedekind-infinite, and there are some set-theoretical niceties involving the validity of certain "optional" axioms that keep this from just being plain "infinite", for all people).

in other words, if you take a finite set "two at a time" you'll out-race me just taking "one at a time", but if both of our sets are countably infinite, we'll both finish at the same time, namely, never.

the bijection between the natural numbers and "half the natural numbers" isn't just restricted to a factor of 2, we could use any other infinite subset we liked, as well:

1↔2
2↔3
3↔5
4↔7
5↔11
6↔13
...
k↔k-th prime

since there's an infinitude of primes, even though we're going through the natural numbers "faster", we'll still never run out of primes, they'll always "keep up" (the well is inexhaustible).

comparing this to a convergent series isn't very apt. a better comparison, is to a divergent series like:

$$\sum_{k=1}^\infty \frac{1}{k}$$

you'd think that as k runs to infinity, since the "height" of our sum is just crawling along and becomes almost imperceptible, that eventually (since we're adding "virtually nothing") the sum would stop growing. but it NEVER DOES. even though it takes many, many terms, the series will eventually pass any limit you set for it. that is the power of infinity, it can do things finite stuff just can't.

the conclusion of the diagonal argument, by the way, is that indeed the list is "too long" to list.

one consequence of this is that the vast majority of real numbers, are numbers we will NEVER know. in reaction to this, some mathematicians (called constructivists), believe that the existence of something should not be taken as "given" unless you can actually produce it somehow. in other word, they feel that the power sets of some sets are "too big" to be called sets.

in fact, there is a theorem that says that every consistent set theory, has a countable model. but normal ZF set theory says that the reals are uncountable. so who's right? in some ways, it's a moot point. in actual practice, mathematicians rarely invoke "global features" of the theory (a notable exception being the use of the axiom of choice), but use "a local piece of the theory" relevant to what they're working on. that is, if you reject "the uncountable set" of real numbers, there's still plenty of "computable real numbers" to use for analysis (the murkiness of the waters does get a bit muddled when talking about measure theory, though, where you often need large collections of sets).

7. Feb 6, 2012

### lavinia

The diagonalization proof does not go down any list. It merely asserts the existence of a number whose n'th digit is different from the n'th digit of the n'th number in the list. Such a number must clearly exist since the sequence of partial decimal expansions forms a Cauchy sequence.

There is one technicality. that is that decimal expansions are not unique. so it is possible to get an alternative expansion for one of the numbers already in the list. So how would you get around this?

8. Feb 6, 2012

### Deveno

i'd disallow all the sequences that are "eventually" all 9's.

9. Feb 6, 2012

### Leucippus

You're problem got "better" with respect to what you were trying to prove.

You were trying to prove that just because a finite problem doesn't work doesn't that doesn't prove that the same situation taken out to an infinite case can't work.

However, your problem got increasingly "better" with every finite step. (with respect to what you were attempting to prove).

You simply wouldn't have been able to prove what you did with a problem that gets increasingly worse with ever finite step. If you think you can, please give an example that illustrates this.

That's what I mean by "better" or "worse".

10. Feb 6, 2012

### Preno

This seems to be the key point that you're missing, Leucippus. We are proving that the list cannot be square even in the infinite case. It's just that we're doing so by contradiction - we suppose that it is square and derive a contradiction. Your objection "but the list isn't square" is therefore absurd, seeing as how that's precisely what Cantor is trying to establish.

Edit: you said it yourself:
That's Cantor's whole point, that such an assumption is false (leads to a contradiction).

11. Feb 6, 2012

### Leucippus

I can see what you're attempting to get at here and I confess that this is the most interesting explanation I've heard thus far.

However, the problem here is that this proposed solution not only doesn't square up the list, but it makes the lists even that much longer. Thus making the slope of the diagonal line even more shallow.

You're claiming that because of these extra zeros permitted in a decimal expansion numerical representation of number we can clearly make up even more numbers than if those zeros weren't available. However, I have problem with that.

I mean, using a number system based on 10 you still only have the numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, available to you as replacements for those extra zeros.

You're claim is that since we have arbitrarily endless strings of zeros to work with this automatically means that we should be able to create arbitrarily endless new numbers.

But I'm not convinced that this is true. On the contrary, it appears to me that the limitation of the numerical system of digits simply won't permit that. In other words, we'd still restricted to only replacing any given zero with a numeral of 1 thru 9. So we'd still be restricted in what new numbers we can create. The fact that we seem to have more zeros to work with wouldn't change that fundamental limitation.

In other words you say, "The proof works because it can act on these 0s just as easily as on any other digit. This means that it can't run out of digits to change before it runs out of numbers in the list."

But is that really true? What do you mean that it can't run out of digits? It would still be stuck with only being able to replace those zeros with a very limited choice of numerals from 1 thru 9.

So I don't see where your proposal solves anything. I don't see where adding extra arbitrary rows of zeros helps.

In fact, as far as I can see I can actually apply your proposed solution to a finite list and demonstrate why it doesn't help a thing.

Let's try it,...

Please remember this example is in binary for brevity sake.
http://users.csonline.net/designer/me/40_digit.gif [Broken]

Well, when we draw our diagonal line through our new additional arbitrary rows of zeros we still come up with a number that's already on the list. There's no way we could do otherwise because of the very nature of numerical lists and the limitations of their numerical digits.

We can demonstrate just as easily that this is also true of a numerical list that is based on ten. We would have more digits to chose from for replacements of zeros, (i.e. 1 thru 9), but we would still always come up with a number that is already on the list. Plus our lists would even be taller yet since we've added arbitrary rows of zeros.

So how is that helping?

I don't see where it helps at all. We'd have very same situation made 'worse'. By worse I simply mean that now the lists are even taller yet thus decreasing the downward slope of the diagonal line resulting in an even slower rate of decent through the list. We could never hope to claim to have created a new number that isn't already on the list.

And that is paramount, because that is precisely what Cantor is claiming to have done. He's claiming to have created a new number that isn't already on the list.

He's assuming that he can take this out to infinity in both the horizontal and vertical directions with equal "speed". In other words, he's assuming that he can do this clear to the bottom of this list of numbers as he moves to the right.

But he can't. The slope of his diagonal line simply isn't steep enough to accomplish this task. The very property of a numerical representation of a list of numbers simply won't permit what he is claiming to be able to do. They can't be used like that because they aren't square lists.

He must have been assuming that since he can take this out to infinity in the horizontal direction that he could also take the same process out to infinity in the vertical direction simultaneously too. But that's a folly. Henri Poincare warned him about this. It's a folly to start thinking of infinity as though it can be completed. But that's precisely what Cantor's proof would require. Cantor's "proof" is not a proof at all, if his diagonal line can't reach the bottom of the list at the same time that it reaches the right side of the list. And clearly that can never be accomplished on lists of numerals because lists of numeral are simply not square lists by their very nature.

So he could never complete his diagonal process to the bottom of his list. The slope of his diagonal line simply isn't steep enough, and it also dynamically takes on a shallower and shallower slope with every digit as he moves to the right. The situation gets increasingly "worse".

He could never hope to prove that he is creating a new number that isn't already on this list.

It's a flawed proof based on the erroneous assumption that lists of number are innately square when in fact they are actually highly rectangular and become increasingly rectangular with ever added digit.

Last edited by a moderator: May 5, 2017
12. Feb 6, 2012

### Leucippus

I'm sorry but you've lost me right there.

I've just demonstrated that even a list of natural numbers cannot even be made to be square. So I don't understand how you can claim that any list of numbers could be square.

It's innately a property of our numerical representational systems that they cannot be made to be square. And this would even apply to a list of natural numbers.

So your claim, "if the reals between 0 and 1 WERE countable, then the list WOULD be square" makes no sense to me. How could any complete list of numbers be square?

Our numerical system of representing numbers simply won't permit that to happen. It's built-in to the very representation of our numbers that these lists cannot be both complete and square at the same time. If they are forced to be square they must necessarily be incomplete, and if they are said to be complete lists they must necessarily rectangular.

This is even true of the natural numbers.

It's an innate property of the very system of numerical notation. Yet it is this system of numerical notation that Cantor is attempting to use in his proof.

13. Feb 6, 2012

### Preno

It's a proof by contradiction. Do you understand how a proof by contradiction works? It works by assuming the thing you're trying to disprove (in this case, that there is a square list of reals).
Yes, that is what Cantor proves. How do you not understand this?

14. Feb 6, 2012

### Leucippus

That doesn't impress me at all, for the following reason:

Stop Cantor's process at any finite point. Chose whatever point you like.

Take this supposedly "new number" that you've created at that point 'truncated' precisely where you had stopped Cantors process and ask yourself, "What do I have?"

You will have a perfectly legitimate rational number. You haven't created anything "new" at all at that point.

And the mere fact that you can do this an ANY POINT in the process, pretty much proves that at no point have you created a "new number" that wouldn't already be on a list of rational numbers.

At NO POINT in this process does Cantor ever create a new number that wouldn't already be a rational number if truncated at that point of the process.

So how does claiming to take this process out to infinity change that?

He's clearly not making any "progress" along the way.

What supposedly 'magically' happens at infinity that changes this?

Cantor's claim seems to be, "Well I will have exhausted every possible number on the list and I obviously have created a new number, so I must have a number that wasn't on the original list"

But as I've shown, he has no reason to make such a claim because he could never make it to the "bottom" on any such list in the first place. His diagonal process simply doesn't have a steep enough slope to make that claim.

He can't claim to ever be able to complete his process (even in theory). Because his diagonal method simply doesn't go down the list fast enough to make such a claim.

There is never a point in this process where he creates a 'new number' that wouldn't already be a rational number if truncated at the point in question.

So how does it follow that this situation would ever change?

Last edited: Feb 6, 2012
15. Feb 6, 2012

### Leucippus

I've already demonstrated that any "complete" list of numbers cannot be square even a complete list of natural numbers.

So that cannot be what Cantor was attempting to show. All complete numerical lists that contain all the numbers allowed by that numerical system of representation must all necessarily be rectangular, including the natural numbers.

16. Feb 6, 2012

### Preno

Well, you haven't demonstrated it, you just provided some sort of intuitive graphical argument, but other than that - that's precisely what Cantor was attempting to show. Cantor's theorem, phrased in your terms, states: there is no square list of real numbers. This is what both you and Cantor are saying.

17. Feb 6, 2012

### Deveno

you don't seem to grasp a key point. the numbers listed are not just rational numbers. the vast majority of them (as it turns out) are aperiodic non-terminating decimals.

if we stop the process after 3 steps, all we have is a decimal representation that differs from the first 3 numbers. if we continue, then after n steps, we just have a decimal that is different from the first n numbers listed.

now, if the real numbers between 0 and 1 COULD be put in a 1-1 correspondence with the natural numbers, at some finite time, we would reach any given real number. that is what it means to be "countable" (eventually you can "count up" to any given element of the set).

but the ever-increasing number we are creating at the bottom, doesn't match any number reached after a finite number of steps. so no matter how long we "count" (how far down the list we go), let's say we go down 3 million places, the number at the bottom is different from the first 3 million numbers.

you could write a computer program, that checked the number at the bottom against the first k numbers at step k, and if it found a match within the first k, instruct it to halt. it will never halt.

if the real numbers between 0 and 1 WERE countable, eventually, at some point, we would find a match, because at some point, we'd get to any real number we cared to name. but the way we are constructing the number (it grows by one digit, as we eliminate the possible matches from the "top down"), it's guaranteed not to match as many numbers as it has digits.

no match = not countable. and we're rigging it so there won't BE a match. which number on the list can it possibly match? the 1st? no. the 2nd? no. the 3rd? no. the 4th? no. the 3,567,281st? no.

all our number on the bottom has to do, to "not match" is be different in ONE digit-place from any given number. how do you think that will fail to happen?

18. Feb 6, 2012

### Leucippus

But I have demonstrated it. You claim that I just provided some sort of intuitive graphical argument. But that is precisely what Cantor's diagonal proof is. It's a graphical argument based on using a numerical representation of numbers.

I'm demonstrating (using precisely the same kind of numerical graphical representations of numbers) why Cantor's graphical diagonalization proof has no merit.

And my points should indeed be extremely easy to intuitively understand.

Let me try to make this as clear as possible using a numerical representation of the natural numbers based on 10.

Here's the simplest case of a 1-digit wide list of number based on ten digits. There are exactly ten of them so in order to list them all the list must be 1 column wide by 10 rows long.

http://users.csonline.net/designer/me/1_dec.gif [Broken]

This is an extremely rectangular list of numbers, necessarily so, forced by the very nature of this numerical representation.

Now, let's move on to a 2-digit complete list of these numbers. And it's important that these list are indeed complete. Incomplete lists would be incomplete and therefore meaningless.

http://users.csonline.net/designer/me/2_dec.gif [Broken]

Now the list has become even vastly more rectangular. Now it has only 2 columns, but it requires a length of almost a hundred rows. It is an extremely skinny tall rectangle, force by the very nature of the numerical representation of these numbers.

Let's do it one more time just to be sure that we have a convincing trend,...

http://users.csonline.net/designer/me/3_dec.gif [Broken]

Yep the trend is pretty convincing. Now we have a list only 3 digits wide yet it's almost a thousand rows long. In fact, it doesn't take a genius to realize that these lists are going to continue getting longer by powers of ten every time we increase their width by a single digit.

So it should be pretty obvious that lists of natural numbers are innately far longer than they are wide.

Therefore to think that we should be able to draw a diagonal line down through a list of numbers represented by these numerical digits and proclaim to have created a new number "that isn't already on the list" is basically a false claim that could never hold true.

Yet this is precisely what Cantor's diagonal proof requires that we must be able to do. We must be able to go down through such a list of number diagonally and actually "reach" the bottom of the list creating a new number in the process, in order to concluded that we have created a new number that isn't already on this list.

That makes absolutely no sense. Cantor is assuming that these lists could somehow be square enabling him to successfully cross over every number on the list diagonally.

That simply isn't possible.

Cantor's diaganoalization proof based on this method is necessarily a flawed proof and doesn't prove what he claims it proves.

It's pretty obvious to me.

~~~~

This isn't to say that the result is necessarily untrue. Maybe the real numbers truly are uncountable. But Cantor's diagonalization "proof" most certainly doesn't prove that this is the case.

It is necessarily a flawed proof based on the erroneous assumption that his diagonal line could have a steep enough slope to actually make it to the bottom of such a list of numerals. That simply isn't possible.

I don't care if he claims to take this out to infinity. How would that change anything? The slope of his diagonal line is only going to get shallower and shallower with ever digit he crosses out. He could never hope to claim to have reached the bottom of the list and created a new number that isn't already on the list.

It's nonsense. What else can I say?

Last edited by a moderator: May 5, 2017
19. Feb 6, 2012

### Leucippus

With all due respect Deveno, your argument here makes no sense to me at all.

Why would you expect to ever get a number that's already on the list that you are generating?

That would never happen in this process. That's obvious.

But that doesn't prove anything in light of my observations.

You're always going to be generating a number that is "further down" the list than where you are currently working. That's a given. Even my finite examples clearly show that this will always be the case. The very process that is being used to generate your "new number" demands that this be the case.

My point is that it doesn't matter.

Numerical representations of numbers like this simply aren't square to begin with. So just because the new number you've created isn't on the list above where you are currently working in this process doesn't mean a thing.

That's my whole POINT.

The numbers above where you are working cannot be a "complete list" with respect to where you are at in the columns, because the slope of the diagonal line simply isn't steep enough to make that guarantee.

And this is why this diagonal method can't prove what Cantor claims it proves.

This diagonal method of creating a "new number" simply isn't anywhere near "fast enough" (i.e. it doesn't have a steep enough slope) to even deal with finite lists of numbers. It will always be "behind" the list. And it just gets further and further behind with every digit that is crossed out.

We can clearly see this in the finite examples I've given.

Why should this change if we try to take this process out to infinity?

The situation is only going to get worse with ever digit we cross out.

20. Feb 6, 2012

### SteveL27

Leu, one point of confusion here is that you are thinking of Cantor's proof as a process. You are adding one row at a time and looking at the antigiagonal and then talking about rectangles.

All of that is erroneous thinking and has nothing to do with Cantor's proof.

Cantor asks us to consider any complete list of real numbers. Such a list is infinite, and we conceptualize it as a function that maps a number, such as 47, to the 47-th element on the list. There's a first element, a 2nd element, and DOT DOT DOT. We assume that ALL of these list entries exist, all at once.

Then we construct the antidiagonal. It's clear that the antidiagonal can't be on the list ... because it differs from the n-th item on the list in the n-th decimal place. (Or binary place if you do the proof in base-2).

Since we started by assuming we had a list of all reals; and we just showed that any such list must necessarily be missing a real; then it follows that there can be no such complete list in the first place.

I suggest that you make an attempt to fully understand this beautiful proof on its own terms. There is nothing here about processes and nothing here about rectangles. You are introducing those irrelevant concepts on your own and losing sight of the proof itself.