Problem with Cantors Diagonalization Proof.

[talanum1]

Look: say you have to prove:

lim_(x->infinty) {f(x)} = 0

one can do this by calculating f(x) for larger and larger x and show that the result goes closer and closer to 0. Also show that this process continues. Then we make the conceptual leap to conclude:

f (infinity) = 0.

Now put the diagonalization process in f (x)'s place, and what the x'th digit changes to in 0's place. Since we cannot specify how the process continues we can't make the conceptual leap.

 

[Brian_G]

Over my head now, but... since I'm invested..

ChatGPT to the rescue (maybe?)



Nice — this is a common confusion, and your forum friend is mixing two different concepts: (1) taking a limit as a variable goes to infinity, and (2) defining an infinite object by a rule. Those are not the same thing, and that difference is exactly why Cantor’s diagonal argument is valid. Below I give a short, direct rebuttal you can use on the forum plus a formal-ish sketch of the diagonal construction that shows the “process” is specified.

---

Short rebuttal (plain English)

• When someone proves (\lim_{x\to\infty} f(x)=0) they are not asserting a value (f(\infty)) exists as a new number — they are showing values (f(x)) get arbitrarily close to 0 as (x) grows. Whether (f(\infty)) is defined or not is irrelevant to the limit statement.
• Cantor’s diagonal argument is not trying to evaluate a process “at infinity” the way your friend imagines. Instead it explicitly defines a new real number by a rule that tells you every decimal (or binary) digit of that number. That rule is a perfectly legitimate definition of an infinite object.
• So the objection “we can’t specify how the process continues” is false: the diagonal proof gives a precise rule that specifies the (n)-th digit for every (n). That is enough to define the real number.

---

Short formal sketch you can paste


Suppose, for contradiction, that the real numbers in ((0,1)) are listed as a sequence
[
r_1,r_2,r_3,\dots
]
and write each (r_k) in decimal (avoid the ambiguous .999... representations by not allowing trailing 9s; one standard trick is to use digits 1–8 only). Let
[
r_k = 0.d_{k1}d_{k2}d_{k3}\dots
]
where (d_{kn}\in{1,\dots,8}).


Now define a new real (r) by specifying its (n)-th digit (c_n) as
[
c_n =
\begin{cases}
1 &\text{if } d_{nn} \neq 1,[4pt]
2 &\text{if } d_{nn} = 1.
\end{cases}
]
This is a completely explicit rule: for every natural number (n) you know exactly how to compute (c_n). The infinite decimal (r=0.c_1c_2c_3\ldots) is therefore well-defined.


By construction (r) differs from (r_n) in the (n)-th digit (because (c_n\neq d_{nn})), so (r) is not equal to any (r_n). That contradicts the assumption that the list contained all reals in ((0,1)). Hence the reals are uncountable.


Note: choosing digits 1 and 2 (or any two digits excluding 9) avoids decimal ambiguity like (0.1999\ldots=0.2000\ldots).

---

Analogy that might help your friend

• Defining an infinite sequence by a rule is everyday math: “let (a_n = 1/n)” completely defines the whole sequence even though it has infinitely many terms. Each term is specified by the rule. Cantor’s diagonal does exactly that for the digits of a new number.
• Limits are statements about asymptotic behavior; defining an infinite object by a formula is a constructive specification. Confusing the two leads to the “you can’t say what happens at infinity” fallacy.

 

[Brian_G]

And...



Excellent — glad that helped.


If you end up posting it and he doubles down with something like “but you still can’t construct the number, it’s only a process that never ends”, the quick comeback is:

“That’s true of any infinite sequence — yet we still define sequences by rules. The diagonal gives an explicit rule for the nth digit, so it’s fully constructive in the same sense as any sequence definition.”

That usually stops the “undefined infinity” argument right there.

 

[Brian_G]

Now put the diagonalization process in f (x)'s place, and what the x'th digit changes to in 0's place. Since we cannot specify how the process continues we can't make the conceptual leap.

Re the edit

The analogy still doesn’t hold. The diagonalization process isn’t about taking a limit or evaluating anything “at infinity.” It defines an infinite sequence by a clear rule — for each position nnn, change the nnn-th digit. That rule fully specifies the infinite object, so there’s no missing “conceptual leap” to make.

Ok, think I'm done

 

[talanum1]

They specify an infinite sequence like "a_n = 1/n", but what is often omitted is proving the formula holds for n = infinity.

True: the diagonalization process describes what happens to the n'th digit since you presumably have the n'th digit, however since we haven't got the digit "at" infinity, the process can't continue and we can't make the leap.

 

[Brian_G]

Your aliens aren't really cleva aliens it seems... here's a rebuttal, anything else you'll need to get directly from the Chatbox. Until the next thread (maybe)..



Ah — perfect, this is the classic misunderstanding of what “infinite” means in mathematics.


Your guy is treating “(n = \infty)” as though it were an actual number that you can plug into a formula — but that’s not what infinity is. The definition of an infinite sequence never requires you to evaluate it at infinity. The definition simply gives a rule for every finite (n). That is what defines the infinite object.


Let’s build you a clear, sharp reply.

---

Forum-ready short reply​

Infinity isn’t a number you ever reach or plug in — it’s a statement that the rule works for all finite n. A sequence like (a_n = 1/n) doesn’t need to be evaluated “at infinity” to be well-defined; it’s defined precisely by what it does for each finite n.

Same with Cantor’s diagonal: for each finite position n, the rule tells you exactly which digit to write. That’s already a complete definition of the infinite number. There’s no missing “digit at infinity,” because infinite decimals don’t end — they’re sequences, not processes with a final step.

---

If you want the more mathematical angle​

In set theory or analysis, we define:
[
a : \mathbb{N} \to \mathbb{R}, \quad a(n) = 1/n.
]
That’s the whole definition. There’s no need or meaning to (a(\infty)); the sequence is the function from naturals to reals. Similarly, Cantor defines:
[
c(n) = 9 - d_{nn} \ (\text{or any digit rule you choose})
]
for all (n \in \mathbb{N}). That’s a fully specified function (c : \mathbb{N} \to {0,\ldots,9}). The “infinite number” is just shorthand for this function — its decimal expansion.


So when your opponent says “the process can’t continue at infinity,” the correct response is:

It doesn’t need to. The rule covers every finite n, which is what an infinite definition means.

 

[kolaval]

What happened to the fun Friday threads?

Also Infinity does not exist in the real world.

 

[Bewlen]

Your aliens aren't really cleva aliens it seems... here's a rebuttal, anything else you'll need to get directly from the Chatbox. Until the next thread (maybe)..



Ah — perfect, this is the classic misunderstanding of what “infinite” means in mathematics.


Your guy is treating “(n = \infty)” as though it were an actual number that you can plug into a formula — but that’s not what infinity is. The definition of an infinite sequence never requires you to evaluate it at infinity. The definition simply gives a rule for every finite (n). That is what defines the infinite object.


Let’s build you a clear, sharp reply.

---

Forum-ready short reply​

---

If you want the more mathematical angle​

In set theory or analysis, we define:
[
a : \mathbb{N} \to \mathbb{R}, \quad a(n) = 1/n.
]
That’s the whole definition. There’s no need or meaning to (a(\infty)); the sequence is the function from naturals to reals. Similarly, Cantor defines:
[
c(n) = 9 - d_{nn} \ (\text{or any digit rule you choose})
]
for all (n \in \mathbb{N}). That’s a fully specified function (c : \mathbb{N} \to {0,\ldots,9}). The “infinite number” is just shorthand for this function — its decimal expansion.


So when your opponent says “the process can’t continue at infinity,” the correct response is:

 

[talanum1]

c(n) = 9 - d_{nn} \ (\text{or any digit rule you choose})
]
for all (n \in \mathbb{N}). That’s a fully specified function (c : \mathbb{N} \to {0,\ldots,9}). The “infinite number” is just shorthand for this function — its decimal expansion.


So when your opponent says “the process can’t continue at infinity,” the correct response is:

But you haven't got d_(nn) for the infinite nn case, so you can't compute c(n) for n infinite. It must be proven to continue to apply "at" infinity, for sequences they omit proving this, but to be rigorous they should.

 

[Brian_G]

What happened to the fun Friday threads?

@epah ran away

Also Infinity does not exist in the real world.

O

 

[Brian_G]

I see you're still hanging around the thread, so, maybe this will help you get it.

But you haven't got d_(nn) for the infinite nn case, so you can't compute c(n) for n infinite. It must be proven to continue to apply "at" infinity, for sequences they omit proving this, but to be rigorous they should.

There’s no “infinite n” to plug in — a sequence is defined for every finite n, and that’s already the rigorous definition of an infinite object; nothing needs to be proven “at infinity.”

You're being stubborn dude, it's already more or less been covered.

 

[talanum1]

We see that even though the diagonalization process is defined for every finite amount of digits, it is NOT defined for the digit "at" infinity. Try that for a paradox!

 

[talanum1]

You're being stubborn dude, it's already more or less been covered.

How can you say that: the digit "at" infinity is not available to be computed with.

 

[talanum1]

"nothing needs to be proven “at infinity.”"

For cases like a_n = 1/n this is true because you always have the n available for computation. But in this case not.

 

[EmmanuelGoldstein]

"nothing needs to be proven “at infinity.”"

For cases like a_n = 1/n this is true because you always have the n available for computation. But in this case not.

It’s Friday bro tantalum… this is an epic fail at a Friday thread
Go have a dop, n tjop and be lekker not exceedingly boring with this garbage

 

[cguy]

Look: say you have to prove:

lim_(x->infinty) {f(x)} = 0

one can do this by calculating f(x) for larger and larger x and show that the result goes closer and closer to 0. Also show that this process continues. Then we make the conceptual leap to conclude:

f (infinity) = 0.

Now put the diagonalization process in f (x)'s place, and what the x'th digit changes to in 0's place. Since we cannot specify how the process continues we can't make the conceptual leap.

The key issue with the above is that you are treating infinity like a number, which in (standard) R, Q, N and Z, it is not. For Cantor's argument, one can claim to have a countable set of reals, but then always construct a new real which isn't in the set. This new number is defined as such, it's not a process - it will always be different to all other elements in a given set. If you try do this with a countable set of rationals, the newly constructed element won't be rational.

 

[cguy]

To say that we must “access the digit at infinity” is to commit a grievous conflation between the construction of an infinite object and the evaluation of a terminal step in a finite process. Infinity, I regret to inform the objector, is not a place one arrives at, nor a digit one retrieves from some metaphysical shelf. It is a conceptual framework—a limit, a horizon, a mode of definition. The diagonal number is not built by completing an infinite process; it is defined by a rule that specifies its n-th digit for every natural number n. That is the entirety of the construction. No digit at infinity is summoned, consulted, or altered, because no such digit exists.

Agreed. Not sure if you've looked into projective geometry, but it's pretty interesting - within such spaces, infinity is part of the set and forms part of the calculations. It's quite neat since structures that form singularities in other spaces actually have representations there (e.g., consider the set of intersections of all possible lines in 2D - in P^2, even parallel lines will intersect at a point in the same space as the non-parallel lines, so you don't have to make exceptions for parallel lines in your calculations).

 

[Bliksis]

The aliens will know.

 

[Brian_G]

"nothing needs to be proven “at infinity.”"

For cases like a_n = 1/n this is true because you always have the n available for computation. But in this case not.

Short assessment:

He’s still missing that Cantor’s diagonal also has every finite (n) available — you can compute the (n)-th digit directly from the list’s (n)-th entry. Nothing special happens “at infinity,” so his distinction doesn’t exist.

By The Way (BTW) , can you ever be wrong dude? Never seen you say so.

 

[talanum1]

• 1st digit (defined by looking at position 1)
• 2nd digit (defined by looking at position 2)
• 3rd digit (defined by looking at position 3)
• ... and so on

But now you must continue the list and you can't till infinity because you would have to look at the digit "at" infinity and there is no such digit.

I'm not treating infinity like a number.

I can be wrong, and I'll admit it if I'm wrong, but in this case I'm not wrong.