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I don't hope to "debunk" Cantor's diagonal here; I understand it, but I just had some thoughts and wanted to get some feedback on this.

We generate a set, T, of infinite sequences, sn, where n is from 0 to infinity.

Regardless of whether or not we assume the set is countable, one statement must be true: The set T contains every possible sequence. This has to be true; it's an infinite set of infinite sequences - so every combination is included. For now, forget countable or uncountable.

As per Cantor's argument, now we define the sequence s - and as a result, we have constructed a sequence that cannot possibly be in the set T. Now there are two conflicting claims:

  1. The set T contains every possible sequence.
  2. The sequence s is not in T.

At this point I'm mainly wondering, why is the possibility and validity of the sequence s not challenged? Why is there no discussion over whether this scenario is symptomatic of inconsistencies in the underlying mathematics (e.g. set theory), etc.? Why did Cantor's diagonal become a proof rather than a paradox?

To clarify, by "contains every possible sequence" I mean that (for example) if the set T is an infinite set of infinite sequences of 0s and 1s, every possible combination of 0s and 1s will be included.

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    It's not at all clear what you are saying at the start. "We generate..." How is it generated? What are the elements of the sequences? Cantor's argument requires clarity in choice of words to get the point across. – 2017-02-20
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    The nature of a proof by contradiction is you make a hypothetical assertion and then show that it produces a contradiction and then conclude that the assertion is false. We have not produced a paradoxical world where your (1) and (2) are true at the same time. We have showed that if (1) were hypothetically true, then (2) would also be true. Since (2) contradicts (1) we must conclude (1) is not true. – 2017-02-20
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    For sake of illustration, just use Cantor's original proof. – 2017-02-20
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    You've simply removed the problematic phrase "diagonal of $T$" and don't tell us how to generate $s$ except "As per Cantor's argument..." But Cantor's argument assumes a map $\mathbb N\to T$, not just a set $T$. – 2017-02-20
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    "it's an infinite set of infinite sequences - so every combination is included" -- That's certainly not true. What if we take all the infinite sequences of 0's and 1's? Then that's an infinite set of infinite sequences but none of the sequences containing 2 is included. – 2017-02-20
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    @user4894 I can see how the wording used might be confusing, but I meant that (for example) if the set T is an infinite set of infinite sequences of 0s and 1s, every possible combination of 0s and 1s will be included. – 2017-02-20
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    The construction of $s$ is sound and valid given you have a listing of elements of $T$. (Here I am not assuming that this list covers every element of $T$. Technically, I am considering any function $\Bbb{N} \to T$.) The crux of the cantor's diagonalization argument is that any listing of elements of $T$ will miss an element of $T$, and this argument supplies an *explicit* construction of an element $s$ that is missed by the given listing. The contradiction arises only when you begin to assume the existence of a 'complete listing', a listing that does not miss any single element of $T$. – 2017-02-20
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    The clarifications make it even harder to understand your question. – 2017-02-20
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    "This has to be true; it's an infinite set of infinite sequences - so every combination is included." -- Nope. – 2017-02-20
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    @Treeguard That's also not true. The sequence of infinite sequences of $0$s and $1$s that contain exactly one $1$ is infinite, but it doesn't contain any sequence with more than one $1$. – 2017-02-20
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    @Treeguard Or perhaps more convincingly to you, if you take one element off of the list, it's still an infinite list, right? But it's missing the element you took off, so it's not complete. The pigeonhole principle is false for infinite sets. You can have a map from an infinite set to itself that is injective but not surjective, or surjective but not injective, which can't happen for finite sets. Sometimes this is used as the definition of "infinite." – 2017-02-20
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    @Treeguard looking at Edit #2, the problem is not with the infinite set of strings you created, that is not where the contradiction is. The issue is that you assume you can label them _1, 2, 3, ... n, ...._ that sets up a bijection with the natural numbers. Cantor's proof is says given that mapping he can find something not on the list so it cannot be a bijection. The mapping is the source of the contradiction not the set of infinite sequences. – 2017-02-20
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    _"This has to be true; it's an infinite set of infinite sequences - so every combination is included." […] "if the set T is an infinite set of infinite sequences of 0s and 1s, every possible combination of 0s and 1s will be included"_ — you're repeating this again and again in different ways, but it's not true (and in fact that's what Cantor showed). (For example, here is an infinite set of infinite sequences of 0s and 1s: the $n$th sequence starts with $n$ zeroes, followed by all $1$s. This is an infinite set of infinite sequences, but it doesn't contain the sequence $1, 0, 0, \dots$.) – 2017-02-21
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    Basically **either** you can take $T$ to be a list of sequences $s_n$ (in which case it can't contain all binary sequences), **or** you can take it to be an unordered set (in which case you can't list out the sequences it contains). You're trying to assume _both_ things are true for $T$ (that it contains all binary sequences _and_ that it can be enumerated as a list of sequences $s_n$), and getting a contradiction. This shows that your assumption was wrong (and this is precisely Cantor's proof). – 2017-02-21
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    "I don't hope to "debunk" Cantor's diagonal here; I understand it..." Clearly you don't. – 2017-02-22

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The "diagonal of $T$" is meaningless.

$T$ is an un-ordered set.

What Cantor says is that any map $\phi:\mathbb N\to T$ misses an element of $T$, and uses the diagonal of $\phi$, not $T$ - that is, he defines the sequence $\{t_n\}$ where $t_n$ is some value other than $\phi(n)_n$.

You can't define the "diagonal of $T$" in general.

Your update to the question simply removes the problematic phrase, "diagonal of $T$", replacing it with the phrase, "As per Cantor,..." But Cantor doesn't construct $s$ from $T$, he constructs $s$ from a map $\mathbb N\to T$.

So your argument is still meaningless. How are you construction $s$ from $T$? It is certainly not possible for it to be the same as the way that Cantor constructs $s$ from $\mathbb N\to T$.

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    What does $\phi(n)_n$ mean? – 2017-02-20
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    $\phi(n)$ is an element of $T$, which is a sequence. $\phi(n)_n$ is the $n$th element of that sequence. @N1ng – 2017-02-20
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    Thanks. I know you meaned $\langle t_n\rangle$ now. – 2017-02-20
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    I have no idea what you mean by $\langle t_n\rangle$, but there is no meaning I can give for those symbols that match my intent. @N1ng – 2017-02-20
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Regardless of whether or not we assume the set is countable, one statement must be true: The set T contains every possible sequence. This has to be true; it's an infinite set of infinite sequences - so every combination is included.

This is simply not correct. Just because a list is infinite does not mean it is complete. For instance, here is an infinite list of integers: $$0,2,4,6,8,10,\dots$$ This list, despite being infinite, does not contain all the integers: it is missing all of the odd integers, as well as the negative integers.

As a result, even though it is easy to come up with a countably infinite set $T$ of sequences, there is no contradiction from the conclusion that such a set does not contain all the sequences.

If you intend to assume that your initial set $T$ contains all sequences, then you get no paradox because you have not justified that assumption! How do you know that there exists a countable set $T$ that contains all sequences? There does exist a set of all sequences, but performing Cantor's diagonal argument to obtain the sequence $s$ requires you to know that $T$ is countable (since it makes use of a specific enumeration of the elements of $T$).

If you don't justify this assumption, then all you have proven is that if you make the assumption, you reach a contradiction. This is a proof by contradiction that the assumption is actually false! That is, the set of all sequences is not countable.

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    Yeah, the phrasing of that section was very odd. I just took OP to mean, "Let $T$ consist of all infinite sequences,..." although OP doesn't say what the sequences are sequences of, and there are obvious problems in the section you singled out. – 2017-02-20
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    @ThomasAndrews: As I interpreted it, that section was the crux of OP's argument: they are assuming that merely because it is possible to find a countably infinite set of sequences, that means the set of sequences must be countable since any infinite list is complete. That is, they don't mean to be defining $T$ to consist of all infinite sequences; rather, they are concluding that as a consequence of $T$ being infinite. – 2017-02-20
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    An infinite list isn't necessarily complete, but in this instance, T is intended to be. For example, say it's every possible combination of binary digits. – 2017-02-20
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    @Treeguard: Aha, but how do you know that a list of every possible combination of binary digits exists at all? – 2017-02-20
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    @Treeguard: You defined $T$ to be the set of all infinite binary sequences. That's fine. But the point is that an infinite set is not necessarily a list (in any sense). That's the whole point of the proof: we have the set $T$. It is the assumption that its elements can be listed (i.e., that there is a surjection from the positive integers to $T$) that leads to a contradiction. Are you familiar/comfortable with proofs by contradiction? If so, it's not clear what you find "paradoxical" here. – 2017-02-20
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    Note also that you list one of the claims as "The sequence $s$ is not in $T$." You don't need to say it that way. We assumed that we had a map from the positive integers to $T$ that mapped to every element of $T$, but then we found an element of $T$ that is not in the image of the map. So we assumed something and then derived that the thing we assumed isn't true. Okay, so the thing we assumed is false. – 2017-02-20
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So now we define the infinite sequence s of complementary numbers to the diagonal of $T$.

This step only works if $T$ is countable. By definition, a sequence can only contain countably many elements.

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    Of course, even if $T$ is countable, "the diagonal of $T$" is meaningless. The only way to talk about "the diagonal" is if we are given a specific map $\mathbb N\to T$. – 2017-02-20
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    @ThomasAndrews I think "the diagonal of $T$" can be made meaningful if $T$ is a list (a sequence) of sequences. (I think this is what the OP intended as well, accompanied by some confusion between sets and lists.) – 2017-02-21
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    Of course, that is exactly what I said. $T$ is described as a set of sequences that contains all possible sequences. It is not described as a sequence of sequences. @ShreevatsaR – 2017-02-21
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    @ThomasAndrews It's not clear how $T$ is described, because the OP is confused. For example, the OP says "We generate a set, $T$, of infinite sequences, $s_n$, where $n$ is from 0 to infinity — this suggests that they're thinking of $T$ as a sequence of sequences, while at the same time also thinking of it as a set. Basically my point is "the diagonal of $T$" can make sense (can be given a sense) if $T$ is a sequence (of sequences), and you also agree. – 2017-02-22
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Cantor's diagonalization argument was taken as a symptom of underlying inconsistencies - this is what debunked the assumption that all infinite sets are the same size. The other option was to assert that the constructed sequence isn't a sequence for some reason; but that seems like a much more fundamental notion. Cantor's argument explicitly constructs a mapping from $\mathbb{N}$ to whatever set the sequences are in; this is what we mean by a "sequence", so it seems bizarre to question whether it exists.

In general, we prefer to question axioms rather than definitions, wherever possible, because definitions are far more fundamental. For example, suppose you believed that all prime numbers were odd, and one day I showed you the number $2$. Would it be more reasonable to decide that (a) for some reason $2$ is either not prime or not a number at all, even though it satisfies the definitions for both, or (b) the axiom "all prime numbers are odd" is just wrong?

Also, nothing becomes a "paradox". Russell's paradox prompted the creation of modern set theory, in which the "paradox" is just a theorem. The "twin paradox" in relativity is just a thought experiment that informs the experimenter about the effects of relativity. Schrodinger's Cat, originally intended as a paradox, is now taken as an illustration of the effects of quantum mechanics. Given the option, no scientist in any field will take an example as an actual paradox; they will alter their assumptions just enough to make it no longer a paradox, rather than throwing everything out and starting fresh. In this case, the solutions to Cantor's "paradox" were to either conclude that some infinities are bigger than others, or that the definition of a "sequence" is inherently flawed in some unknown way. The first option was clearly simpler, so we chose that.

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    I like this answer, but just to address one point: I wouldn't mean to suggest that a sequence doesn't "exist" but rather, could it be argued that it is invalid? One can write an equation dividing a number by zero, but it would be considered undefined; similarly, could it be said that while you can write a sequence, it too could potentially be 'undefined'? – 2017-02-20
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    @Treeguard An equation is "undefined" if the number it specifies doesn't exist - there is no number $x$ so that $x \cdot 0 = 1$, so $1 / 0$ is undefined. In other words, there is no distinction between "is undefined" and "does not exist". Critically, also, the definitions of operations like division explicitly state when the value is undefined; if we're prepared to accept that some things that look like sequences aren't "really" sequences, we need to add something to the definition of "sequence" to let us tell when that happens - and there's no straightforward way to do that. – 2017-02-20
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    You are wrong about Schrödinger's cat. It is still a paradox. – 2017-02-20
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    @TonyK Its interpretation is up to debate, but no one considers it a true paradox (which is to say, a condemnation of the principles it's founded on). According to the Copenhagen interpretation, Schrodinger's Cat is just a straightforward description of what happens (requiring us to abandon the axiom that a cat can't be both alive and dead). According to other interpretations (which as far as I know are more standard) Schrodinger's Cat is an oversimplification of more abstract processes, and it's the simplification that creates the apparent paradox. – 2017-02-20
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    _Was_ there any "assumption that all infinite sets are the same size" to debunk prior to Cantor? My understanding (from secondary sources) is that both Cantor and his contemporaries regarded it as _obvious_ that you can't have a bijection between $\mathbb N$ and $\mathbb R$ -- if there were anything surprising about the result, it was how clever one needed to be to produce an actual proof of this fact, not that it was true. (On the other hand, that $\mathbb R$ and $\mathbb R^2$ _are_ equinumerous surprised Cantor and everyone else). – 2017-02-20
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    @HenningMakholm In point of fact, it was originally considered "obvious" that (for example) the set of primes was not the same size as $\mathbb{N}$. Cantor's early results debunked that by showing that all unbounded subsets of $\mathbb{N}$ were the same size, and so was $\mathbb{N}^2$. After that, though, there was briefly a sense of "infinite is infinite", because no examples of differently-sized infinities were known. (So, I guess, what the assumption was depends on the particular point in time you're talking about.) – 2017-02-20
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    A paradox does not have to be contradictory, it can also simply be counterintuitive. The dictionary says "a seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true." So these counterintuitive results still qualify as paradoxes even if there are no flaws. – 2017-02-20
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Regardless of whether or not we assume the set is countable, one statement must be true: The set T contains every possible sequence. This has to be true; it's an infinite set of infinite sequences - so every combination is included.

I know that others have already given counterexamples to this, you could actually come up with counterexamples for quite a few infinite sets. For example, the set of all real numbers between 5 and 10 is clearly uncountably infinite, but it's still a proper subset of the real numbers.

You may want to research the Infinite Hotel Paradox. Suppose that you have a hotel with a countably infinite number of rooms, all of which are booked. One night, a bus pulls up with an infinite number of numbers looking for rooms. The paradoxical thing: the hotel can accommodate all of them in spite of being at "full occupancy." All you have to do is move everyone to even-number rooms (i.e. if you occupy room $n$, you would have to move to room $2n$).

One point to emphasize here from the previous paragraph is that, when reasoning about infinite sets, your intuition can steer you badly wrong if you're not careful. A lot of people accidentally import assumptions about how finite sets work and then assume that infinite sets will work the same way, and there's no requirement that they will.

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You already got a lot of answers about your errors. But let me add what Cantor's argument actually says.

Ultimately it is about functions from a set $X$ to a set $Y$, where $Y$ has more than one element. And it says that there are more functions from $X$ to $Y$ than there are elements in $X$. Or, more formally, the set of functions from $X$ to $Y$ has greater cardinality than $X$.

The most famous application of Cantor's diagonal element, showing that there are more reals than natural numbers, works by representing the real numbers as digit strings, that is, maps from the natural numbers to the set of digits. And the probably most important case, the proof that the powerset of a set has larger cardinality than the set itself, is brought to that form by identifying a subset with a function from the set to $\{0,1\}$, with $0$ meaning "this element is not in the subset" and $1$ meaning "this element is in the subset".

Note that there are always two sets involved in Cantor's diagonal element. The first one is the set I called $X$ above, and the second one is the set of functions from $X$ to $Y$ (or a set that can be described by such functions).

You are talking about sequences. Sequences are, bu definition, functions from the natural numbers to whatever objects you form sequences of. So here $X=\mathbb N$. It doesn't matter what $Y$ is, as long as it has at least two elements. Since you assert that $T$ is an infinite set, we can assume that this is indeed the case, as if $Y$ had only one element, there would be only one sequence, and from one sequence you cannot make an infinite set. And obviously with empty $Y$ you cannot form any sequence at all, and the empty set is finite.

So what you would prove with Cantor's diagonal proof is not that $T$ is incomplete. Rather, you prove that if you have a bijection $\mathbb N\to T$, then there's a sequence that's not in $T$. Which in reverse means that if $T$ does contain all sequences (that is, there's no sequence that's not in $T$), then there is no bijection $\mathbb N\to T$, which in turn means that $\mathbb N$ and $T$ don't have the same number of elements. Since it is easy to see that the set of all sequences must be at least as large as $\mathbb N$ (simply consider the sequences which are almost constant, but have exactly one element different), this implies there are more sequences than natural numbers.

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Of course you can define such an infinite set T. As you said

Regardless of whether or not we assume the set is countable, one statement must be true: The set T contains every possible sequence.: The set T contains every possible sequence.

However if it is not countable then you CANNOT use the Cantor's arguments on this set. Therefore the problem arises.

If you look into the Cantor's theorem you will see that the numbering of the sequences matters. Therefore the countability of the set T is first assumed and then after the contradiction is reached, all the blame is put in on our single assumption which is the countability.

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These two sentences are not related the way you think:

Regardless of whether or not we assume the set is countable, one statement must be true: The set $T$ contains every possible sequence.

As per Cantor's argument, now we define the sequence $s$ - and as a result, we have constructed a sequence that cannot possibly be in the set $T$.

As per Cantor's argument, you define the sequence $s$ term-by-term (digit-by-digit in Cantor's proof), chosing each single term against some sequence chosen from the set $T$. And as $s$ is a sequence, it has only countably infinite set of terms, so you used at most countably infinite set $S$ of sequences.

Of course, $S \subseteq T$, but you have not shown $S = T$, even for countable $T$ (note that infinite $T$ can have, and even has, proper subsets equipollent with itself, so equipollency of $S$ and $T$ does not imply their equality).
This is the important deviation from the Cantor's proof, which assumes countability of $T$ and explicitly shows a bijection between terms of $s$ and elements of $T$ (and that implies $S=T$).

The more, you have not shown (and in fact, you're unable to show), that countable $S$ exhausts $T$, if $T$ is uncountable.

You can conclude, as Cantor did before, that your newly constructed $s$ certainly does not appear in $S$, but it's not enough to infer $s\notin T$.

The whole 'paradox' is just a result of a mistake in your reasoning.

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Cantor's argument says that you can't enumerate the elements in $T$, i.e. define a bijection between $\mathbb N$ and $T$. He does so by showing that whatever the enumeration, let $E$ (such that there is a bijection between $\mathbb N$ and $E$), you will miss some element.

You have two nonconflicting claims:

  • $T$ contains all sequences,

  • for any enumeration $E$, one can build a sequence $s$ such that $s\in T$ but $s\notin E$.

Hence, $E$ is a proper subset of $T$.

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Let's put it this way.

Let $T = \{\text{the set of all sequences}\}$.

Let $S \subset T$ so that $S$ can be sequenced.

Let $s_S$ be the "impossible sequence" so that $s_S \not \in S$.

As $s_S \not \in S$ but $s_S \in T$, we know $S \subsetneq T$.

But $S$ was any arbitrary subset of $T$ that could be sequenced. But $S$ can not be $T$. So $T$ can not be sequenced.

So $T$ is uncountable.

Q.E.D.

That's all there is to it. No paradox.

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"Regardless of whether or not we assume the set is countable, one statement must be true: The set T contains every possible sequence. This has to be true;"

It has to be true because that is how we defined it; not for the arguments you give (which are naively wrong). But that's irrelevant.

"As per Cantor's argument, now we define the sequence s - and as a result"

Whoa!!!!! Hold on! HOW did we define the sequence $s$ on the set $T$? We can only use the Cantor diagonal argument on $T$ if $T$ is a sequence. We don't know that $T$ is a sequence; only that $T$ is a set.

And that is the entire point. IF $T$ could be made into a sequence, THEN we could create a sequence $s$, that is not in $T$ which would be a contradiction. So we must conclude our hypothesis, that $T$ can be made into a sequence, is false.

So now we know that $T$ can not be made into a sequence. And if we can't make $T$ into a sequence, we have no way of creating $s$. And we have failed to create a sequence not in $T$.

So there is not paradox because we didn't do anything. We know $T$ can not be sequenced. Since $T$ can not be sequenced we can't and didn't create an $s$.

If we tried to create $s$ we'd have to take a sub set of $T$, $S \subsetneq T$ that can be sequenced. We construct $s$. We conclude $s \not \in S$ and therefore $s \in S^c \cap T$. Which is .... just honky-dory.

What we did conclude was this statement: $T = \{\text{ all infinite sequences}\}$ can not be ordered into a sequence.

And that is the core of the diagonal argument.

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Thomas Andrew's answer is the best, but I thought it is easier to swallow when we make things extremely concrete, so here goes:

Instead of 'sequences' we take Sequences of binary digits, and we add an ornamental '0.' in front of each one to turn in into a real number.

Now instead of the abstract set $T$ we can use the unit interval $[0, 1]$

Note that every sequence of 0s and 1s appears as the binary expansion of some number in $[0,1]$ (with an ornamental '0.' attached to it at the beginning) so $[0, 1]$ fits all the requirements of the question, while at the same time it is a very concrete and familiar object. Also, the elements already have an ordering on them.

Now, to paraphrase fleablood above: 'Your move. How are you gonna construct the forbidden sequence of digits (or equivalently: number) $s$?'

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"We generate a set, T, of infinite sequences, s(n), where n is from 0 to infinity. Regardless of whether or not we assume the set is countable..."

You do realize that you just defined T to be countable, don't you? You associated every member of T with a natural number n, which is the definition of "countable."

Cantor's proof is often misrepresented. He assumes only that (1) T is the set of all binary strings, and that (2) S is a subset of T; whether it is proper or improper is not addressed by this assumption.

Let A be the statement "S is countable," and B be the statement "S is equal to T; that is, an improper subset."

Diagonalization proves the lemma "If A, then not B" by exploiting the way S can be counted to produce a string that is in T, but not in S.

By contraposition, this also proves "If B, then not A." That is, "If S is all of T, then it is not countable."