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Cantor's diagonalization argument says that given a list of the reals, one can choose a unique digit position from each of those reals, and can construct a new real that was not previously listed by ensuring it does not match any of those digit position's place values. The conclusion is that $| \Bbb{R}| > |\Bbb{Z}|$. Let's call the function that produces digits of the unlisted real $D$.

The proof by contradiction starts by assuming a list $X$ containing all reals, therefore $|X|=|\Bbb{R}|$. Let's say that $x_{i,j}$ is the $j$-th digit of the $i$-th element of $X$ and $j \in \Bbb{Z}$. Any digits not present are assumed to be zero-valued. (Correct me if I make any invalid assumptions along the way.) For the moment let's only say of $i$ that it can take on positive and negative values, which is easily justified by $X$ containing positive and negative reals.

Let's call $y$ the unlisted real produced by diagonalization, and $y_k$ are the digits of the unlisted real. $k \in \Bbb{Z}$. Though many ways to produce place values exist, lets settle for:

$$D(k) = y_k = (x_{k,k} + 1) \mod 10$$

This means the domain of $D$ is $\Bbb{Z}$.

However, by definition, the diagonalization process must consume all elements of $X$, and so D must be a function that can be invoked with a unique input the same number of times as there are elements in $X$. Again, we assumed that $|X|=|\Bbb{R}|$. This means that the cardinality of the domain of $D$ is $| \Bbb{R} |$. If we accept that $| \Bbb{R}| > |\Bbb{Z}|$, then we have:

  • the domain of $D$ is $\Bbb{Z}$.
  • the cardinality of the domain of $D$ is $| \Bbb{R} |$
  • $| \Bbb{R}| > |\Bbb{Z}|$

A contradiction is formed before we even attempt to produce a $y$. My interpretation of this contradiction is that, If we accept that $| \Bbb{R}| > |\Bbb{Z}|$, then we must also accept that the function $D$ cannot exist, and therefore diagonalization is impossible.

To my question, shouldn't the impossibility of diagonalization mean that it is not useful for proving $| \Bbb{R}| > |\Bbb{Z}|$?

I keep thinking that $D$ is similar (what's the precise word?) to some $D'$ where the input is some element of $X$, and the output is an unique $i$ from $y_i$. That would be $D': \Bbb{R} \rightarrow \Bbb{Z}$, a demonstration of a bijection all on its own. Perhaps that a list can contain the reals, and that $D$ exists, are the same assumption, and by that, we can reach the contradiction articulated by Cantor.

Brent
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  • The usual proof goes like this: you assume there is a surjective map from $\mathbb{N}$ to $\mathbb{R}$, and then use diagonalization to produce a real number not in the image of that map, which contradicts the surjectivity. Therefore the assumption must be false: there is no surjective map from $\mathbb{N}$ to $\mathbb{R}$. This implies there is no injective map from $\mathbb{R}$ to $\mathbb{N}$ (actually these two statements are equivalent under AC). – xXF Dec 26 '21 at 19:04
  • Isn't that the same as a list of the reals? – Brent Dec 26 '21 at 19:13
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    For what it's worth, proof by contradiction can be avoided. One simply proves that for each function $f:{\mathbb N} \rightarrow {\mathbb R},$ there exists (in fact, one can construct explicitly) a real number that does not belong to ${f(x): x \in {\mathbb N}}.$ – Dave L. Renfro Dec 26 '21 at 19:14
  • Yes, I'm just trying to state it in a concise way; also for me it's more natural to use $\mathbb{N}$ instead of $\mathbb{Z}$, although they are basically the same. I don't fully understand you question, but it seems you are confused about the distinction between the map $f:\mathbb{N}\rightarrow\mathbb{R}$ (which is assumed to be surjective) and the diagonalization process. After fixing such a map $f$, you recursively construct a real number $x$: choose the tenth digit of $x$ to be different from tenth digit of $f(1)$, hundredth digit of $x$ different from that of $f(2)$, etc. – xXF Dec 26 '21 at 19:18
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    While the common expositions of the proof often begins by assuming you have a surjection, this is not necessary. One can avoid contradictions completely, and simply show how to construct, for a given function $f\colon\mathbb{N}\to[0,1]$, a number $s_f\in [0,1]$ with the property that $\forall n\in\mathbb{N}$, $f(n)\neq s_f$. There are no "arguments by contradiction" embedded there. I usually object to the "proof by contradiction" casting because the assumption of surjectivity is unnecessary. I call these proofs "fake proofs-by-contradiction"; similar issues with infinity of primes. – Arturo Magidin Dec 26 '21 at 19:25
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    @xXF: I really wish you wouldn't assert that one assumes the function is surjective. That is a completely unnecessary assumption that is never used in the argument until you have explicitly produced a number not in the range, and then you trot out the assumption from the wings to center stage to say "Ta-da! Contradiction!" A completely unnecessary complication, when removing that initial assumption and that final line gives you a perfectly fine proof of "for every $f\colon\mathbb{N}\to\mathbb{R}$, there is $r_f\in \mathbb{R}\setminus\mathrm{Im}(f)$". – Arturo Magidin Dec 26 '21 at 19:29
  • @ArturoMagidin I agree starting the proof by assuming surjectivity is making things worse. It seems all we need is the equivalence between $\forall x\lnot\phi(x)$ and $\lnot\exists x\phi(x)$. Does this equivalence involve any law of contradiction? – xXF Dec 26 '21 at 19:32
  • @xXF: I don't think you need any such equivalence. We are proving "there does not exist a surjection"; we do this by showing "for every function, there is an element of the codomain that is not in the image". – Arturo Magidin Dec 26 '21 at 19:34
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    I give a proof here with no argument by contradiction showing that there is no surjection from $\mathbb{N}$ to $2^{\mathbb{N}}$; it is an easy matter to establish a bijection between $\mathbb{R}$ and $2^{\mathbb{N}}$, e.g. using Cantor-Bernstein, and so there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$. – Arturo Magidin Dec 26 '21 at 19:37
  • @ArturoMagidin "Any list is not complete" is not literally the same as "there does not exist complete list". I think the former translates to $\forall x\lnot\phi(x)$ where $\phi(x)$ says "the list $x$ is complete" while the latter translates to $\lnot\exists x\phi(x)$. Sometimes people take both $\forall$ and $\exists$ to be primitive symbols in first order logic, and sometimes people define $\forall x$ as $\lnot\exists x\lnot$. In any case I feel that at least we need something like $p\leftrightarrow\lnot\lnot p$, which I believe requires some extent of law of contradiction. – xXF Dec 26 '21 at 19:38
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    "A contradiction is formed before we even attempt to produce a $y.$" Setting aside the question of whether contradiction is necessary to prove this particular fact, it seems to me that in every proof by contradiction, a contradiction is formed before we even attempt to "produce a $y$" or whatever it is that the proof attempts to do. At some point we say "assume $P$" and at that instant the contradiction is already implicit. The rest of the work to "produce a $y$" is simply to demonstrate a contradiction that already exists. – David K Dec 26 '21 at 19:38
  • @xXF: $p\iff \neg\neg p$ is excluded middle, not "law of contradiction". In any case, you don't need an equivalence, you need an implication. But at that level, it's really irrelevant. This argument, built atop standard logic, is a direct argument of a stated proposition, not a proof by contradiction. – Arturo Magidin Dec 26 '21 at 19:58
  • @ArturoMagidin I'm surprised that there's a bijection between $\Bbb{R}$ and $2^\Bbb{N}$. I'm thinking of scientific notation, where you have a mantissa that's a $\Bbb{Z}$ and an exponent that's a $\Bbb{Z}$, but that only gives us $\Bbb{Q}$. I'll have to come to grips with your linked answer though, since at first glance it seems you might be using a different meaning of $2^\Bbb{N}$ than the power set. – Brent Dec 26 '21 at 20:04
  • @Brent: "where you have a mantissa that's a $\mathbb{Z}$..." I have no idea what it is you are thinking. There is a very easy bijection between $(0,1)$ and $(-\pi/2,\pi/2)$: $f(x) = -\pi/2+x\pi$. There is a very easy bijection between $(-\pi/2,\pi/2)$ and $\mathbb{R}$: just use $f(x)=\arctan(x)$. And there are easy injection going each way between $2^{\mathbb{N}}$ and $(0,1)$: from $(0,1)$, write in binary, picking one representation uniformly when required; and from $2^{\mathbb{N}}$, send the sequestion $\sigma$ to the real number with $5$ in position $n$ if $\sigma(n)=0$, and $6$ otherwise. – Arturo Magidin Dec 26 '21 at 20:08
  • @Brent: (cont) Cantor-Bernstein then tells you (without arguments by contradictions) that from the injections $(0,1)\to 2^{\mathbb{N}}$ and $2^{\mathbb{N}}\to (0,1)$ you can construct a bijection $2^{\mathbb{N}}\to (0,1)$. Not just compose the bijections $2^{\mathbb{N}}\to(0,1)\to(-\pi/2,\pi/2)\to\mathbb{R}$. – Arturo Magidin Dec 26 '21 at 20:09
  • @DaveL.Renfro Would you point me to an example? – Brent Dec 26 '21 at 20:18
  • The answer I link to above provides such a proof. – Arturo Magidin Dec 26 '21 at 20:30

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