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The classic proof of the Cantor set start with the assumption that the set

$$B=\{x\in A:x\notin f(x)\}$$

exists, where $f: A\to\mathcal P(A)$ is a bijective function. I understand the proof but I dont understand the assumptions where you start to make this proof.

To be clear, why a set $B$ can be constructed? How you can justify this assumption? To me the proof of the Cantor theorem is far to be clear or complete if there are not an explanation about why $B$ must be possible.

Then, can someone explain to me or justify, via other theorems if possible, why $B$ must exist? Thank you in advance.

P.S.: can someone explain to me the downvotes in the question?

Masacroso
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    It's well defined. It must exist. It might be empty but it must exist. Existing doesn't mean any elements exist. – fleablood Jul 05 '16 at 02:37
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    Your "explanation" is very unsatisfactory @fleablood – Masacroso Jul 05 '16 at 02:40
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    No where in the proof is it assumed B is non empty. If B is not assumed to nonempty there is no reason not to refer to it as a well defined set. So basically I simply don't understand your concern. – fleablood Jul 05 '16 at 02:42
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    all conceivable sets exist. That is not an issue. The issue is whether the set is empty. Let G={prime numbers that have seven distinct integer divisors} is a set. It exists. I don't have to justify it. However if I ever refer to any element in it, I have to justify that any elements actually exist. Which I can't. – fleablood Jul 05 '16 at 02:47
  • @fleabood About the claim that "all conceivable sets exists" I understand this is an axiom. Then the answer to my question is: their existence is defined axiomatically. – Masacroso Jul 05 '16 at 02:49
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    Why did you want till you got to Cantor's theorem to raise this doubt? Why not when you first saw the interval $(a,b)$ defined as ${x:a\lt x\lt b}?$ What is the justification for the existence of that set? – bof Jul 05 '16 at 02:59
  • @bof Because the definition of $B$ is more complex. I cant express my reason now but I see it more complex. I can explain the case for the odd numbers or $(a,b)$ but the definition of $B$ seems to me very strange. – Masacroso Jul 05 '16 at 03:01
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    Maybe, depends on what sort of mathematician one is. Sets and math doesn't "exist" as such. We can talk about any concept we want. I can talk about {unicorns in my garden}. That doesn't mean there are unicorns in my garden. We can say {unicorns in my garden} doesn't "exist". But if so, the proof never says that B "exists" at all. My understanding is that Cantor defines the set but never assumes it "exists" in the sense it is non-empty. That is the * conclusion* that is systematically proven. – fleablood Jul 05 '16 at 03:09
  • What about G={odd integers evenly divisible by 2}. Do you have any problem with its existence? – fleablood Jul 05 '16 at 03:14
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    @fleablood: Not all conceivable sets exist. It's simply not possible. For example, take a Reinhardt cardinal and all the sets given by the axiom of choice. It is a theorem of ZF that they cannot all exist. – user21820 Jul 06 '16 at 03:39
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    @Masacroso: Although technically the answer you accepted has answered your question, the reason why you feel differently about the set in Cantor's theorem compared to the set of odd numbers and the interval $(a,b)$ is explained only by constructive viewpoints. Among the subsystems of second order arithmetic, RCA0 gives you the set of odd numbers, ACA0 gives you nearly all real analysis, but not even full Z2 gives you sets of sets of natural numbers, not to say any sets of arbitrary sets as the axiom schema of specification in ZF gives you. – user21820 Jul 06 '16 at 03:43
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    By the way, I saw your deleted answer. Just to make it clear, the axiom of choice has nothing to do with this issue. – user21820 Jul 06 '16 at 04:05
  • Thank you for the clarifications @user21820, it is really useful. I realized that I had a lot of mistakes in my "autoanswer", so I deleted it. The point is that it is related (how I feel it) about constructivity, by example the set $B$ is ad hoc. Maybe it is a miss-perception of the matter, or not. With time and knowledge I will answer myself correctly about this question. – Masacroso Jul 06 '16 at 04:09
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    @Masacroso: Then I think you'll be interested in reading https://www.sas.upenn.edu/~htowsner/prooftheory/ReverseMath.pdf and my post at http://math.stackexchange.com/a/1808558. At least you will then be able to classify more finely the kinds of sets that you see, or the kinds of foundational systems you come across. – user21820 Jul 06 '16 at 04:13

4 Answers4

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The existence of the set $B$ is an immediate consequences of one of the basic axioms (technically, an axiom schema) of set theory, called variously the axiom of subsets, or separation, or specification, or comprehension. It says that, given any set $A$ and any condition (technically, a condition expressed by a first-order formula in the language of set theory), there is a set $B$ containing exactly the elements of $A$ that satisfy the condition.

The axiom of subsets is the basis for almost every set defined in mathematics. You don't like it, fine. But why drag Cantor's diagonal set into the argument, you might as well ask, what is the justification for the set of odd numbers.

bof
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  • Thank you, this is was I was searching for... some justification, maybe via the axioms of ZF. Sorry but my knowledge of ZF is small, this is the reason for this question. – Masacroso Jul 05 '16 at 02:57
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    +500 for that last part. – Asaf Karagila Jul 05 '16 at 04:19
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    Nah your last sentence is misleading. Most constructivists will have no problem with recursive subsets of the natural numbers, and certainly odd numbers are recursive! It's a completely different issue if you want to construct collections of arbitrary sets, not of natural numbers. – user21820 Jul 05 '16 at 10:55
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    @user21820: Your comment is misleading. Any set theory should have some reasonable justification for the existence of sets. True, there are some constructive systems in which Cantor's theorem fails, but that drifts into the "it's only so because it obeys the definition, not because it makes any sense", at least from the point of view of any "classically trained mathematician". – Asaf Karagila Jul 05 '16 at 19:47
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    @AsafKaragila: Nope. RCA0 corresponds to computable mathematics and only permits recursive subsets of natural numbers. ACA0 has full arithmetical comprehension but has no such thing as collections of sets. This has nothing to do with whether Cantor's theorem fails; my point stands that it is a strawman argument to bring up the set of odd numbers, since the set used in the diagonalization for set theory only makes sense if you assume that the axiom schema of specification has some meaning, whereas the set of odd numbers makes sense at a much lower level of RCA0. – user21820 Jul 06 '16 at 03:27
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    @user21820: But last time I checked neither one was a set theory. Or at least a topos of some sort. Right? I mean, we don't ask about Cantor's theorem in vector spaces, because linear algebra is not supposed to actually deal with sets and power sets. I don't see the relevance of your comment, then, brining arithmetic into the discussion. – Asaf Karagila Jul 06 '16 at 04:41
  • @AsafKaragila: They are indeed not full set theories, but predicative viewpoints might reject anything beyond ACA. Perhaps I was not clear enough. For anyone who assumes ZF, all these sets are essentially on par with one another, since they all come from the same specification axiom. However, for those who do not subscribe to ZF, they may accept arithmetical comprehension (based on an intuitive 'understanding' of natural numbers) but may reject even the notion of power-set, in which case Cantor's theorem has to be interpreted more constructively to make sense. – user21820 Jul 06 '16 at 04:52
  • @user21820: And how do you interrupt Cantor's theorem in SOA, where the notion of power set does not exist? – Asaf Karagila Jul 06 '16 at 04:57
  • @AsafKaragila: The constructive version is that for any list of functions from $\def\nn{\mathbb{N}}$$\nn$ to $\nn$ there is a function that disagrees with every function on that list. Technically the list of functions is of type $\nn \to ( \nn \to \nn )$, which we don't natively have even in ACA0, but we can interpret $\nn \to \nn$ as all formulae $φ$ that represent a function. Each formula can be coded by a single natural, and doing it twice gives what we want, even in PA. Then we have $\forall f \in (\nn\to(\nn\to\nn))\ \exists g \in (\nn\to\nn)\ \forall k\ \exists n\ ( f(k)(n) \ne g(n) )$. – user21820 Jul 06 '16 at 05:40
  • @user21820: That is kind of cheating here. And you wonder why things don't work! :-P – Asaf Karagila Jul 06 '16 at 05:53
  • @AsafKaragila: Cheating is in the eyes of the beholder. =) Set theorists who believe in the existence of the full power-set will of course say that this notion of functions is too restrictive, but even set theorists who like $V=L$ would admit that this notion is reasonable. As for me, I haven't yet seen with my eyes a function that cannot be represented by a formula, but I'll change my mind if I see one. =P – user21820 Jul 06 '16 at 05:59
  • @user21820: IDDQD, IDKFA, IDCLIP. Now run through the levels of Doom. :-) – Asaf Karagila Jul 06 '16 at 06:05
  • @AsafKaragila: Haha! You're trying to make me see stars right... But I don't play FPSs or RPGs; the only games I play are abstract games like Weiqi or puzzle games. =) – user21820 Jul 06 '16 at 06:18
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Actually one should not assume in advance that the function is bijective; rather one should prove that it's not, by showing that the set $$ B = \{ x \in A : x\notin f(x)\} $$ is not in its image.

Let's try an example: $A=\{1,2,3\}$ and $\begin{cases} f(1) = \{1,2\}, \\ f(2) = \{1,3\}, \\ f(3) = \{2,3\}. \end{cases}$

Then $B = \{ x\in A:x\notin f(x) \} = \{2\}$, since $\Big( 1\in f(1) \text{ and } 2\notin f(2) \text{ and } 3\in f(3)\Big)$.

For each element $x$ we see that either $x\in f(x)$ or $x\notin f(x)$. Those for which the latter alternative holds are members of the set $B$.

The above is not a logical argument to the conclusion that I would like to support, but is presented in case setting forth a concrete instance might clarify your thinking about this definition of the set $B$.

Michael Hardy
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Here is an example on a finite set: $S = \{1,2,3\}$, $P\left(S\right)=\{\phi,S,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\}\}$ take for example: \begin{eqnarray} f\left(1\right)=\{1,2\}\\ f\left(2\right)=\{1,3\}\\ f\left(3\right)=\{1\} \end{eqnarray} $B=\{x\in S; x\notin f\left(x\right)\}$ \begin{eqnarray} 1\notin B\\ 2\in B\\ 3\in B \end{eqnarray} $B=\{2,3\}\in P\left(S\right)$

ryaron
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Another example, from the infinite set of natural numbers: \begin{eqnarray} S=\mathbb{N}\\ P\left(S\right)=\{\phi, S, \forall U\neq\phi;U\subset S\}\\ f\left(x\right)=\{x\} \end{eqnarray} Then f is 1-1.

$B=\{x\in S;x\notin f\left(x\right)\}$
Then $\forall x\notin B$
$B=\phi \in P\left(S\right)$
Which exists and again is not in the image of $f$.

ryaron
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