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I am taking my first analysis class, and I am really enjoying it. I have recently stumbled upon Cantor's Theorem which states that there exists no surjective map $f$ of a set $A$ to its powerset, $P(A)$. The proof is pretty straight forward, that is not the issue, but I was wondering if anyone could speak to the existance of the Cantor Diagonal Set, $B = \{x\in A:x\notin f(x)\}$. It seems a little too convenient to define a set where a mapping is restricted when one is proving that there is a set which cannot be mapped to.

I am looking for a little intuition. Thanks.

EDIT: I would like to thank those who provided answers/comments to this question. The Cantor Diagonal Set is now much more clear to me. I withdrawal my statement that the proof to Cantor's Theorem is "convenient." Thanks so much!

Mark Watson
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  • Read this answer: http://math.stackexchange.com/questions/162/why-is-the-set-of-all-sets-a-paradox/909772#909772 – Asaf Karagila Oct 04 '15 at 22:27
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    It should be said, $B$ is defined in terms of $f$ - that is, it should really be written as $B_f$, for any $f:A\to P(A)$. So the set $B_f$ is entirely defined in terms of $f$ and $A$. – Thomas Andrews Oct 04 '15 at 22:36
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    "t seems a little too convenient to define a set where a mapping is restricted" - we're not restricting the mapping in any way. It's not like we're saying "define $B$ such that it's not in the range of $f$". Maybe that $\notin$ sign is confusing you. – user2357112 Oct 05 '15 at 00:11
  • @user2357112 that is exactly how I have been interpreting it! – Mark Watson Oct 05 '15 at 04:47

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Better to consider it as $B_f$ - we take a function $f:A\to P(A)$ and define $B$ in terms of it.

In particular, we are stating exactly when $x\in B_f$. That is:

$$x\in B_f\iff x\in A \text{ and } x\notin f(x)$$

That entirely determines $B_f$, given an $f$.

And what the argument shows is that no matter what $f$ you give men, $B_f$ is not in the range of $f$.

So no $f$ is onto.

Thomas Andrews
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  • So you see absolutely no issue with $B_f$? – Mark Watson Oct 04 '15 at 22:50
  • It seems to me that we answered completely different questions. This is an answer for why $B$ works for a given $f$, rather than to all $f$. My answer is why $B$ even exists in the first place. – Asaf Karagila Oct 04 '15 at 22:54
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    @MarkWatson No, and more importantly, you haven't stated what your issue is with it. If you give me an $f$, have I not told you how to determine which elements of $A$ go into $B_f$? – Thomas Andrews Oct 04 '15 at 22:55
  • I'm answering why $B$ exists, too, @AsafKaragila - but emphasizing in notation that $B$ is defined in terms of both $f$ and $A$. But yeah, I'm not using axiomatic reasons for why $B$ exists, only intuitively that we are defining it in terms that are non-circular. – Thomas Andrews Oct 04 '15 at 22:58
  • I lack the mathematical sophistication to state things formally, but it seems a little too convenient to state that there is no $f:A \to P(A)$ because we define $B_f$ such that $f$ is not allowed. I know I am obviously wrong to think this way: this is why I am looking for a better intuitive understanding. I must also say that I appreciate the efforts made by you and others for your attempts to elaborate. – Mark Watson Oct 04 '15 at 23:03
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    @Mark: Nobody says that $f$ is not allowed. The proof shows that if you give me a function $f\colon A\to\mathcal P(A)$, then I can define a set $B_f$ such that for every $a\in A$, $f(a)\neq B_f$. Since $B_f\in\mathcal P(A)$, it shows that $f$ wasn't surjective. – Asaf Karagila Oct 04 '15 at 23:06
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    I think it helps to imagine that $B_f$ does not exist until after $f$ has been defined (i.e. selected arbitrarily). This before and after relationship is not quite rigorous, but it does correspond to the order of the proof, and the order of the relevant dependencies. The idea is that, for each possible $f$, we can construct a $B_f$ counterexample, so $f$ cannot be onto no matter which one it happens to be. – Kevin Oct 05 '15 at 01:26
  • @Kevin: Do you really think the moon isn't there if you aren't looking at it? – Asaf Karagila Oct 05 '15 at 07:01
  • @AsafKaragila Was that in reply to a comment from Kevin that was deleted? It seems a non sequitur and insulting. – Thomas Andrews Oct 05 '15 at 14:00
  • @Thomas: No, it's actually a reply to the idea that you should imagine that $B_f$ does not exist until $f$ is chosen. And I didn't mean to be insulting in any way, I just quoted Einstein... :-) – Asaf Karagila Oct 05 '15 at 14:42
  • @Asaf: "the moon" (lowercase) only exists once you specify a planet. – Kevin Oct 06 '15 at 02:43
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Sets are the mathematical attempt to formalize the idea of an object which is itself a collection of other mathematical objects.

As such we expect collections to have certain behaviors.

For example, if I have a collection of marbles, I expect that the collection of all the white marbles also to be a collection. If I have a collection of bottles of liquor, I expect the collection of all bottles of whisky to be a collection as well (not to be confused with the collection of all the bottles of whiskey!).

This is formalized in the idea that if $A$ is a set, and $\varphi$ is a first-order property in the language of set theory, then all those elements of $A$ which satisfy the property $\varphi$ also forms a set.

Since properties might refer to other existing objects, this might allow parameters. And so if $f$ is a function we can write the property $\varphi(x)$ to be $x\notin f(x)$. So if $A$ is a set, then $\{a\in A\mid a\notin f(a)\}$ should also be a set.

This is known as comprehension, or rather bounded comprehension (also known as separation and subset). And this is the crux that allows us to ensure that $B$ is a set.

There are set theories like Quine's New Foundations, in which not every property defines a set. In that set theory there is, in fact, a universal set, and Cantor's theorem is not quite provable in its easy formulation because the restrictions on which properties define sets disallow the diagonal set to exist.

Asaf Karagila
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  • Heheh hello there! Honestly, if I have an infinite collection of marbles, I don't expect to be able to obtain the collection of all the white marbles, unless I also have infinite time or infinite speed. =P In some sense the Cantor theorem is not even about infinite sets if one views the given function as an oracle function, since the output of the proof is a finite object. Please don't whack me! =P – user21820 Oct 05 '15 at 01:30
  • See, the axiom of choice says agrees with you sort of and answers "yeah, obtaining white marbles takes infinite time so let's make it an axiom you can actually obtain all the white marbles". – chx Oct 05 '15 at 03:01
  • @user21820: I think you lost your marbles... :-P – Asaf Karagila Oct 05 '15 at 06:57
  • @chx: Well the axiom of choice is even more than what I was referring to. Technically if we view the collection of marbles as simply a decider (from computation theory), equivalently an indicator function on the set-theoretic universe, then whether or not it is infinite doesn't matter when forming subcollections, if the specifying property is decidable. However, from this perspective the top type is decidable but yet not a collection. But my actual point is that Cantor's proof is an effective one; given any map $f : S \to P(S)$, $(x \mapsto x \notin f(x) )$ decides a subset not in $Ran(f)$. – user21820 Oct 06 '15 at 04:57