Just reading Terence Tao's book. If we consider Cantor's theorem which states that given any set X, there does not exist any bijection between that set and its power set, P(X).

But why should we accept this theorem philosophically? To me, this seems quite reasonable: Take a ball. Cut the ball into pieces. From the pieces, we can construct the ball again.

I understand the proof. It is quite unsettling philosophically, though.


closed as unclear what you're asking by Alex Kruckman, Lord Shark the Unknown, Aqua, José Carlos Santos, cmk Jul 10 at 12:01

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 8
    $\begingroup$ I do not see the connection between the stuff about the ball and Cantor's theorem. $\endgroup$ – spaceisdarkgreen Jul 4 at 3:54
  • $\begingroup$ @spaceisdarkgreen please see The_Sympathizer he is exactly has what I have in mind. $\endgroup$ – Learner Jul 4 at 4:10
  • 1
    $\begingroup$ Do you believe the theorem for finite sets? Try it out for infinite sets using the positive integers as a test case. Each subset of the positive integers corresponds to a unique sequence of 0's and 1's and the power set has the same cardinality as the set of all sequences of 0's and 1's. Is this set countable? I think you can use a Cantor diagonal argument to show it cannot be. $\endgroup$ – Chris Leary Jul 4 at 4:10
  • 3
    $\begingroup$ If you understand a proof but think you nonetheless have a counterexample, the course of action is clear: apply the proof to the specific case of your counterexample. You will either find an error in the proof or discover your counterexample is not actually a counterexample. We accept the theorem because we accept the axioms it is based on and accept the reasoning from those axioms as valid. Any refutation of the theorem will need to result in finding a flaw in the argument. $\endgroup$ – spaceisdarkgreen Jul 4 at 4:13
  • 1
    $\begingroup$ I just wanted the way of looking at it from the right angle. The_Sympathizer provided this way. I wanted to be satisfied philosophically though. Not only based on the axioms. $\endgroup$ – Learner Jul 4 at 4:13

I think I see what you're getting at. It is this: a ball contains an infinite number of points, but if we dice it up into pieces, which are subsets of those points, we will always have fewer pieces than we do points in the ball. At most, we can atomize it down into an infinitude of pieces, each of which are the single, individual points constituting said ball. Hence, we can only divide the ball into no more subsets than there are constitutive points, and so why should we say that the power set - the set of all subsets - contains more pieces than there are points in the ball?

The problem with that reasoning is this: The power set doesn't just contain subsets obtained by a specific cutting-up of the ball. It contains all possible subsets you can make by every possible cutting up of the ball. Imagine having infinite copies of the ball, or reassembling the ball perfectly after a cutting, and then cutting them or it up over and over and over in every conceivable way you can, and beyond, and keeping, or keeping track of, every piece you can form, separately. That is, it doesn't just, say, contain the division of the ball across its diameter planes into halves, or into quarters, or any sequence of pieces formed by repeated such divisions. It also contains, say, pieces formed by stripping it like an onion. Pieces formed by cutting letters and hearts out of it. Pieces formed by cutting thin slivers of various kinds. In fact, these are actually the tamest pieces: the vast majority of ways look like clouds of infinitely fine dust - effectively, set-theoretic static.

And it's a general principle of things (this, I suppose, you could class as "philosophical", as it's not definable formally, at least I wouldn't know how) that "there are many more ways to make something that look like noise, than there are to make something that look like signal". For example, it's mind-bogglingly more likely that if you take a bitmap, and fill it up with random pixels, it will look like static, versus its looking like a legible photograph. It's overwhelmingly more likely that if you shuffle a card deck, it'll end up with the cards in no legible or easily-describable order. It's overwhelmingly likely that if you strike a weak building with a tornado, it becomes a jumbled mess of shattered wood and fasteners, than another, neatly-constructed building. It's "far easier to destroy than create". This is also the reason behind the physical law of increasing entropy (the second law of thermodynamics).

And each different instance of a frame of static, gets its identity from the exacting details of every single pixel. There isn't any other simpler way to talk about them than to just painstakingly specify each and every one (formally, this is called "Kolmogorov randomness"). Likewise, here, the sets in $P(X)$, where $X$ is a ball, are overwhelmingly formed by the exacting details found by taking now literally infinite pains to specify, one by one, the presence or absence of every single one of its infinitude of points, with no other simpler description available.

Intuitively, do you now feel that there must be a LOT of sets in $P(X)$? Perhaps even more than in $X$ itself, even if $X$ is infinite, i.e. $P(X)$ is "even more infinite"?

  • $\begingroup$ Yeah exactly this is precisely what I have in mind in regards to the ball argument. $\endgroup$ – Learner Jul 4 at 4:10
  • $\begingroup$ @Learner : thanks! :) #mehhr. Do you find this answer useful? $\endgroup$ – The_Sympathizer Jul 4 at 4:11
  • $\begingroup$ Yeah very useful thanks for the insight. $\endgroup$ – Learner Jul 4 at 4:12
  • $\begingroup$ @Learner : you're welcome :) ❤ #Meep $\endgroup$ – The_Sympathizer Jul 4 at 4:13
  • 3
    $\begingroup$ The first paragraph is an argument for accepting the axiom of choice. $\endgroup$ – Asaf Karagila Jul 4 at 8:41

Not the answer you're looking for? Browse other questions tagged or ask your own question.