I think it is still intuitively surprising that some infinite sets somehow have more elements than other infinite sets. The powerset operator is very special because it can create this difference.
That assumption is unnecessary. Cantor's theorem works just as well for finite sets: it's simply the statement that 2ⁿ > n.
Edit: Actually maybe this is just the first concern mentioned :/.
Edit2: Maybe clearer explanation: you can produce an infinite sequence of digits from many real numbers, but it doesn't seem to me to be valid to consider that infinite sequence to actually be the real number.
When we talk about "an infinite sequence of digits", what are we talking about? Are we talking about ACTUALLY having an infinite sequence of digits in front of us? Or are we talking about having a RULE that will in theory produce them?
Platonists believe that they actually exist in a reality beyond human conceptual limits. (Maybe in the mind of God?) Formalists side step the question by making everything an abstract symbol manipulation game. (Though the rules of the game are chosen to result in something matching platonic intuition.) And Constructivists take the rule approach, and not just that, but a rule that we can write down and actually evaluate.
Cantor's diagonalization argument obviously works according to Platonists. Formalists choose their rules to make it work as well. But Constructivists, well, it doesn't work there!
Why not? Well when you get down to it, how do you define a rule and how do you prove that it works? Well, a "rule" can be thought of as just a computer program. And now we have the problem that, thanks to the Halting Problem, it is impossible in general for one computer program to predict what another program will do!
You can, of course, attempt to write Cantor's number down. You can implement the rule that he describes using a theorem prover to filter a list of all programs to just the ones that we can prove are numbers. We can write a perfectly well-defined program that produces Cantor's number. BUT THERE IS NO CONTRADICTION?
Why not?
Because THAT program is one that the theorem prover CAN'T prove works! (In fact the theorem prover will prove what it is doing, and prove that if it always works, then its set of axioms is consistent. And now you're up against Gödel's Incompleteness Theorem!)
The result is that in the two most popular philosophies of math, Cantor's argument works. But in the third, it doesn't, it can't, and the reals are countable. There the complications of diagonalization are simply another illustrations of the challenges of self-reference, and not the assertion that more numbers exist than can be written down!
Isn't that a bit of an overreach? It seems that in Constructivism, Cantor's diagonalization doesn't work as a proof. But that doesn't mean that the reals are countable, just that this proof isn't valid.
However, to understand that, we need to answer several questions:
* What is a real number, after all? How do we even define it?
* When we have a sequence of digits like 0.12345678..., and when we say this sequence represents a number X, what exactly does that mean?
* (...which leads to): What is a sequence? What does it mean for an infinite sequence to converge?
So, it's definitely not "Duh, it's obvious": some work is needed to understand all this. I recommend reading the first several chapters of any good book on analysis.
What's the last number you've written after two minutes? Is it the last digit of pi?
http://math.stackexchange.com/questions/409658/can-every-rea...
1 = 0.9999999999..... for example
Definitions: 2 is the set {0,1}. N is the set of natural numbers {0,1,2,...}. An infinite sequence of 0's and 1's c_0,c_1,c_2,... can be thought of as a function c : N -> 2 (that is, the values in a sequence are a function of their position: c_i = c(i)). A set X is countably infinite if there is an invertible function f : N -> X (which, in other words, is a sequence f_0,f_1,f_2,... where each element of X appears exactly once). Let's write A -> B for the set of functions from A to B, rather than the usual B^A.
Suppose for sake of contradiction (N -> 2) is countably infinite, so there is an invertible function f : N -> (N -> 2). Let F : (N -> 2) -> N be the inverse, so I can save having to type f^-1. Let g : N -> 2 be defined by g(n) = 1 - f(n)(n). Then, g(F(g)) = 1 - f(F(g))(F(g)) = 1 - g(F(g)), so g(F(g)) must be 1/2, contradicting the fact that it ought to be 0 or 1 instead.
* Apply Downward Lowenheim-Skolem to get a countable model of set theory
* Construct the reals using whichever technique you want.
* Since the base set theory is countable, so is the set of constructed reals.
The same technique can give you countably many groups, countably many rings, countably many points in the plane, etc. Everything is countable.
It's a rather limited edge-case in mathematics, it's misleading to simply assert that "everything is countable."
Any particular theory, capturing some aspect of the reals, can be given a countable model. Those "constructed reals" are countable, but they miss out almost all of the reals.
Think about it this way: a theory is like an interface, a theory of the reals provides an API to access the reals. The API is necessarily limited; not least because there are only countably many ways we can combine the operations.
Now consider a mock implementation of that interface: rather than passing around reals, we invent some (countable) dummy objects to use instead. Since the API is limited, we can always come up with a mock implementation which cannot be distinguished from the real implementation using that API/theory.
That doesn't mean that the reals are countable. It does mean that uncountability is an implementation convenience; regardless of what result we're after, we could get it using countable objects, but we might have to spend a lot of effort "out smarting" the algorithms we're using (i.e. coming up with our "mock" set).
cantors diagonalization argument is proof of that. you can't pull some silly trick to make them countable. there are many properties of R that are countable, but that doesn't make R itself countable.
Here is a definition of the reals to consider. A real number is a computer program which implements a function f from positive integers N to the rationals such that |f(n) - f(m)| < 1/n + 1/m. This is a reasonable definition of real numbers, and all real numbers that you're likely to hear of are real numbers under this definition.
But now consider. There are a finite number of symbols that we build programs out of. Therefore for every N, there are a finite number of possible computer programs of length N. Only some of which represent real numbers under this definition. Therefore there is a countable sequence which contains all real numbers in it!
What is the key philosophical difference between this system of mathematics and the usual one? Quite simply this. Classical mathematics asserts the existence of things that cannot be written down or calculated by humans. This system of mathematics denies the existence of things which we cannot write down.
Now about a hundred years ago there was a major debate between these two philosophical camps. In the end the classical school won simply because most mathematicians don't care about philosophy, and classical mathematics is easier to work with. But the purportedly inescapable logical conclusions that you're taught about are not actually as inescapable as you've been lead to believe.
If we open a book on set theory, we will find a proof of Cantor’s theorem which shows explicitly that for every map e : A → P(A) there is a subset of A outside its image, namely S = { x ∈ A ∣ x ∉ e(x) }
What about e(x) = { x }? In that case S would be the empty set, wouldn't it? Does map imply some additional properties in this case? Am I misunderstanding the statement and it does not try to imply that S is non-empty for every e?
Consider decimal numbers between 0 and 1 in binary.
Here's how I am going to synthesize this set.
Step 1: Take non-decimal binary numbers and consider them to be padded with an infinite of zeros at the left.
...0000000
...0000001
...0000010
...0000011
...0000100
...0000101
...0000110
..........
Do we agree that this will contain all the non-decimal non-negative binary numbers?
In particular, is the following number in the above set? ...111111 (all ones, not zero-padding on the left). If not, why not. And if not, this seems to be a matter of definition to me. If yes, move on.
Step 2: Place a decimal at the end of each line above and flip left and right.
0.0000000...
0.1000000...
0.0100000...
0.1100000...
0.0010000...
............
Do we agree that this contains all the decimal positive binary numbers between zero and one: [0, 1).
Let's now apply diagonalisation on this.
It says that the number 0.11111111... will not be present in the above set.
Perhaps someone can see my confusion and enlighten me. :-) Thanks!
Then you flip them over to the other side,in an operation that you yourself do not fully understand.
The first number that you flip (0.1) turns into 0.5 decimal
The second number that you flip (0.01) turns into 0.25 decimal
The third is 0.75
0.125
and so on. You are creating a subset of the rational numbers, which is obviously countable. It is countable because you constructed it to be countable. You constructed something that was countable and then tried to prove that it is countable.
Happens all the time :)
I did understand before how flipping changed the numbers into 0.5, 0.25, etc., but had missed that this process would only create rational, thereby missing the irrationals altogether. There's enough food for thought for me now. :-)
Maybe another base will help. In base 3:
0.0... 0.10... 0.20... 0.010...
Once again, a naive diagonalization yields 0.1... but this time at least we have not extended past 1. Is there a way to avoid that convergence? Sure, we can change the 0 digit to either 1 or 2 randomly, or based on some pattern or enumeration, instead of just 1. So now we can take many diagonals, and they all look like:
0.121212... 0.122122... 0.121121... 0.22212221... ...
Augh! What happened? Our diagonalization appears to have revealed an infinity of missed reals! This is similar to the construction of the Cantor set (https://en.wikipedia.org/wiki/Cantor_set) and hopefully illustrates the problem with your enumeration of the reals.
Step 2 fails because you can't create a 1-1 correspondence between the real numbers between zero and one, and the integers. That is what you are trying to prove. I'm too tired to figure out if your proof is valid or not, but your conclusion is correct.
Make up your mind.
decimal: Not base 10, but fractional, those with a decimal point binary: Not just 0 or 1, but base 2.
0.01 is a binary positive decimal number which is 0.25 in base 10.
It depends on which majority is at question. For that majority that is irrational, we can say any member can be approximated arbitrarily closely by a rational number.
These are also merely countable...
