I don't understand the way cantor's diagonal argument work. I suppose the origin of the problem is an assumption that I think must be made for the proof to be correct. Personally I really thinks that cardinality of R is greater than N. The problem is in the proof.
In every cantor diagonal proof we have the following: 1) A "complete" list of reals. 2) a real sk defined this way: sk[n] = 0 if a[n][n] = 1 or sk[k] = 1 if a[n][n] = 0.
if the list is complete, sk must be in the list. So by assumption there must be a n where sk[k] = 0 if a[k][k] = 1 and sk[k] = 1 if a[k][k] = 0; This number is not correctly defined since it is not equal to itself. So the definition of sk is not a real if we assume that the list is complete and we assume that any real is equal to itself.
does it make sense to take an incomplete list of reals? I personally think if it is not complete it is useless to demonstrate that you have a real that is not in the list since by definition the list is incomplete. for example 1) lets take the subset of natural numbers of the real set as "the list of reals" 2) lets define a real sk as a fraction for example "sk = 0.5".
you can demonstrate that sk is not in the set but the set of rational numbers is countable so the set of "natural numbers with 0.5" must also be countable.
First thanks a lot for all your comments. I'm sorry if I have been so "informal" expressing my point of view.
My main concern with the proof is that for me it seems that the definition of the mathematical object sk (the real number that is not in the list) is self contradictory. And from a contradiction we can deduce everything in classical logic.
first we start from this assumptions:
- Q="L is a complete list of the reals".
- P1="sk is a member of the real set".
We want to demonstrate this sentence F="sk is a member of the list L". Without any other assumption (only P1 and Q) F follows from Q and P1.
Now we are going to add some structure (property P2) to the number sk. First I'm going to explain two suppositions to introduce my idea.
P2="sk equals 2".
P1 and P2 are true. The definition of sk is no contradictory with Q. In this case F is true.
P2="sk equals i".
P1 and P2 are not true since sk can not be defined has a real and a complex. In this case The definition of sk is at least ambiguous. We trust in P1 so P2 is false. Q still true and F is true since the only property that holds for sk is our first assumption P1.
Suposition3: (cantor diagonal proof)
We should remember that we trust in P1 and in Q so sk is a real and it is in the position k in the list. lets express sk like s[k] to be able to express the other real numbers of the list like a[n] (double index a[n][n] represents the n digit of the n number). Now we want to add this property to sk:
P2="s[k][n] equals 0 if a[n][n]=1 and equals 1 if a[n][n]=0". For n != k there is no problem. But for the k index we could express P2 in this terms "s[k][k] is not equal to s[k][k]". This is in contradiction with P1 since all real numbers are equal to themselves. like in our previous example (with complex property) since we trust in P1 then P2 must be false. Q is still true and F is true since the only property that holds for sk is our first assumption P1.
Now lets take it the other way. We start from this assumptions:
- Q="L is a complete list of the reals"
- P2="sk is a mathematical object with this structure: sk[n] equals 0 if a[n][n]=1 and equals 1 if a[n][n]=0"
From this assumptions, F (sk is in the list of reals) is not obviously true since we have first to demonstrate that sk is a real in the realm where Q and P2 holds.
We want to add the property P1: "sk is a member of the real set". from P1 and Q it follows that sk is in the list so sk has a position k in the list. Again if P1 and P2 are true then s[k][k] is not equal to s[k][k]. So in a realm where Q and P2 hold sk can not have the property P1. Since sk can not be a real the list is complete without sk so Q is still true and F (sk is in the list) is simply false.
It is important to understand that P2 can only be expressed in a realm were Q is taken as an assumption since P2 needs the list of reals to be correctly defined. This is probably one of the confusion of this issue since it seems that P2 is simply expressing a binary infinite sequence (so a real number) but the construction is dependent of the existence of Q. So if an object exist with P2 it must be in a realm where Q holds and a complete list exist.
We can resume all this feelings like this:
- If we belief in Q we can express and belief in P2 but an object that have the property P2 has not the property P1. (an object with property P2 is not a real in Q's realm).
- If we belief in Q and we belief in P1 an object that have the property P1 can not have property P2. (a real can not have P2 has a property in Q's realm).
- If we don't belief in Q we are not able to express P2 since we need the complete list of reals to define P2.
The only way to refute Q (is what Cantor Diagonal proof do) is to assume that there exist an object with both properties in the realm where Q holds. When I informally said that in the cantor diagonal proof the number sk is defined has "I'm a real not equal to myself" I wanted to express the fact that in a realm where Q holds there is no mathematical object that have both properties P1 and P2.
Another way of seeing this is: Since in the realm where Q holds, P1 express that sk is a real but P2 express that sk is not a real, an object with both properties (P1 and P2) in this realm seems to be a contradiction (a paradox) and we can derive anything from a paradox.
Probably what is so confusing about this issue is that it seems to express the idea than an object that is build from a complete list of reals is itself not a real.
That is the reason why in my first attempt to express this issue I was concerned with the completeness of the list. if the list is not complete P1 and P2 can both be true but I think that in that case the diagonal argument does not prove what it wants to prove.