A new point of view on Cantor's diagonalization arguments

In summary, the conversation is discussing a new perspective on Cantor's diagonalization arguments and thanking individuals for their contributions. The conversation also delves into the topic of alephs and the differences between conventional mathematics and the speaker's own system. The speaker's system claims to be more expansive than Cantor's transfinite universes, with different relationships between aleph0 and 2^aleph0. The conversation also touches on the concept of magnitude in relation to the binary tree representation.
  • #176
I don't see the arguement

Cantor's Diagonalization Argument

Theorem-The interval [0,1] is not countably infinite.
Proof:-The proof by contradiction proceeds as follows:

Assume (for the sake of argument) that the interval [0,1] is countably infinite. We may then enumerate all numbers in this interval as a sequence, ( r1, r2, r3, ... ) We already know that each of these numbers may be represented as a decimal expansion. We arrange the numbers in a list (they do not need to be in order). In the case of numbers with two decimal expansions, like 0.499 ... = 0.500 ..., we chose the one ending in nines. Assume, for example, that the decimal expansions of the beginning of the sequence are as follows:

r1 = 0 . 5 1 0 5 1 1 0 ...
r2 = 0 . 4 1 3 2 0 4 3 ...
r3 = 0 . 8 2 4 5 0 2 6 ...
r4 = 0 . 2 3 3 0 1 2 6 ...
r5 = 0 . 4 1 0 7 2 4 6 ...
r6 = 0 . 9 9 3 7 8 3 8 ...
r7 = 0 . 0 1 0 5 1 3 5 ...
...
We shall now construct a real number x in [0,1] by considering the kth digit after the decimal point of the decimal expansion of rk.

r1 = 0 . 5 1 0 5 1 1 0 ...
r2 = 0 . 4 1 3 2 0 4 3 ...
r3 = 0 . 8 2 4 5 0 2 6 ...
r4 = 0 . 2 3 3 0 1 2 6 ...
r5 = 0 . 4 1 0 7 2 4 6 ...
r6 = 0 . 9 9 3 7 8 3 8 ...
r7 = 0 . 0 1 0 5 1 3 5 ...
...

The digits we will consider are indicated in bold and illustrate why this is called the diagonal proof. From these digits we define the digits of x as follows.
if the kth digit of rk is 5 then the kth digit of x is 4
if the kth digit of rk is not 5 then the kth digit of x is 5
For the example above this will result in the following decimal expansion.

x = 0 . 4 5 5 5 5 5 4 ...

The number x is a real number (we know that all decimal expansions represent real numbers) in [0,1] (clearly). Hence we must have rn = x for some n, since we have assumped that ( r1, r2, r3, ... ) enumerates all real numbers in [0, 1]. However, because of the way we have chosen 4's and 5's as digits in step (6), x differs in the nth decimal place from rn, so x is not in the sequence ( r1, r2, r3, ... ). This sequence is therefore not an enumeration of the set of all reals in the interval [0,1]. This is a contradiction.

Hence the assumption that the interval [0,1] is countably infinite must be false.

Q.E.D.

It is a direct corollary of this result that the set R of all real numbers is uncountable. If R were countable, we could enumerate all of the real numbers in a sequence, and then get a sequence enumerating [0, 1] by removing all of the real numbers outside this interval. But we have just shown that this latter list cannot exist. Alternatively, we could show that [0, 1] and R are the same size by constructing a bijection between them.
 
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  • #177
Cantormath,

Welcome to Physics Forums! And you're right, there is no argument.

What you've stumbled upon here is a piece of crackpottery from the old days when we adopted an "anything goes" attitude in the Theory Development Forum. We've since tightened things up so that we only allow things that make sense. :smile:

So feel free to stick around and enjoy the Forums. Don't worry about this thread, because the orginal poster isn't even here anymore.
 

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