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October 9th, 2018, 06:34 AM   #181
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The natural numbers don't end.
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October 9th, 2018, 06:39 AM   #182
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Quote:
Originally Posted by zylo View Post
333....... occurs in the list of natural numbers no matter how many digits you have because the list of natural numbers doesn't end.
Each string in that list is, by definition, identified by a finite integer, whereas the non-terminating string 333....... isn't. As zylo has just posted, the list of natural numbers doesn't end, and that's true even if non-terminating strings are explicitly excluded, so there's nowhere where non-terminating numbers can exist in that list, and there is no "end" of that list after which they could be added.
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October 9th, 2018, 07:13 AM   #183
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It/s really quite simple

Quote:
Originally Posted by zylo View Post
ZA!: All natural numbers are on a list increasing by 1. (Peano)

ZA2: Any real number [0,1) is on the list because it is a natural number with a radix point.

That serves for all of math except set theory, which is really the mathematics of words and symbols, which can only be defined in terms of words and symbols.
A real number is one that can be specified to any number of digits.

I can calculate (specify) sqrt2 to any number of digits. That's the definition of a real number. The end of the calculation doesn't exist.

The real numbers only exist to arbitary n digital places. To the extent they exist they can be counted.

Hence my remark that you can't count a set of things that don't exist.

EDIT
This thread exhibits the futility of dealing with the real numbers beyond the above quote.

Last edited by zylo; October 9th, 2018 at 07:33 AM.
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October 9th, 2018, 08:46 AM   #184
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So Zylo is a finitist and Cantor’s argument doesn’t apply. Glad we cleared that up.
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October 9th, 2018, 09:00 AM   #185
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Quote:
Originally Posted by zylo View Post
This thread exhibits the futility of dealing with the real numbers beyond the above quote.
No, it exhibits the futility of trying to help you actually understand something.
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October 9th, 2018, 09:26 AM   #186
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Quote:
Originally Posted by zylo View Post
A real number is one that can be specified to any number of digits. . . . This thread exhibits the futility of dealing with the real numbers beyond the above quote.
There are some details of real numbers that can be confusing to some people. Cantor's diagonalization of a list was done using an arbitrary list of non-terminating binary sequences (strings), without mentioning reals.

Most non-terminating sequences (or reals) don't have any convenient definition, but that doesn't mean that they don't exist. It's not required that there is a known method of determining their digits.

Any list of strings can be diagonalized if each string has a position in the list that can be specified by a finite integer. Without that restriction, it isn't possible to prove that the list is countable. That restriction is therefore crucial. It wasn't met by your list, as the non-terminating string 333..... can't have a position identified by a finite integer.

Diagonalization allows a string to be specified by reference to the list in such a way that it can't be in the list. This means that knowing the content of the list is irrelevant. It doesn't even matter whether two strings in the list are the same. It doesn't matter if nobody knows what any of the strings are. All that matters is that a finite integer suffices to denote the position of each digit in a string, regardless of whether the value of that digit is known or defined in some way, and that a finite integer suffices to denote the position in the list of each string, regardless of whether the digits in the string are known or defined in some way.

Quote:
Originally Posted by zylo View Post
A real number is one that can be specified to any number of digits.
That's incorrect. It's also irrelevant in relation to diagonalization.
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October 9th, 2018, 09:40 AM   #187
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To 2 places:
0'0
10'
01
11
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October 9th, 2018, 10:16 AM   #188
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Quote:
Originally Posted by zylo View Post

A real number is one that can be specified to any number of digits.
That sounds more like a computable number: https://en.wikipedia.org/wiki/Computable_number

Quote:
Originally Posted by zylo View Post
I can calculate (specify) sqrt2 to any number of digits. That's the definition of a real number.
That's not the definition of a real number, no. Again, with a computable real number such as $\sqrt{2}$, you can calculate it out to however many digits you want.

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Originally Posted by zylo View Post
The end of the calculation doesn't exist.
Perhaps I'm being bias, but I thought my example was a decent way of contemplating this. You didn't seem interested and just kept repeating the same old stuff as though, if you said it enough, it would be true.

I asked you if $\bigcup B = \bigcup B'$. I was basically asking if you think there is an element of $f(k)$ that doesn't exist in $\bigcup f([0,1)_2)$. It's a logical proposition because $f(k)$ is unique and $f([0,1]_2)$ is pairwise almost disjoint. The only problem is that we can't solve for whatever element of $f(k)$ isn't in $\bigcup f([0,1)_2)$ because such an element would be the 'last' element of $f(k)$. I then showed a couple of funky results if we accept that interpretation, such as the reals being countable and the ability to show a difference in cardinality between sets that in fact have equal cardinality. So, either there is a problem with induction and self-referential arguments, or we can't assume that there is an element of $f(k)$ that isn't in $\bigcup f([0,1)_2)$.

Quote:
Originally Posted by zylo View Post
The real numbers only exist to arbitary n digital places. To the extent they exist they can be counted.
You keep using the term "digital" in a weird way. Anyways, real numbers are independent of their binary, ternary, decimal, etc., representations. Where $\frac{1}{3} = 0.1_3 = 0.\overline{3}_{10}$, how can you assert that $\frac{1}{3}$ only exists to "n digital places" (whatever that means)?

Quote:
Originally Posted by zylo View Post
Hence my remark that you can't count a set of things that don't exist.
Why not? I can count the number of unicorns in my office right now (zero).

The set of natural numbers doesn't 'exist' outside of the theoretical, so if the very set we're using to do the counting doesn't exist, why can't we use it to count other sets that don't exist?

Quote:
Originally Posted by zylo View Post
This thread exhibits the futility of dealing with the real numbers beyond the above quote.
In my example, Cantor's diagonal equated to what would have been the last element of an infinite sequence. You asserted that the sequence didn't end while simultaneously asserting that Cantor's diagonal appeared in the sequence (a contradiction). I tried to expand on this by asking how $f(k)$ could possibly be a subset of the union of $B$ when Cantor's very argument assures that it is different from every element of $B$. I personally could view the proof by induction that $f(k) \subset \bigcup B$ as being flawed, but when I do, the results are inconsistent so, in my view, we are choosing between complete and inconsistent versus incomplete and consistent. What do you think about that last sentence of mine Zylo (anybody)?
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October 9th, 2018, 11:53 AM   #189
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Quote:
Originally Posted by zylo View Post
To 2 places:
0 0
1 0
0 1
1 1
So you want to see CDA with infinite binary sequences?:

0) 0 0 0 0 0 0 0 0 ...........
1) 1 0 0 0 0 0 0 0 ...........
2) 0 1 0 0 0 0 0 0 ...........
3) 1 1 0 0 0 0 0 0 ...........
4) 0 0 1 0 0 0 0 0 ...........
5) 1 0 1 0 0 0 0 0 ...........
6) 0 1 1 0 0 0 0 0 ...........
7) 1 1 1 0 0 0 0 0 ............
......................................

Cantor's diagonal string is 1 1 1 1 1 1 1 1 ....... which is the last number in a sequence of natural numbers.
So CDA proves that a sequence of natural numbers has no last member. I knew that.
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October 9th, 2018, 12:06 PM   #190
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Quote:
Originally Posted by zylo View Post
Cantor's diagonal string is 1 1 1 1 1 1 1 1 ....... which is the last number in a sequence of natural numbers.
That's incorrect. There's no such thing as a last natural number. If your particular list doesn't contain all natural numbers, the last one that it does contain cannot possibly contain just '1's and no '0's.
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