October 9th, 2018, 03:40 PM  #191 
Senior Member Joined: Jun 2014 From: USA Posts: 479 Thanks: 36  It takes Cantor's argument to assert that the string of all 1's is not in the sequence. Otherwise, it could hit all 1's, reset back to 0's, and then keep going for all we know. This would be like going from $0.\overline{1}$ to $1.\overline{0}$ and then continuing down the real line towards $\infty$. 
October 9th, 2018, 03:48 PM  #192  
Senior Member Joined: Jun 2014 From: USA Posts: 479 Thanks: 36  Quote:
1 0 0 0 1 1 0 0 1 1 1 0 . . . Cantor's diagonal is 1 1 1 1 ... In the above patternbased sequence, Cantor's diagonal shows that there is no last element of the sequence perhaps. I've always viewed Cantor's argument that way to be honest. Reaching that last string would imply the end of an infinite sequence, so naturally we should be able to construct a theorem saying we can't do that. Personally, if the bolded elements result in the sequence 1 1 1 1 ..., then I don't see how we can claim that 1 1 1 1 ... isn't in the list without Cantor's argument. The moment we have enough items in our list to ensure that Cantor's diagonal is complete, we would simultaneously have enough elements in our list to know that the next element is 1 1 1 1 ... .  
October 9th, 2018, 06:22 PM  #193  
Global Moderator Joined: Dec 2006 Posts: 20,302 Thanks: 1972  Quote:
Cantor's diagonal exists if the original list's $n$th string has at least $n$ digits for each finite value of $n$ for which the list contains an $n$th string, and every string in the list has a finite position number in the list. There's no list that contains some strings that have an infinite position number rather than a finite position number, because the definition of "list" makes that impossible.  
October 11th, 2018, 04:30 AM  #194  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
Quote:
The set T, of all possible such strings, is denumerable if (A) you can define a function s(n) that maps every n in N to a string in T, and (B) you can prove that every string in T is mapped this way. If what you meant to say was "But you can't define a function s(*) that returns a function t(*) in T for a random n in N," then you are wrong:
Again, there is no n in N where this does not define an infinite string. This s(*) is not a complete listing of T of course, because that IS impossible. +++++ Your continued assertion "you can't do XXX for any n in N, because you can't get to the end of N" is not the absolute truth you seem to think it is. It is an axiom you choose to accept, but cannot prove. Cantor chose to accept the Axiom of Infinity, which says you can if XXX(1) is defined, and any XXX(n+1) can be constructed from XXX(n). Cantor's conclusion was that there are infinite sets which are not denumerable. It wasn't about R, but does apply to R. It is true in his mathematics. It may or may not be true in yours  you'd have to define that mathematics for us to tell. Oh, wait; N does not exist in yours, either. Which, ironically, means that you can't put R into a bijection with N. Gee, isn't that what you think Cantor said?  
October 11th, 2018, 05:11 AM  #195  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
Accepting the gauntlet thrown, here is Cantor's proof in modern language. You can compare it to the text at Cantor's Diagonal Argument. If zylo still can't understand it, he can't claim to disprove it. But he can ask for a clarification, and I will provide it. I will use the characters '0' and '1' instead of 'm' and 'w', and change some names.
+++++ Notes:
Last edited by JeffJo; October 11th, 2018 at 05:35 AM.  
October 11th, 2018, 05:21 AM  #196  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
You provided a wellformed definition, that can identify the character at any position n in N. Zylo would call this a "random position," and seems to include "n=infinity" as a random selection that we must reach, but obviously can't. I have no hope that we ever will, but to change his mind we need to dispel that notion.  
October 11th, 2018, 05:28 AM  #197  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
That doesn't mean that we can't define the endless string "333...". We can. As you are fond of pointing out, its end cannot ever be reached, so it does not represent a natural number. Just like "infinity" or "aleph0" are not natural numbers.  
October 15th, 2018, 11:11 AM  #198  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
0 . 3 . 9 10 . 33 . 99 100 . 333 . 999 1000 . 3333 . 9999 .. 100000....... .. 333333...... .. 999999........ The last segment is simply symbolic for infinity where you are going but never get to. Note 999999....... ("Cantor String")  
October 15th, 2018, 12:18 PM  #199 
Global Moderator Joined: Dec 2006 Posts: 20,302 Thanks: 1972 
Every string in that list you quoted contains "endless" zeros, so any other endless strings are missing.

October 15th, 2018, 02:38 PM  #200  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
Cantor's Diagonal Argument Assume the set of all infinite binary sequences can be enumerated: S1) 0 0 1 0 1 1 0...... S2) 0 1 0 0 1 0 0....... S3) 1 1 0 1 0 0 1...... .............. CDS = 1 0 1............ CDS is an infinite binary sequence that is different from every member of the enumeration and therefore not in the enumeration. Contradiction. Therefore The set of infinite binary sequences is not enumerable, PROBLEM CDS has no predecessor in the list and therefore can only be the last member of the list, which doesn't exist because the list of infinite binary sequences has no last member. There is no contradiction and the sequence of infinite binary strings is enumerable.  

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binaryexpressed, cardinality, continuum hypothesis, diagonal argument, numbers, real, set 
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