My Math Forum Cantor's Diagonal Argument. Infinity is Not a Number

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 March 24th, 2016, 09:32 AM #41 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 For any enumeration of a countably infinite set of non-terminating binary sequences, Cantor used a "diagonal" construction to produce a non-terminating binary sequence s that is not in that enumeration. If you think that part of his argument contains a flaw, explain in this thread what the flaw is.
March 24th, 2016, 10:12 AM   #42
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Quote:
 Originally Posted by skipjack If you think that part of his argument contains a flaw, explain in this thread what the flaw is.
Define infinite as more than countably infinite. Endless doesn't rule out countably infinite.

Let T be the set of all infinite binary sequences. T is obviously uncountable, so you can't assume, as Cantor does, that it is countable.

Given: T={1,2,3,4}
Assume T={1,2,3} -> 4 does not belong to T
Conclusion T={1,2,3.4}

If you assume infinite means countably infinite, Cantor's argument leads to the conclusion that the natural numbers are uncountable.

Last edited by skipjack; March 24th, 2016 at 01:10 PM.

 March 24th, 2016, 10:25 AM #43 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 That doesn't identify a flaw in the part of Cantor's argument that I specified, in which the enumeration considered is countably infinite by definition, and each binary sequence is countably infinite by the definition of a non-terminating sequence. Can you identify a flaw in that particular part of Cantor's argument?
March 24th, 2016, 01:38 PM   #44
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Quote:
 Originally Posted by skipjack Can you identify a flaw in that particular part of Cantor's argument?
Cantor's s is infinite. Infinity is not a natural number.

Last edited by zylo; March 24th, 2016 at 01:51 PM. Reason: add: first sentence.

 March 24th, 2016, 02:34 PM #45 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 Given that the sequence s is intended to be infinitely long, as is each of the sequences used, why is that a flaw? Last edited by skipjack; March 25th, 2016 at 05:46 AM.
March 25th, 2016, 03:35 AM   #46
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Quote:
 Originally Posted by zylo Cantor's s is infinite. Infinity is not a natural number.
So what? S is a sequence, not a number.

March 25th, 2016, 10:53 AM   #47
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Quote:
 Originally Posted by skipjack Given that the sequence s is intended to be infinitely long, as is each of the sequences used, why is that a flaw?
Correct. Herewith the flaw:

The argument of post #42 holds no matter how you interpret infinite.

Given an infinite binary sequence, let the last non-zero binary digit be in the nth place. Then this sequence is a natural number.

It follows that all the sequences with finite n are a countable subset of the set of all infinite binary sequences.

The argument of post #42 now follows. You can't assume T is countable to arrive at a contradiction. Given T={1,2,3,4}, you can't assume T={1,2,3}.

EDIT:
The above seems to prove that the set of all infinite binary sequences is uncountable, since the natural numbers are a subset. Without Cantor's diagonal argument. The resolution depends on the answer to the question:

Is a countably infinite binary sequence always a natural number?

If the answer is yes, and I suspect it is, then "Cantor's s is infinite. Infinity is not a natural number" may still be valid.

Last edited by skipjack; April 4th, 2016 at 11:40 AM.

March 25th, 2016, 12:43 PM   #48
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Quote:
 Originally Posted by skipjack Given that the sequence s is intended to be infinitely long, as is each of the sequences used, why is that a flaw?
Let T be the set of all countably infinite binary sequences.
Assume T is countably infinite and list all members:
There are countably infinite many digit places.
There are always more rows than digit places:
0 0
1 0
0 1
1 0
for example.
s has a digit for each row.
Therefore s has more than countably infinite many digits and is therefore not a member of T and there is no contradiction. Cantor's argument fails.

EDIT
This replaces previous post.

Last edited by skipjack; April 4th, 2016 at 11:41 AM. Reason: add countably and binary

March 25th, 2016, 01:00 PM   #49
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Quote:
 Originally Posted by zylo Herewith the flaw: The argument of post #42 holds no matter how you interpret infinite.
I asked you to identify a flaw in the first part, which I described, of Cantor's proof. In post #42, you are not considering that part of Cantor's proof. You're not identifying a flaw in Cantor's argument if you say "I'll do some thing similar to, but not the same as what Cantor did, and draw attention to some consequence of what I've done (which isn't what Cantor did)." That's effectively what you did.

Quote:
 Originally Posted by zylo Given an infinite binary sequence, let the last non-zero binary digit be in the nth place. Then this sequence is a natural number.
If, say, n = 1, is the natural number you're referring to 1, 10, 100, 1000 or what?

Quote:
 Originally Posted by zylo It follows that all the sequences with finite n are a countable subset of the set of all infinite binary sequences.
They are countable, but why is that worth demonstrating? In the part of Cantor's argument that I'm referring to, he specifies that he is considering a countable set of binary sequences, but they aren't necessarily sequences that contain only a finite number of non-zero digits.

Quote:
 Originally Posted by zylo The argument of post #42 now follows. You can't assume T is countable to arrive at a contradiction. Given T={1,2,3,4}, you can't assume T={1,2,3}.
You're now clearly going beyond the scope of my request. I referred to the first part of Cantor's proof. In that part, Cantor isn't making the supposition you mention and isn't obtaining a contradiction; he is just constructing a sequence s that is derived from the sequences he's using and is different from every one of them.

 March 25th, 2016, 02:57 PM #50 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Post #47 is obsolete- replaced by my previous post, #48. A countably infinite binary sequence exists for every value of n and is therefore always a natural number. Therefore the set of all countably infinite binary sequences is a set of natural numbers and therefore countable. The flaw in Cantor's proof to the contrary is given in my previous post, #48. _______________________________________ 0 1 1 0..... =0x2^0+1x2^1+1x2^2+........

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