My Math Forum Help me to understand Cantor please

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March 13th, 2016, 04:19 AM   #21
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Quote:
 Originally Posted by Jimbo . . . I can pull the same trick for any infinite set - including the Natural numbers, . . .
When you do that for a list of the natural numbers, each sequence has a finite number of non-zero digits, but the "trick" defines a sequence with an infinite number of non-zero digits - which is a sequence not already listed, but not a sequence that corresponds to a natural number.

March 13th, 2016, 04:33 AM   #22
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Quote:
 When you do that for a list of the natural numbers, each sequence has a finite number of non-zero digits, but the "trick" defines a sequence with an infinite number of non-zero digits - which is a sequence not already listed, but not a sequence that corresponds to a natural number.
We're definitely getting there.

Remove the words "natural numbers" from the sentence.

Any sequence indexed by the natural numbers mustbe finite, and therefore must have a finite number of non-zero elements. Using the sequence to represent the natural or real numbers don't even come into it at this point.

There are an infinite number of natural numbers, so a sequence can have an infinite number of elements but it can't be infinitely long if it is indexed on the Natural numbers since all Natural numbers are finite - there is no Natural number available to represent the sequence's length.

This statement of course directly contradicts itself, and I think that is where my problem is. Or possibly yours

 March 13th, 2016, 04:42 AM #23 Senior Member   Joined: Mar 2016 From: UK Posts: 101 Thanks: 2 In fact I can go further - the indices used for the elements of the set of sequences is a tuple - two Natural numbers together. Tuples are countable. Therefore the complete set of elements of all of the set of sequences is countable. That's a real problem if the set itself is not countable. Last edited by Jimbo; March 13th, 2016 at 04:51 AM.
March 13th, 2016, 06:19 AM   #24
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Quote:
 Originally Posted by Jimbo Any sequence indexed by the natural numbers mustbe finite[B]
If this were true, the sequence of natural numbers 1, 2, 3, … would be finite. Is that what you believe?

March 13th, 2016, 06:21 AM   #25
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Quote:
 Originally Posted by Jimbo Therefore the complete set of elements of all of the set of sequences is countable.
Yes it is.

March 13th, 2016, 07:30 AM   #26
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Quote:
 If this were true, the sequence of natural numbers 1, 2, 3, … would be finite. Is that what you believe?
No, I think it's a contradiction.

Yes it is.

Then the complete set of sequences must also be countable, since it must be smaller than the set of all the elements of all the sequences.

We can always find a new value but the resultant set is still countable.

 March 13th, 2016, 07:41 AM #27 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,229 Thanks: 2411 Math Focus: Mainly analysis and algebra It's not a contradiction. Refer to the post you read yesterday which demonstrated that every natural number is finite but that the set of natural numbers is of infinite size. The correct conclusion is not that the complete set of sequences is countable, but that any list of sequences is countable 'this is the definition of "countable"). But the diagonal argument shows that no list of sequences is complete. Since no countable set of sequences is complete, the complete set of sequences cannot be countable. Last edited by v8archie; March 13th, 2016 at 07:49 AM.
March 13th, 2016, 07:49 AM   #28
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Quote:
 Originally Posted by Jimbo The proof is NOT that you can always find another sequence, the proof is that the complete set of Reals is not representable by a set of sequences.
Change the third from last word "set" to "list" (which means "countable set") and you are correct.

March 13th, 2016, 09:56 AM   #29
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Quote:
 Since no countable set of sequences is complete, the complete set of sequences cannot be countable.
And yet the complete set of all elements of all sets is countable.

Which directly implies that the complete set of sets is countable.

Quote:
 But the diagonal argument shows
Only if the diagonal argument is valid. (removed duff remark regarding counter-example in binary sets)

Last edited by Jimbo; March 13th, 2016 at 10:10 AM.

March 13th, 2016, 10:25 AM   #30
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Quote:
 Originally Posted by Jimbo And yet the complete set of all elements of all sets is countable.
No. The set of all elements of a list of sequences is countable.

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