My Math Forum Cantor's Infinite Binary Sequence

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March 9th, 2018, 07:01 PM   #41
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Quote:
 Originally Posted by AplanisTophet Is this the part where you guys explain that zylo’s arbitrary selection process is no different than asserting the reals can be well ordered given the axiom of choice, but despite that, they still aren’t countable?
No, I don't think that's part of the thread at all, at least I hope it's not.

But it's a very good question.

Consider: On the one hand, we can not make a list of the reals. Cantor proved that. On the other hand, we can well-order $\mathbb R$ -- and well-ordering was very well known to the early set theorists -- and make a "long list." In fact this is a nice visualization of what an uncountable well-ordered set looks like. It's a really long list.

Now, how do we square this with Cantor's beautiful and revolutionary proofs that the reals may not be listed?

The answer is that a list by definition is a map $\mathbb N \to X$ where $X$ is some set. Note two things: One, a list is countable by definition. And two, a list has order type $\omega$. It's not some exotic countable ordinal. It's plain old familiar $\omega$.

So when we use the word "list," there is an equivocation of meaning:

* The professionals hear the word list and they think of a map whose domain is the positive integers. Like all good definitions, it has the theorems baked into it.

* The amateurs hear the word list and they think of everyday common sense meanings, or perhaps they imagine wild transfinite meanings that are not supported by the technical definition.

This I believe is one of the sources of the pervasive Cantor confusion.

Regarding your other remarks, if you don't like a poster, don't read their stuff. Or write a mathematical response to Zylo. Not a long ad hominem diatribe.

If you are lacking in the ability to frame an argument that Zylo can agree with; that does not grant you the right to be uncivil. It means you have an opportunity to become more crystal clear in your own thinking and writing, so that readers may learn something.

My personal preference would be for you to post more math and less personal shit about people you don't like.

I speak for myself and nobody else.

Last edited by Maschke; March 9th, 2018 at 07:28 PM.

 March 9th, 2018, 07:25 PM #42 Senior Member   Joined: Jun 2014 From: USA Posts: 525 Thanks: 40 I like zylo, which is the only reason I posted what I did. Take that to heart.
March 9th, 2018, 07:30 PM   #43
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Quote:
 Originally Posted by AplanisTophet I like zylo, which is the only reason I posted what I did. Take that to heart.
LOL Well ok then!

March 10th, 2018, 05:23 AM   #44
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Quote:
 Originally Posted by zylo answered in posts #4 and 7.
And my response to posts #4 and #7 is post #10.

You claim (incorrectly) in post #14 that I asked for an interpretation which you provided in post #4 and which you say (also in post #14) is irrelevant to post #1.

So, I return to my original question (from post #2) in the limiting case what is the second element in your list? And I ask that you give an answer that you believe is relevant to your post #1.

March 10th, 2018, 07:21 AM   #45
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Quote:
 Originally Posted by zylo How do you know you've put all the members of a set, such as points on a line or set of all infinite binary sequences, into a list? The answer is easy. The set is empty. Say you've made a list, ie, enumerated, from a set. Is the list complete- is there still a member of the set? Add it to the list, the natural numbers are endless. Conclusion: The contents of any set of distinguishable elements is countable. Zylo's Theorem. Ok, Zylo-Skipjack Theorem for boxing me into it. If he likes.
Quote:
 Originally Posted by AplanisTophet Quoted for posterity. Delusional.
Delusional is not a mathematical argument. Neither is undefined jargon unless you understand it. So thanks for confirmation due to inability to find an intelligent refutation.
Personally, I don't consider memorization of uncomprehended jargon an education.

Quote:
 Originally Posted by skipjack It's certainly possible to put some points in a list, zylo, but you haven't defined a process that puts all the points from a given line segment into a list. Until you've done that, you can't support your argument by asserting that an all-inclusive list is possible.
To assign a number to a point on a line segment:
Divide the line into 10 parts (or 2 or whatever the radix is). That gives the first digit.
Divide the previous partition containing the point into 10 parts. That gives the second digit.
Divide the previous partition containing the point into 10 parts That gives the third digit.
Repeat the process indefinitely to get a natural number, or a fraction if you put a period before it.

To find the position of a natural number, or a fraction, on a line segment, invert the process.
Divide the line into ten parts. The point is in the first digit partition.
Divide the first digit partition into ten parts. The point is in the second digit partition.
Divide the previous partition containing the second digit by 10, and choose the partition of the third digit.
Eventually you get the point corresponding to the given number.

Numbers based on a unit segment of a line.
Assign numbers to the unit segment per above. Any other point on the line has a number corresponding to a number in the unit segment plus the number of unit segments, so it has 2 parts: A,B. A is the number of unit segments and B is the number of the point in the segment.

The best way to understand it is to draw a line segment. Try it with a radix of 2.
Place a point on a line and divide the line in half. Then the first digit is 9 or 1, depending on which half your in. After 3 partitions you should have a number like 101 (or .101). If your point falls on a partition line, the rest of the digits are 0.

March 10th, 2018, 07:33 AM   #46
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Quote:
 Originally Posted by zylo To assign a number to a point on a line segment:
Zylo, there's no disagreement that we can assign a number to a point.

But there is no list that contains all of them.

 March 10th, 2018, 07:46 AM #47 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 There's a problem with "Repeat the process indefinitely to get a natural number", because though each repetition results in a natural number, endless repetition results in an endless string of digits,which isn't a natural number. If you try to circumvent that by putting a "." at the beginning, you need to change your description to "Repeat the process indefinitely to get a fraction", but there's a still a problem with endless repetition, because a dot followed by an endless string of digits needn't correspond to a fraction.
 March 13th, 2018, 10:02 AM #48 Senior Member   Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25 One of the reasons Cantor specifically and delibrately did not use the real numbers $\mathbb R$, as the example for his Diagonalization proof, may have been the possibility that doubters could find endless (no pun intended) ways to misrepresent it. Regardless, any attempt to disclaim it using $\mathbb R$ is invalid, since it isn't about $\mathbb R$. Zylo's attempt in this thread started out addressing the actual example, but the discussion keeps digressing towards $\mathbb R$. I want to take it the other direction. This addresses the same issue he asked in his original post, without obfuscation. Let Jn be the n-place binary sequence with exactly (n-1) "0"s followed by one "1" : J1 = "1" J2 = "01" J3 = "001" etc. What is $\displaystyle \lim_{n \rightarrow \infty}$Jn? The only difference is that Zylo's set Ln of length-n strings is replaced by a single length-n string. But now v8archie's question is the same as Zylo's, as I believe was the intent. And the answer is that you can't define the end of the infinite-length string that Zylo implies as a limit, and that I reduced to one string of infinite "0"s, in order to put the "1" at its end. This is the same issue as the fact that every element of $\mathbb N$ is a finite number, while the cardinality of $\mathbb N$ is not. The "infinity" that n approaches in a limit is not the infinity that Cantor uses as a cardinality. This is a fundamental fact that all Cantor doubters misrepresent in some way. Thanks from topsquark Last edited by JeffJo; March 13th, 2018 at 10:04 AM.
 March 13th, 2018, 11:14 AM #49 Senior Member   Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25 Axiom of Infinity: There is a set $\mathbb N$ that:Contains the number 1, and If it contains the number n, it also contains the number n+1 The Mathematics of Cantor's proof accepts this as a postulate. You don't have to accept it if you don't want to, but then anything you infer cannot say anything about what Cantor proved. But the existence of $\mathbb N$ is not all that this establishes. It also establishes a way we can define infinite sets. There is a set $\mathbb M$ that:Contains the string M0 of infinite "0"s, Contains "1"+M0, If X is a finite string of "0"s and "1"s, and $\mathbb M$ contains X+"1"+M0, then it also contains X+"11"+M0, and If it contains Mn, it so a contains a string that reverses every character in Mn. I believe that this establishes, in Cantor's Mathematics, the existence of the set he used as the example in Diagonalization. But the proof is almost always misrepresented. What he proved (and I'm sorry, I don't know how to format this) was "If the function C:$\mathbb N$->$\mathbb M$ exists, then C is not surjective." Note that such functions do exist, like the Jn I defined earlier. The proof does not assume the function is surjective, which is what most doubters seem to object to. You can't invalidate the proof by raising objections to the assumption which is not made. What it does do, directly and not as proof by contradiction, is prove that there cannot be a bijective function between $\mathbb N$ and $\mathbb M$. If you further assume that C is injective, it proves that the cardinality of $\mathbb M$ is greater than that of $\mathbb N$. And if you think you have a bijective function, please re-examine the statement Cantor proved. It can't be subjective.
March 13th, 2018, 11:15 AM   #50
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Quote:
 Originally Posted by zylo Delusional is not a mathematical argument. Neither is undefined jargon unless you understand it.
Pot calling the kettle black.

If you don't learn what technical terms mean, it's no surprise that you can't understand anybody's proofs or explanations. Neither is it any surprise when your ideas come across as nonsense.

If you can't communicate properly, that's your own problem that you have to solve.

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