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May 9th, 2017, 08:00 AM   #11
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Quote:
 Originally Posted by AplanisTophet They sometimes act like animals though...
I believe 'they', act like mathematicians.

May 9th, 2017, 09:35 AM   #12
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Quote:
 Originally Posted by InkSprite I think I can prove $A$ is just as large as real numbers.
Can't you use Cantor–Schröder–Bernstein theorem to prove their Cardinal number is the same and that's it?

May 9th, 2017, 10:39 AM   #13
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Quote:
 Originally Posted by JeffM1 We imagine the real numbers in the interval [0, 1). Numbers do not have decimal points. What do have decimal points are numerals in the decimal system
This is a very useful point to remember.

Quote:
 Originally Posted by JeffM1 Now let us think of infinite strings in the form $0.x_1x_2... \text { where } x_i \in \{0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9\}.$ Then think of the infinite strings in the form $x_1x_2...$ Is it not easily demonstrable that those two sets can be placed in 1-to-1 correspondence and so have the same number of elements?
It's much less obvious how to put the real numbers (in $[0,1)$) into an intuitive 1-1 correspondence with those strings as some pairs of the strings represent a single real number under the usual system.

May 9th, 2017, 11:09 AM   #14
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 Originally Posted by Joppy What is all the fuss about Cantor's diagonal argument exactly? Not only here do I see plenty of attempts to refute it, but elsewhere too. I have not studied the ideas formally or informally (hence the ignorance in what follows), but I have to ask... Is it simply the nature of the topic that allows for very subtle changes in usage of language that produce misconstrued definitions and results? Thus, poor interpretations by those lacking rigor. Because that's what it always seems like, a back and forth of "what do you mean by this". Or is there actually some funny business going on... i.e., some unfinished business left by Cantor.
A math professor wrote a paper about that. It's here in ps form. http://www.math.ucla.edu/~asl/bsl/0401/0401-001.ps

It's called, An Editor Recalls Some Hopeless Papers, by Wilfred Hodges. It begins:

Quote:
 Originally Posted by 'Wilfred Hodges I dedicate this essay to the two-dozen-odd people whose refutations of Cantor’s diagonal argument (I mean the one proving that the set of real numbers and the set of natural numbers have different cardinalities) have come to me either as referee or as editor in the last twenty years or so. Sadly these submissions were all quite unpublishable; I sent them back with what I hope were helpful comments. A few years ago it occurred to me to wonder why so many people devote so much energy to refuting this harmless little argument—what had it done to make them angry with it? So I started to keep notes of these papers, in the hope that some pattern would emerge. These pages report the results. They might be useful for editors faced with similar problem papers, or even for the authors of the papers themselves. But the main message to reach me is that there are several points of basic elementary logic that we usually teach and explain very badly, or not at all.

My own opinion is that Cantor's uncountability results are the very first example anyone ever encounters of a result in higher math that's deeply counterintuitive, yet accessible to beginners. So a lot of people get stuck there.

 May 9th, 2017, 11:46 AM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,550 Thanks: 2551 Math Focus: Mainly analysis and algebra For me, it's not so much that the idea is counter-intuitive. I don't think that many people have such deeply held beliefs about infinite sets as to be repulsed by the idea that they can't all be listed I think that many (such as Zylo) struggle to cope with the shattering of the idea that "infinity" isn't a number, the idea that arbitrarily large finite numbers do exist and that there is something even bigger than the "bigger than all the numbers" that they always thought of as a number. I'll read the article with interest though. Thanks from Joppy
May 9th, 2017, 01:13 PM   #16
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Regarding showing that the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$, you wrote:

Quote:
 Originally Posted by InkSprite I did so with a space filling infinite-dimensional fractal curve. I was quite proud of that. I guess winning an award in mathematics and pocketing some come-up money isn't going to be that easy. I'm really glad this place exists.
I confess that I have no idea what a “space filling infinite-dimensional fractal curve” is. I sincerely hope that you do if you are using it in a proof.

Let me walk you through a textbook method of proving that $|A| = |\mathbb{R}|$. You'll need a very firm grasp on this sort of stuff in general if your goal is to disprove Cantor's Theorem.

Cardinality arguments are fairly simple in that when two sets have equal cardinality, you can show that a bijection exists between the two sets (if not both a surjective function from one set onto the other as well as an injective function from that set into the other).

The way to show that $|A| = |\mathbb{R}|$ is to show that there exists a function $g$ where $g : A \rightarrow \mathbb{R}$ is bijective.

To do this, I suggest you first find a bijective function $h$ from $A$ onto [0, 1). This shouldn’t be too hard given the definition of $A$, where $f^{-1} : A \rightarrow \{x \in [ \,0,1) \,: x \text{ is a non-terminating decimal} \}$ is bijective and gets you most of the way there.

Then, you can biject [0, 1) with $\mathbb{R}^+$. Let $j(x) = \frac{x}{1-x}$. (These proofs are tedious: e.g., does $\mathbb{R}^+$ include 0?... I point this out to show how easy it is to make a small error, even one that is easily corrected).

Now, you need only biject $\mathbb{R}^+$ with $\mathbb{R}$. I’ll leave that to you (call it function $b$).

Then, returning to our initial goal for a function $g$ where $g : A \rightarrow \mathbb{R}$ is bijective, we have:

$g(x) = b(j(h(x)))$ implies $|A| = |\mathbb{R}|$.

If you thought this was somewhat difficult, then good perhaps. As I told you earlier, if you really want to disprove Cantor's theorem, you'll need to show that $|\mathbb{N}| = |A|$.

Just as we formalized what the term "infinitely large integer" means earlier in this thread, you now need to formalize what you mean by the term "converge" as you are using it in the other thread.

Quote:
 Originally Posted by zylo 101010.... converges in the sense that it converges to a very specific natural number
V8archie is correct in that the term "converge" has a very specific meaning in mathematics already, so you should use a different term. Any mathematician would require that you give a precise definition of what you mean by "converge" before they can take you seriously in this context. I would help you with this definition, but I frankly have no idea what you mean.

May 9th, 2017, 01:42 PM   #17
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 Originally Posted by AplanisTophet You'll need a very firm grasp on this sort of stuff in general if your goal is to disprove Cantor's Theorem.
I know the world of education is generally very liberal and we like to think that everybody's opinion is valid, but this is mathematics. It requires proof not opinion. Anyone who wants to disprove Cantor is a crank, which means that their opinion is worth less than everybody who understands that we already have proofs (plural) that the theorem is correct.

Quote:
 Originally Posted by AplanisTophet given the definition of $A$, where $f^{-1} : A \rightarrow \{x \in [ \,0,1) \,: x \text{ is a non-terminating decimal} \}$ is bijective
No it isn't. At least, the bijection you are talking about fails somewhere, namely the point where you stop considering digit strings as sequences and start considering them at representations of real numbers. This is because the reals in any interval do not all have a unique decimal (or binary, or whatever base we are working in today) representation. So the map from numbers onto digit strings is one-to-many.

May 9th, 2017, 01:57 PM   #18
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Quote:
 Originally Posted by v8archie I know the world of education is generally very liberal and we like to think that everybody's opinion is valid, but this is mathematics. It requires proof not opinion. Anyone who wants to disprove Cantor is a crank, which means that their opinion is worth less than everybody who understands that we already have proofs (plural) that the theorem is correct. No it isn't. At least, the bijection you are talking about fails somewhere, namely the point where you stop considering digit strings as sequences and start considering them at representations of real numbers. This is because the reals in any interval do not all have a unique decimal (or binary, or whatever base we are working in today) representation. So the map from numbers onto digit strings is one-to-many.
You can ignore this post by V8. He's simply wrong and, given how much time he spends here, he should know better.

V8 is trying to point out that in any base we have equalities such as (base 10):

0.1 = 0.0999999...
0.25 = 0.24999999...

In base 10, any rational of the form $\frac{a}{10^b}$, where $a, b \in \mathbb{N}$, will have two expansions: one finite and one infinite. In going from $A$ to [0, 1) using $f^{-1}$, we are not hitting any rationals of the form $\frac{a}{10^b}$, hence my comment that $f^{-1}$ gets you "most of the way there."

 May 9th, 2017, 02:15 PM #19 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,550 Thanks: 2551 Math Focus: Mainly analysis and algebra Your definition of A isn't very clear then. And there is no element of A that maps to zero, so you still don't have the bijection you were looking for. Perhaps a little more consideration and a little less aggression?
May 9th, 2017, 02:34 PM   #20
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Quote:
 Originally Posted by AplanisTophet Regarding showing that the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$, you wrote: I confess that I have no idea what a “space filling infinite-dimensional fractal curve” is. I sincerely hope that you do if you are using it in a proof. Let me walk you through a textbook method of proving that $|A| = |\mathbb{R}|$. You'll need a very firm grasp on this sort of stuff in general if your goal is to disprove Cantor's Theorem. Cardinality arguments are fairly simple in that when two sets have equal cardinality, you can show that a bijection exists between the two sets (if not both a surjective function from one set onto the other as well as an injective function from that set into the other). The way to show that $|A| = |\mathbb{R}|$ is to show that there exists a function $g$ where $g : A \rightarrow \mathbb{R}$ is bijective. To do this, I suggest you first find a bijective function $h$ from $A$ onto [0, 1). This shouldn’t be too hard given the definition of $A$, where $f^{-1} : A \rightarrow \{x \in [ \,0,1) \,: x \text{ is a non-terminating decimal} \}$ is bijective and gets you most of the way there. Then, you can biject [0, 1) with $\mathbb{R}^+$. Let $j(x) = \frac{x}{1-x}$. (These proofs are tedious: e.g., does $\mathbb{R}^+$ include 0?... I point this out to show how easy it is to make a small error, even one that is easily corrected). Now, you need only biject $\mathbb{R}^+$ with $\mathbb{R}$. I’ll leave that to you (call it function $b$). Then, returning to our initial goal for a function $g$ where $g : A \rightarrow \mathbb{R}$ is bijective, we have: $g(x) = b(j(h(x)))$ implies $|A| = |\mathbb{R}|$. If you thought this was somewhat difficult, then good perhaps. As I told you earlier, if you really want to disprove Cantor's theorem, you'll need to show that $|\mathbb{N}| = |A|$. Just as we formalized what the term "infinitely large integer" means earlier in this thread, you now need to formalize what you mean by the term "converge" as you are using it in the other thread. V8archie is correct in that the term "converge" has a very specific meaning in mathematics already, so you should use a different term. Any mathematician would require that you give a precise definition of what you mean by "converge" before they can take you seriously in this context. I would help you with this definition, but I frankly have no idea what you mean.
I don't have beef with cantor's theorem I just jumped on zylo's train because I saw his comment about .9 repeating not being 1. Which I'm sure is related to the same issue, but I'll save that for another day. Right now I'm 100% focused on the infinite dimensional stuff.
I jumped in way to fast. I thought [0-1) was clever, but Maybe not.
What I could do is write a function that gives you every combination of [0-1) minus the decimal and how it can match up to one.

The "space filling curves" is an idea I got from someone else, so it's just my personal conjecture that it's possible in infinite dimensions. So, when defining an alphabet you can treat each string character as a dimension. When you run the function to get a particular repeating digit string you can just enter a natural number to line right up with what that function returns.

In other words, every natural number can be a parameter that is fed through this thing that generates a unique string. Since the plots in each dimension determine the characters used, the only thing you have to prove is that you fill space with this fractal curve at every "argument = Natural number". I'm a novice making a bold claim, but as long as space filling curves exist, I'm convinced it can be done. I'm having fun learning. I'll try to pull up that fractal curve stuff and see if I can ask that guy about some examples. It was somewhere on this forum I believe. I think it might have been Joppy

... And yes. I found that hard.
Sorry for being long winded and redundant. I'm excited.

Zylo said the converging thing. I'm just the other crazy guy that jumped in for no apparent reason because I don't like .9repeating==1.
Speaking of converging. can you converge on an irrational number?

I think I do have a bijection function. I think that's what I just tried to describe, but I will have to do more research about that space filling curve thing.
I'm not interested in disproving cantor. I'm interested in seeing if that curve fractal thing does though.

What comes to mind when you think about representing each character-slot as a dimension? The space in those dimensions would be discrete, but can still be filled by a continuous function as long as the function returns a discrete value for a discrete/natural number.

Last edited by InkSprite; May 9th, 2017 at 02:47 PM.

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