May 9th, 2017, 07:00 AM  #11 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,600 Thanks: 546 Math Focus: Yet to find out.  
May 9th, 2017, 08:35 AM  #12 
Newbie Joined: Jul 2016 From: Israel Posts: 24 Thanks: 0  
May 9th, 2017, 09:39 AM  #13  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra  Quote:
Quote:
 
May 9th, 2017, 10:09 AM  #14  
Senior Member Joined: Aug 2012 Posts: 1,956 Thanks: 547  Quote:
It's called, An Editor Recalls Some Hopeless Papers, by Wilfred Hodges. It begins: Quote:
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, 10:46 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra 
For me, it's not so much that the idea is counterintuitive. 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. 
May 9th, 2017, 12:13 PM  #16  
Senior Member Joined: Jun 2014 From: USA Posts: 364 Thanks: 26 
Regarding showing that the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$, you wrote: Quote:
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 $f^{1} : A \rightarrow \{x \in [ \,0,1) \,: x \text{ is a nonterminating decimal} \}$ is bijective and gets you most of the way there. Then, you can biject [0, 1) with $\mathbb{R}^+$. Let $j(x) = \frac{x}{1x}$. (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.  
May 9th, 2017, 12:42 PM  #17  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra  Quote:
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 onetomany.  
May 9th, 2017, 12:57 PM  #18  
Senior Member Joined: Jun 2014 From: USA Posts: 364 Thanks: 26  Quote:
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, 01:15 PM  #19 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 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, 01:34 PM  #20  
Member Joined: Dec 2016 From: United States Posts: 53 Thanks: 3 Math Focus: Abstract Simulations  Quote:
I jumped in way to fast. I thought [01) was clever, but Maybe not. What I could do is write a function that gives you every combination of [01) 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. Let me finish reading. 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 characterslot 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 01:47 PM.  

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