My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree13Thanks
Reply
 
LinkBack Thread Tools Display Modes
March 2nd, 2018, 07:23 AM   #1
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,462
Thanks: 106

Cantor's Infinite Binary Sequence

Let Ln be the list of n-place binary sequences:
00......00 n-binary digits
00......01
00......10
00......11
00....100
............
1111111 n-ones

What is the difference between $\displaystyle \lim_{n \rightarrow \infty}$Ln and Cantor's list* of infinite binary sequences?

* https://en.wikipedia.org/wiki/Cantor...gonal_argument

Last edited by skipjack; March 3rd, 2018 at 01:37 PM.
zylo is offline  
 
March 2nd, 2018, 08:28 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,403
Thanks: 2477

Math Focus: Mainly analysis and algebra
In the limiting case, what do you think the second element in your list is?
Thanks from Maschke, topsquark and zylo
v8archie is offline  
March 2nd, 2018, 09:02 PM   #3
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 444
Thanks: 254

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by zylo View Post
Let Ln be the list of n-place binary sequences:
00......00 n-binary digits
00......01
00......10
00......11
00....100
............
1111111 n-ones

What is the difference between $\displaystyle \lim_{n \rightarrow \infty}$Ln and Cantor's list* of infinite binary sequences?

* https://en.wikipedia.org/wiki/Cantor...gonal_argument
$\displaystyle \lim_{n \rightarrow \infty}$ is nonsense as limits are notions defined in metric spaces, and I have no idea how to parse Ln as a point in a metric space or even discuss such a limit. Thus, this limit is the same thing as 3 + purple or $\frac{10}{97 Honda Civic}$ as they are each equally meaningless.
Thanks from Maschke and topsquark

Last edited by skipjack; March 3rd, 2018 at 01:39 PM.
SDK is offline  
March 3rd, 2018, 06:07 AM   #4
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,462
Thanks: 106

Quote:
Originally Posted by v8archie View Post
In the limiting case, what do you think the second element in your list is?
Always 1 (one). In the limit as n approaches infinity, think of it as counting the first point after 0 on a line.

I gave my definition of lim in OP. There are other definitions. So what?

$\displaystyle \lim_{n \rightarrow \infty}$Ln is a list (representation) of the natural numbers in binary notation. If you add a period before each sequence you get a list (representation) of [0,1) in binary notation.

How can a list of the natural numbers not be countable?

Does anyone care to answer OP?

EDIT
On a line label a point 0 and another point 1. Then divide the line in half and label the points 0,1,2 respectiveley. Divide each half in half and then label the points 0,1,2,3,4. Keep repeating. In the limit you will have labelled each point on the segment with the natural numbers, the first after 0 always being 1.

Last edited by zylo; March 3rd, 2018 at 06:27 AM.
zylo is offline  
March 3rd, 2018, 07:28 AM   #5
SDK
Senior Member
 
Joined: Sep 2016
From: USA

Posts: 444
Thanks: 254

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
Originally Posted by zylo View Post
I gave my definition of lim in OP. There are other definitions. So what?
No you didn't. You defined the sequence of lists and then threw the letters "lim" onto the front of it and assumed it was meaningful. It isn't. It's nonsense. Much the same as "3" and "purple" are both meaningful things, until I try to add them and ignore that nobody knows what that means.

Quote:
Originally Posted by zylo View Post
$\displaystyle \lim_{n \rightarrow \infty}$Ln is a list (representation) of the natural numbers in binary notation. If you add a period before each sequence you get a list (representation) of [0,1) in binary notation.

How can a list of the natural numbers not be countable?
It is countable.

Quote:
Originally Posted by zylo View Post
EDIT
On a line label a point 0 and another point 1. Then divide the line in half and label the points 0,1,2 respectiveley. Divide each half in half and then label the points 0,1,2,3,4. Keep repeating. In the limit you will have labelled each point on the segment with the natural numbers, the first after 0 always being 1.
This assertion is simply false. You haven't proved this and you CAN'T prove it because it isn't true. For example, the number $\frac{1}{\pi}$ is in the interval $[0,1]$ but will NEVER be labelled at any step of your iteration. Spend more time reading; it's embarrassing.
Thanks from topsquark and v8archie

Last edited by skipjack; March 3rd, 2018 at 01:42 PM.
SDK is offline  
March 3rd, 2018, 08:34 AM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,403
Thanks: 2477

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zylo View Post
Always 1 (one). In the limit as n approaches infinity, think of it as counting the first point after 0 on a line.

On a line label a point 0 and another point 1. Then divide the line in half and label the points 0,1,2 respectiveley. Divide each half in half and then label the points 0,1,2,3,4. Keep repeating. In the limit you will have labelled each point on the segment with the natural numbers, the first after 0 always being 1.
You say the second point is "one". So what is the point $100\dots000$? You claim it's a natural number. Which one? How would I write it as a summation of powers of 2?

There is no first point after the end-point of a line. By definition, if you claim that you have that first point, I can always show you one that comes before it.

Again, there is no limit with every point labelled. If you think you have labelled them all, you could give me two adjacent points, and I will show you a point between them.

My question was not about the "value" of your second element. I want to know what you think it consists of. It's clearly some zeros followed by a 1. How many zeros? It can't be an infinite number of zeros because that means that the zeros never end and thus can have nothing following them.

So my position is that you are unable to construct your limiting list and therefore it is a fallacy to assume that it exists.

You first have to show that the list exists before you can draw any conclusions. If you do manage to show that the limiting list is well defined, I'll give you the conclusion that you draw from the assumption that is does.

Good luck with that, it's already been proved not to exist. So you are going to need a different mathematical system.

Last edited by v8archie; March 3rd, 2018 at 08:39 AM.
v8archie is offline  
March 3rd, 2018, 11:10 AM   #7
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,462
Thanks: 106

0 0 0 1 numerical value 2^0
0 0 1 0 numerical value 2^1
0 1 0 0 numerical value 2^2
1 0 0 0 numerical value 2^3,
no matter, in general, how many zeros precede the first 1. Google "binary number".

Since no one seems able or willing to address or answer OP,

Hint:
Infinite binary sequence:
$\displaystyle \lim_{n \rightarrow \infty} a_{1}a_{2}......a_{n}$, $\displaystyle a_{i}$ = 0 or 1

Last edited by skipjack; March 3rd, 2018 at 01:44 PM.
zylo is offline  
March 3rd, 2018, 11:42 AM   #8
Senior Member
 
Joined: Jun 2014
From: USA

Posts: 366
Thanks: 26

Quote:
Originally Posted by zylo View Post
Since no one seems able or willing to address or answer OP
Someday you'll look back on this and realize how patient and willing everyone has been with you... well, maybe. I hope so anyways. I can personally relate to being both wrong and stubborn for what that's worth too. It's when you cross over into being delusional and obsessed that it's cause for concern.

I personally don't have much time to spend leisurely here at mymathforum so I'm not here often, but I decided to browse last night and there you were making the same old arguments yet again.

I will refrain from responding to you when it comes to 'Cantor stuff' from now on simply because I think attempting to do so is doing you a disservice. I hope you answer the questions that I posed for you in your other thread and I hope others help you in that regard, but beyond that, I strongly suggest others refrain from entertaining your attempts to assert what is essentially the same thing over and over and over again.
AplanisTophet is offline  
March 3rd, 2018, 11:52 AM   #9
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,462
Thanks: 106

Yet another off-topic reply. I answered the first two off-topic questions out of courtesy, the others should be obvious from what I have said. If not, start your own thread.

Anyone with an answer to OP?
zylo is offline  
March 3rd, 2018, 01:29 PM   #10
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,403
Thanks: 2477

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zylo View Post
0 0 0 1 numerical value 2^0
0 0 1 0 numerical value 2^1
0 1 0 0 numerical value 2^2
1 0 0 0 numerical value 2^3,
no matter, in general, how many zeroes precede the first 1.
And so, again: what is 100…000 in your limiting case? How do you write it as a power of 2?

Quote:
Originally Posted by zylo View Post
Since no one seems able or willing to address or answer OP
I've done nothing but address you original post. But you've now, once again, resorted to ignoring points made and questions posed in favour of repeating nonsense.

Last edited by v8archie; March 3rd, 2018 at 01:52 PM.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
binary, cantor, infinite, sequence



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Is an Infinite Binary Sequence a Natural Number? zylo Real Analysis 45 July 18th, 2017 03:11 PM
Cantor's Diagonal Argument and Binary Sequences zylo Topology 6 May 23rd, 2016 06:57 PM
Real Number as Binary Sequence zylo Topology 8 April 8th, 2016 08:19 PM
Cantor's Diagonal Sequence is in the list zylo Topology 10 March 8th, 2016 02:16 PM
help with the calculus of infinite sequence 1210 Real Analysis 3 September 20th, 2007 08:06 AM





Copyright © 2018 My Math Forum. All rights reserved.