October 8th, 2018, 07:48 AM  #161  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
You can't get to the end of N. Your set of characters {'0','1'} IS the set of natural numbers, regardless of any word games. You can't operate on infinite things because they don't end. List the natural numbers as a column of increasing binary numbers on the left and {'0,'1'} on the right. They are identical. I'll simplify your argument: I say there are sequences of binary digits on the right that are not on the left. Or even easier: Cantor says the real numbers are not countable. It's like Relativity. If you push a physicist up against the wall on the speed of light is constant, the last thing he says is: it's an axiom. Last edited by skipjack; October 8th, 2018 at 11:58 AM.  
October 8th, 2018, 11:13 AM  #162 
Senior Member Joined: Jun 2014 From: USA Posts: 493 Thanks: 36  Ok, for the sake of argument, let's assume that $k \in A$ (where $k = 0.\overline{1}_2$ and $A = (\frac{2^i1}{2^i})_{i=0}^\infty$ ). Since we're assuming that $A$ contains the elements $a_0$ through $a_\omega = k$, we should be able to have a similar sequence that contains all of the elements of $A$ except $k$. Let $A'$ contain $a_0$ through $a_{\omega1}$ then and revisit the table from my original post so I can ask you a question: Does $\bigcup B = \bigcup B'$? $$\begin{matrix} \underline{A \text{ & } A'} & \underline{\text{function } f} & \underline{B \text{ & } B'} & \underline{\{ x \in f(k) : x \notin \bigcup_{j=0}^i b_j \}} \\ a_0 = 0.\overline{0}_2 & \rightarrow & b_0 = \{ 0.0, 0.00, 0.000, \dots \} & \{ 0.1, 0.11, 0.111, \dots \} \\ a_1 = 0.1\overline{0}_2 & \rightarrow & b_1 = \{ 0.1, 0.10, 0.100, \dots \} & \{ 0.11, 0.111, 0.1111, \dots \} \\ a_2 = 0.11\overline{0}_2 & \rightarrow & b_2 = \{ 0.1, 0.11, 0.110, \dots \} & \{ 0.111, 0.1111, 0.11111, \dots \} \\ a_3 = 0.111\overline{0}_2 & \rightarrow & b_3 = \{ 0.1, 0.11, 0.111, \dots \} & \{ 0.1111, 0.11111, 0.111111, \dots \} \\ \\ \text{Standard sequence } \uparrow & \vdots & \text{Nonstandard 'end of sequence' } \downarrow & \text{Is } f(k) \subset \bigcup B' \text{ ?}\\ \\ a_{\omega2} = 0.\overline{1}00_2 & \rightarrow & b_{\omega2} = \{ 0.1, 0.11, 0.111, \dots, 0.\overline{1}0, 0.\overline{1}00 \} & \{ 0.\overline{1}1, 0.\overline{1}11 \} \text{ or } \{ 0.\overline{1}1 \} \text{ ?} \\ a_{\omega1} = 0.\overline{1}10_2 & \rightarrow & b_{\omega1} = \{ 0.1, 0.11, 0.111, \dots, 0.\overline{1}1, 0.\overline{1}10 \} & \{ 0.\overline{1}11 \} \text{ or } \{ \text{ } \} \text{ ?} \\ a_\omega = k \in A, k \notin A' & \rightarrow & b_\omega = f(k) \in B, f(k) \notin B' & \{ \text{ } \} \\ \end{matrix}$$ What about my original question for you Zylo?: In closing, I will point out that $[0,1]$ would be enumerable if summary item #3 were true. That is, where $\bigcup f([0,1]_2)$ is an enumerable set of finite strings, a listing of $f([0,1]_2)$ could be created via an enumeration of $\bigcup f([0,1]_2)$ because the union of the listing of $f([0,1]_2)$ would not equate to $\bigcup f([0,1]_2)$ unless the listing contained all of the elements of $f([0,1]_2)$. Do you see why this is Zylo (anybody)? 
October 8th, 2018, 12:34 PM  #163  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
The set of all natural numbers, N, is something else entirely. And it exists in ZFC because an axiom says it does. Quote:
The axiom that defines N does not "end," and has no need to. We simply accept that an endless definition is a valid way to describe a complete set. If you don't want to utilize the axioms that Cantor's proof does, don't claim that you have disproved it. Proofs are based in axiomatic systems, because mathematics has no absolute truths. We can define an example of the function t(*) by:
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October 8th, 2018, 01:40 PM  #164  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  You can't compare infinite strings unless you can get to the end. Quote:
Last edited by zylo; October 8th, 2018 at 01:57 PM.  
October 8th, 2018, 01:59 PM  #165  
Senior Member Joined: Jun 2014 From: USA Posts: 493 Thanks: 36  Quote:
$$P = \{x \in \mathbb{N} : x \text{ is even} \} \subset \mathbb{N}$$ $$P = N$$  
October 8th, 2018, 02:11 PM  #166  
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  Quote:
Quote:
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By definition, there isn't a "last decimal place" to add 1 to if your .3333.... is endless. If it's not endless, the result of changing the last '3' to '4' is rational, not irrational.  
October 8th, 2018, 04:07 PM  #167 
Senior Member Joined: Jun 2014 From: USA Posts: 493 Thanks: 36  I picked a sequence that turned an infinite string of zeros into an infinite string of ones, onebyone. The sequence has a clear beginning, that of all zeros, and a clear end, that of all ones, given the nonstandard assumption that it could end. The elements of the sequence seem quite "comparable" to me. Induction tells us that the first element of $f(k)$ is in $b_1$, the second in $b_2$, and so on. That seems reasonable and is the standard way of showing that $f(k) \subset \bigcup B$. It doesn't seem possible to challenge this view. We might quibble with it though. Asserting that $f(k) \subset \bigcup B \iff f(k) \in B$ would be to throw out the proof by induction and in doing so be left with the interpretation that $f(k) \not\subset \bigcup f([0,1)_2)$ because $f(k) \notin f([0,1)_2)$ (equivalently, there is an element of $f(k)$ that is not in $\bigcup f([0,1)_2)$). A function $t$ from $\bigcup f([0,1]_2)$ to $f([0,1]_2)$ would be surjective simply by stipulating that $x \in t(x)$ for all $x \in \bigcup f([0,1]_2)$, thus the reals would be countable. We would have other issues too. For example, if $v$ was a function from $f([0,1]_2)$ to $\mathcal{P}(\bigcup f([0,1]_2))$, then we could define a set $C$: $$\text{Let }C = \bigcup \{ x \in f([0,1]_2) : x \not\subset v(x) \} \in \mathcal{P}(\bigcup f([0,1]_2))$$ $$\text{Let } v(q) = C, \text{ where } q \in f([0,1]_2)$$ $$q \subset v(q) \implies q \not\subset C = v(q)$$ $$q \not\subset v(q) \implies q \subset C = v(q)$$ $$\not\exists q \text{ such that } v(q) = C \implies \text{ function } v \text{ is not surjective}$$ This is a problem because the cardinality of the two sets are equal, so there must exist a surjection from $f([0,1]_2)$ onto $\mathcal{P}(\bigcup f([0,1]_2))$. 
October 9th, 2018, 12:38 AM  #168 
Global Moderator Joined: Dec 2006 Posts: 20,370 Thanks: 2007  
October 9th, 2018, 03:32 AM  #169  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
The end of increasing n and beyond are figments of the imagination. To compare two strings, they have to be of the same length. Infinity is not a length.  
October 9th, 2018, 03:53 AM  #170  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  Quote:
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The functions I called t(*)? I defined an example. An infinite example. Any real number in [0,1) can be used to define another, using binary representation. Did you miss the part where the Axiom of Infinity makes it acceptable to define an object via an algorithm that has no end? The string "01010101..."? Did you miss the part where that is just a way to represent the function on paper? That has nothing to do with whether the function exists? Something derived from the set N? That is the set of natural numbers. It is not a string, and contains no strings. Every number in N is finite. Yes, there are binary representations of those numbers; and yes, every one has a finite length. Do you think Cantor ever claimed otherwise? The set itself is endless (i.e., infinite), so its cardinality is not a natural number. You seem to think that Aleph0 is what Cantor imagined he would find at the end of the endless process that defines N. The "limit of n as n approaches infinity." IT IS NOT. Did you mean I can't define a function s(*) as a function whose domain is N and whose codomain is T? I can define one as easily as I defined the example of t(*), but I don't need to. What Cantor's Diagonal Proof shows, is that for any such function s(*) that anybody can define, it cannot be a surjection. The proof doesn't need to define the function s(*), it says that one can't be defined as a surjection. It proves it by using the same, valid technique where an endless process is applied to a set of objects that is known, or assumed, to exist.  

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binaryexpressed, cardinality, continuum hypothesis, diagonal argument, numbers, real, set 
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