July 26th, 2016, 11:37 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Rudin Theorem 2.15 and Countability of the Real Numbers
2.15. Theorem "Let A be a countable set and let B$\displaystyle _{n}$ be the set of all ntubles (a$\displaystyle _{1}$....a$\displaystyle _{n}$), where a$\displaystyle _{k}$ $\displaystyle \varepsilon$ A {k=1,...,n) and the elements a$\displaystyle _{1}$....a$\displaystyle _{n}$ need not be distinct. Then B$\displaystyle _{n}$ is countable." Proven by induction, ie, valid for countable infinity. Let A=(0,1,2,..,9) Consider the nplace decimals in [0,1): .a$\displaystyle _{1}$....,a$\displaystyle _{n}$ $\displaystyle \equiv$ (a$\displaystyle _{1}$....a$\displaystyle _{n}$)=B$\displaystyle _{n}$. By 2.15 the Reals are countable: for finite n B$\displaystyle _{n}$ corresponds to the rationals and is finite and as n approaches countable infinity B$\displaystyle _{n}$ corresponds to the reals and is countably infinite. Or is there some kind of mysterious quantum leap beyond countable infinity in the size of B$\displaystyle _{n}$ as n approaches countable infinity? 
July 26th, 2016, 12:35 PM  #2  
Senior Member Joined: Aug 2012 Posts: 1,888 Thanks: 525  Quote:
From one point of view there is a mysterious state change. The set of all finite tuples is indeed countable. But the set of all infinite tuples is uncountable, where an "infinite tuple" is a function from the natural numbers to the set of digits. From another point of view there is no state change at all. Is it true that $n < 2^n$? For example $0 < 1$, $1 < 2$, $2 < 4$, $3 < 8$, and so forth. For every set of finite cardinality $n$, its power set has strictly larger cardinality. And this holds true for sets of infinite cardinality too. All of this is subsumed under Cantor's theorem, which says that if $X$ is any set whatsoever, finite or infinite, then the cardinality of $X$ is strictly less than the cardinality of its powerset, $\mathcal P(X)$. This is a simple and beautiful proof that does not involve diagonalization, avoiding pedagogical pitfalls. Right? Last edited by Maschke; July 26th, 2016 at 01:02 PM.  
July 26th, 2016, 01:28 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra  No. Induction only proves statements for $n \in \mathbb N$. That means for finite $n$ only. $n \in \mathbb N$ does not and cannot approach "infinity" of any type. Every natural number $n$ is finite and the quantity of natural numbers less than or equal to $n$ is a vanishingly small proportion of the quantity of finite natural numbers greater than $n$. No matter what $n$ you pick, you will never make that proportion even a millionth of a billionth of a trillionth of a percent. The phrase "as $n$ approaches infinity" is meaningless in the way you are trying to use it. In the mouths of mathematicians it has nothing whatever to do with "infinity" of any sort. 
July 26th, 2016, 08:06 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420 
Framing sensible answers to these posts is amusing but quite fruitless. Quote:
This is his standard argument that the set {0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9} is finite and exhausts the rational numbers. No doubt the set is finite, but it does not quite exhaust the rational numbers. The argument gets no stronger if we add more finite digits. Quote:
 
July 26th, 2016, 08:20 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra 
Yes, more or less. He is thinking of $\lim \limits_{x \to \infty} f(x)$, but he simply doesn't understand what that notation means. In particular, he doesn't understand that "infinity" is nothing at all to do with it.

July 26th, 2016, 08:49 PM  #6  
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420  Quote:
But I agree with you that he does not understand what the notation means.  
July 26th, 2016, 09:10 PM  #7  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
* This "the" means exactly what it says: all the reals.  
July 26th, 2016, 11:08 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra 
Do you not read the refutations posted by others? Do you not understand them? Is that why you ignore every comment that says that you are wrong? Are you ashamed that you don't understand what people tell you? Or do you just imagine that everybody else is wrong because you have spent two minutes guessing some stuff?

July 27th, 2016, 02:54 AM  #9 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
Every irrational number in [0,1) is given numerically by a countably infinite decimal, which, along with the rationals in [0,1), are countably infinite by 2.15. I have yet to see a correct proof the reals are uncountable. Another point of view is, since the reals are defined by the natural numbers, there can't be more of them than you can define with the natural numbers. That is off the cuff. 
July 27th, 2016, 05:33 AM  #10  
Senior Member Joined: May 2016 From: USA Posts: 1,029 Thanks: 420  Quote:
EDIT: I should have acknowledged Archie's point here. A proof by induction is a proof of a statement that is logically equivalent to a statement beginning "For every natural number n." Quote:
Quote:
Last edited by JeffM1; July 27th, 2016 at 05:40 AM.  

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