My Math Forum Rudin Theorem 2.15 and Countability of the Real Numbers
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 July 26th, 2016, 11:37 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 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 n-tubles (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 n-place 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
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Quote:
 Originally Posted by zylo Or is there some kind of mysterious quantum leap beyond countable infinity in the size of B$\displaystyle _{n}$ as n approaches countable infinity?

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
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Quote:
 Originally Posted by zylo Proven by induction, ie, valid for countable infinity.
No. Induction only proves statements for $n \in \mathbb N$. That means for finite $n$ only.
Quote:
 Originally Posted by zylo as n approaches countable infinity
$n \in \mathbb N$ 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
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Framing sensible answers to these posts is amusing but quite fruitless.
Quote:
 Originally Posted by zylo Let A=(0,1,2,..,9) Consider the n-place decimals in [0,1): .a$\displaystyle _{1}$....,a$\displaystyle _{n}$ $\displaystyle \equiv$ (a$\displaystyle _{1}$....a$\displaystyle _{n}$)=B$\displaystyle _{n}$. ... For finite n B$\displaystyle _{n}$ corresponds to the rationals and is finite
We seem to start from the premise that n is finite as are the rational numbers.

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:
 and as n approaches countable infinity
The language of "n approaches infinity" is indeed commonly used, but it creates the false impression that there is an end of the line to the natural numbers. I suspect he really thinks that there is some finite number that is very close to infinity and then his set will be close to capturing all the rationals. He just has a few more to go, and then his set will capture not only all the rationals but all the reals in the interval [0, 1) as well.

 July 26th, 2016, 08:20 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 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
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Quote:
 Originally Posted by v8archie 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.
Actually I suspect he is thinking of $\displaystyle \lim_{x \rightarrow \infty} f(x) = \infty.$

But I agree with you that he does not understand what the notation means.

July 26th, 2016, 09:10 PM   #7
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Quote:
 Originally Posted by zylo ..... 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.
the is a typo I missed. A collection of n-place decimals wouldn't be the rationals, just rationals.

* 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: 6,965 Thanks: 2282 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: 1,155 Thanks: 90 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
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Quote:
 Originally Posted by zylo 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 accept that each irrational in [0, 1) is represented by a countably infinite decimal. I do not accept the validity of 2.15 as you use it. The n in that theorem is a natural number, not infinity of any sort.

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:
 I have yet to see a correct proof the reals are uncountable.
There is a difference between denying the premises of a proof (such as denying ontologically the existence of transfinite numbers) and denying the validity of a proof.

Quote:
 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.
It is off the wall. As you are using the word "defined," the natural numbers are defined using 10 (or 2) digits. It does not follow that there are only 10 (or 2) natural numbers.

Last edited by JeffM1; July 27th, 2016 at 05:40 AM.

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