One to the power of infinity

[Nathan]

Thanks for the answer Basic. I have another question now.

With Cantor's proof, the amount of digits precision must be larger than the number of set members. For a given maximum precision, p, the number of set members would be 2^p. As p tends to infinity, I would have thought 2^p tends to a much bigger inifinity.

 

[Simon F]

Thanks for the answer Basic. I have another question now.

With Cantor's proof, the amount of digits precision must be larger than the number of set members.

I don't understand what you are saying. Cantor's diagonal proof deals with a (countably) infinite set of real numbers. Each real in your set can have an infinite number of digits.

For a given maximum precision, p, the number of set members would be 2^p. As p tends to infinity, I would have thought 2^p tends to a much bigger inifinity.

Err well there are different classes of infinity. For example, the natural numbers, integers, and rational numbers are all "countably infinite", but the reals and complex numbers are not.

 

[Nathan]

I'll try to expand on my previous post.

Each member of the set has infinite precision, and there are an infinite amount of set members (cardinality?). If the cardinality of the set is less than the precision, Cantor's proof works. If it is greater, it is impossible to form a diagonal that crosses all the set members, so his proof fails.

My argument was that for a given precision, the cardinality is greater than the precision. I guess that if [0, inf) has the same cardinality as [0, inf+), then its fair that [0, inf) and [0, 2^inf) have the same cardinality. It just seems a bit strange that the 2^p is larger than p as p tends to inf, yet inf and 2^inf are the *same*.

 

[Nathan]

[edit]Stupid double posts.[/edit]

 

[Fred]

[0, inf) and [0, 2^inf) ....

eep you're getting confused. Inf is not a number! You would need to write down the generator of the 2 sets rigorously in order to figure if its countable or not.

The best way to understand this is to pick up a book on analysis and go through the rigourous mathematical definitions, its tedious but it gets the job done.

Look, the idea is if you can find an equivalence relation between a set and the integers (or subset of the integers), then it is countable.

In fact its stronger than that, since you can take the the union of a *infinite sequence* of countable sets, and that set will still be countable.

After this, you start getting into basic point set topology, where there are notions of infinite amounts of points around some neighborhood (eg dense, compact, etc).

You can try the following as exercises.

prove the rationals are countable

Prove that the set of all sequences whose elements are 1 and 0 are uncountable.

 

[Simon F]

you're getting confused. Inf is not a number!

Except in IEEE-754 where Inf is not "not a number"

[Snipped some good comments]

You can try the following as exercises.

prove the rationals are countable

Prove that the set of all sequences whose elements are 1 and 0 are uncountable.

The second is trivial given the earlier discussions.

As for the first, are we allowed to have a many-to-one mapping of N->Q? If so it's also easy.

 

[embargiel]

Basically, a brilliant mathematician named Cantor found out about this (and other stuff) and went crazy dying in a sanitarium, but the basic theory is around the idea of lining up numbers one after another, and analyzing that there is a way of lining up a diagonal in a certain way to prove that (insert interesting infinity here) > (normal infinity).

Wait.... This bothers me. If one type of infinity can be greater than some others, then 1 can't be equal to .9_ (like what Nathan suggested), can it?

 

[Simon F]

There is nothing inconsistent. 0.9999.. recuring equals 1.0.

IIRC, when we covered Cantor's proof in 1st year Uni', to avoid inconsistencies none of the numbers in the list 'ended' in recuring 9s (since they were equivalent to another number).

 

[embargiel]

Using the same logic, we can prove that a line is twice as long as itself.


http://www.mathpages.com/home/kmath063.htm

 

[Fred]

'Cantor went crazy dying in a sanitarium'

You know, I imagine there was probably other reasons, other than math, that stuck him there.

But its like the old mathematician saying.

'Never let the facts get in the way of a good anecdote.'

 

[Simon F]

Using the same logic, we can prove that a line is twice as long as itself.
http://www.mathpages.com/home/kmath063.htm

I don't see that it's the same logic at all.


A not too dissimilar thing is to compute the integral of F(x) from x=0 to x=1 where F(x) = 1 when X is rational and F(X)=0 when X is irrational.

 

[embargiel]

I might be wrong, but in the paradox of limit, as the number of jaggies goes bigger and bigger, the smaller the triangle is. And yet, the total length is still twice as long as AC's. If we take a limit as the number of jaggies goes to infinity, then the sidesof the triangle will approach zero. If it's zero, then it's perfectly coincide with the line AC, making the line that is equal to AC twice as long as the length of AC.

 

[Basic]

Yes, you've understood the jaggie-thingy right. But that hasn't anything to do with wether 0.999... == 1.0... , and the link you gave doesn't say so either.

... entities in any given sequence
generally possess multiple properties, and an entity that possesses
the limiting value of one of those properties doesn't necessarily
possess the limiting value of ALL those properties (many of which
may not even converge).

The difference here is that in the case of 0.999... == 1.0..., we're not talking about different properties, the only property we're talking about is the real value.

 

[Simon F]

I might be wrong, but in the paradox of limit, as the number of jaggies goes bigger and bigger, the smaller the triangle is. And yet, the total length is still twice as long as AC's. If we take a limit as the number of jaggies goes to infinity, then the sidesof the triangle will approach zero. If it's zero, then it's perfectly coincide with the line AC, making the line that is equal to AC twice as long as the length of AC.

In that paradox, it claims that the limit, as n->Infinity, of the sequence of (jaggy) piecewise linear functions, Gn(x), is the line F(x)=0 (0<=x <=1). I'm going to prove that it isn't.

Consider the first derivative of F(x). This is 0 everywhere.

Now consider the first derivative of Gn(x). This can be a choice of sqrt(3), -sqrt(3), OR non-existant at (2^(n+1) - 1) points.

Now let's consider the derivative of the limit function - we end up with a countable infinity of points where the derivative does not exist (i.e. each one can be given a unique integer index). Furthermore, by Cantor's argument there must still be real numbers that have derivatives which are either sqrt(3) or -sqrt(3) !

Therefore the limit of Gn(x) is not the line F(x) = 0.

QED.


AFAICS, what you are showing is that for any random point, r, the limit of Gn(r) is 0, which is a different thing to claiming the whole set of mappings is a line.