April 12th, 2016, 11:31 AM  #21  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
I can see that your principal 'conversation' is with v8archie. You should definitely listen to his words, they are usually wise. Furthermore I both acknowledged and tried to build on what archie said, when saw that you were facing the same difficulties that beset so many studying this subject. That is why I limited my input, whilst acknowledging his. However sometimes a change of viewpoint helps and this is what I offered you. The new view hopefully offers enlightenment, after which you can return to the formal path. That does not require a long conversation, but an acknowledgement is appreciated.  
April 12th, 2016, 11:54 AM  #22  
Member Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2  Quote:
 
April 12th, 2016, 12:04 PM  #23  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
So anytime you need amplification, just holler. Meanwhile just follow the path. It is always best if you can figure the details out for yourself, once pointed in the right direction.  
April 12th, 2016, 01:44 PM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra  Do you know the difference between potential infinities (as used in real analysis) and the actual infinities of settheory (e.g. $\aleph_0$, $\omega$, etc.).

April 12th, 2016, 08:20 PM  #25 
Member Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2  
April 12th, 2016, 09:04 PM  #26 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
No. Potential infinities are used in analysis  in particular in limit expressions such as $\sum \limits_{n=0}^{\infty} {1 \over 2^n}$. Although the rigorous definition avoids the use of any infinity entirely (as outlined about), we talk about the limit as $n$ goes to infinity. This is a potential infinity because there is no suggestion that $n$ actually attains any infinite value. One reason it never does so is because there are no infinite natural numbers and in the context of an infinite sum, $n$ is a natural number. Another reason is that the definition of a limit never actually uses the value of the expression at that limit. (It is this allows us to define the continuity of a function in terms of limits). The actual infinities of set theory do, however exist (for some definition of existence). There are an infinite number of natural numbers and we denote this quantity as $\aleph_0$. Then there are the ordinal numbers which are particularly applicable to the infinite series we've been looking at. Unfortunately, I'm far from an expert on this stuff, but we can describe the length of the series above as being $\omega$. This doesn't mean that it has $\omega$ terms (it has $\aleph_0$ terms). But $\omega$ is called a "limit ordinal" because it describes the length of the series in the limit. We can write the sum as $\sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n+1}}$. This is just a reordering of the terms (all those with an even index, followed by all those with an odd index). And, because the original series is absolutely convergent, we should get the same result. $$\begin{aligned}\sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n+1}} &= \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over 2^{2n}} \\ &= \sum \limits_{n = 0}^{\infty} {1 \over \left(2^2\right)^n} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over \left(2^2\right)^n} \\ &= \sum \limits_{n = 0}^{\infty} {1 \over 4^n} + \frac12 \sum \limits_{n = 0}^{\infty} {1 \over 4^n} \\ &= {1 \over 1  \frac14} + \frac12 \cdot {1 \over 1  \frac14} \\ &= \frac43 + \frac12 \cdot \frac43 = \frac63 = 2\end{aligned}$$ But looking back at that derivation, we have two infinite sums. There are still $\aleph_0$ terms (the number of terms doesn't change just because we write them in a different order) but now the length of the sum is $\omega + \omega = \omega \cdot 2$. In other words, it is twice as long as before. Freaky. So, all these infinities do work together in a logical way, but each has its own domain and its own meaning. And you can't mix them. Analysis doesn't need (or want) to get bogged down in the mechanics of the sum it is calculating, especially when things get freaky and apparently contradictory. Instead, it's much less bother to avoid the actual infinities all together. We also avoid them because you can't get the these infinities by adding another terms to the sum which is what we do when we say $\sum \limits_{n=0}^{\infty} {1 \over 2^n} = \lim \limits_{n \to \infty} \sum \limits_{k = 0}^n {1 \over 2^k}$. They only exist "in the limit" and the limit expression doesn't actually use that limit at "infinity". Another reason is that neither $\aleph_0$ (the quantity of terms) nor $\omega$ (the length of the sum) are numbers that can be manipulated like $n$ in the sum. $\aleph_0$ doesn't submit to any normal rules of arithmetic and $\omega$ isn't commutative under addition or multiplication, so we definitely don't want them in our series anywhere. Probably I've just confused you more. It's difficult stuff to grasp, and very strange, but also has a practical sort of logic to it. I apologise for any errors in the above, it's my interpretation of what I've been reading. As I said, this is not in my field. Last edited by v8archie; April 12th, 2016 at 09:49 PM. 
April 12th, 2016, 10:39 PM  #27 
Member Joined: Jun 2014 From: Alberta Posts: 56 Thanks: 2 
This is just cool; I don't have any other way to describe it. I feel like I am fortunate enough to know that I am confused. My Background is just firstyear calculus, firstyear linear algebra, and one semester of analysis. But I am obsessed with this stuff anyways. 

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contradiction, natural, numbers, set 
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