January 19th, 2019, 10:18 AM  #21  
Senior Member Joined: Oct 2009 Posts: 867 Thanks: 330  Quote:
It is easier to construct the set of reals, than to construct specific real numbers. As a (very close) analogy, assume that natural numbers $\mathbb{N}$. The power set $\mathcal{P}(\mathbb{N})$ kind of exists by axiom. In either case, it is a definable set. That doesn't mean we can actually exhibit all elements of $\mathcal{P}(\mathbb{N})$ easily. We can't, since many of these elements are undefinable. Much of the same considerations hold for the real numbers.  
January 19th, 2019, 11:10 AM  #22 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
How do you specify a cut for Pi, like you do for 2$\displaystyle ^{1/2}$? How do you know Pi = C/d is even a real number? Edit: If the set of reals are defined by cuts, and you can't construct all real numbers with them, they are meaningless. You can do the same thing by repeatedly dividing a line by 10 to come up with a decimal representation of 2$\displaystyle ^{1/2}$. If you knew a correct series for Pi, Then you could construct a cut for Sn and then show the cuts approach a limit which is also a cut. Ahha! for each n Sn is a cut. Therefore Sn is a cut for all n. Last edited by zylo; January 19th, 2019 at 11:32 AM. 
January 19th, 2019, 11:56 AM  #23 
Senior Member Joined: Oct 2009 Posts: 867 Thanks: 330  
January 19th, 2019, 12:23 PM  #24 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  
January 19th, 2019, 12:45 PM  #25 
Senior Member Joined: Jun 2014 From: USA Posts: 574 Thanks: 44  
January 19th, 2019, 01:50 PM  #26  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra  Quote:
On the other hand, if you believe that noncomputable real numbers exist, there must exist monotonically increasing sequences of which they are the limit. Thus following the procedure on all such sequences must construct all real numbers. I think Microm@ss understands the concept in his post above: #21. Last edited by v8archie; January 19th, 2019 at 01:57 PM.  
January 21st, 2019, 06:06 AM  #27 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
There are an endless number of irrational numbers. How do you construct or define them all with a Dedekind Cut unless you have a definable property (ie, 2$\displaystyle ^{1/2}$) to compare with rational numbers? An arbitrary Dedekind Cut is meaningless. It doesn't define anything. Edit: You could do the following. If Sn is the sum after n terms of an endless decimal, you could construct a cut for each n, therefore for the endless decimal, which is kind of academic, because it doesn't prove the cut is unique. Last edited by zylo; January 21st, 2019 at 06:18 AM. 
January 21st, 2019, 08:34 AM  #28  
Senior Member Joined: Jun 2014 From: USA Posts: 574 Thanks: 44  Quote:
 
January 21st, 2019, 09:31 AM  #29 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
If I can construct a cut for each n of Sn, I can construct it for all n, ie, for the endless decimal, ie, for the real number defined by the endless decimal; In principle, not in practice, because I could never get to the end. Just like I can define a sum for .333.....3 to n decimal places, for all n, but could never come up with an actual number because I could never get to the end. There are an endless number of irrational numbers. How do you construct or define them with a Dedekind cut unless you have a definable property to compare with rational numbers? For an epsilon delta sum, epsilon is always specified as >0 (if the author is honest) because you can't find an n for epsilon = 0 because the sum never ends. Each term increases the sum, no matter how far you go. Sn of .999..........9 increases with every addition of a 9 for all n but is bounded above by 1. It's so much easier to accept that the real numbers are defined by decimal sequences, limited or unlimited, and derive the consequences for analysis from that. The terms of a standard series such as for 2$\displaystyle ^{1/2}$ are rational so Sn is rational no matter how many terms you take. EDIT By the way, it follows from my decimal definition of real numbers that they are complete, ie, there is no infinite decimal that is not in the definition. Obviously, there are other numbers beside the rationals, which are either repeating or finite decimals. Last edited by zylo; January 21st, 2019 at 09:38 AM. 
January 21st, 2019, 09:38 AM  #30 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra  

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