January 17th, 2019, 02:52 PM  #11  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra  Quote:
The construction is similar to the definition of the limit. For each $a < x$ we can find an $N$ such that $a < c_n$ for all $n > N$ precisely because for each $\epsilon > 0$ there exists an $N$ such that $(x  c_n) < \epsilon$ for all $n > N$. (Note that $(xc_n)$ is nonnegative because the sequence $(c_n)$ is monotonically increasing with limit $x$). Thus, for every $a < x$ we can pick $\epsilon < (xa)$ and thus find our value of $N$. "All sufficiently large $n$" then refers to "all $n > N$".  
January 17th, 2019, 07:44 PM  #12  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  Quote:
 
January 17th, 2019, 08:10 PM  #13  
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706  Quote:
The problem is that we don't yet know that such a thing exists. You are assuming that which must be proven. Given the rationals we have to show that we can embed them in a complete metric space. Until we do that, there is no such $x$. Last edited by Maschke; January 17th, 2019 at 08:43 PM.  
January 17th, 2019, 10:55 PM  #14 
Senior Member Joined: Oct 2009 Posts: 783 Thanks: 280  It's hopelessly circular too. Since a complete METRIC space already uses R. Indeed, the very definition of a metric is a function with codomain R. So to define completions or complete metric spaces uses R. And so defining Q as the completion of R, is circular.

January 18th, 2019, 03:08 AM  #15  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,657 Thanks: 2635 Math Focus: Mainly analysis and algebra  Quote:
Your cut for $\sqrt2$ used $\sqrt2$ as much as mine used $x$. You only picked your definition for $A$ and $B$ because the polinomial $x^22$ has a root equal to $\sqrt2$. Quote:
Remember that the original question was not about proving any properties of the reals, it was about how one might define Dedekind cuts for the noncomputable numbers. My answer is that you don't need to worry about how to define them individually if you have a plan that includes all of them. Last edited by v8archie; January 18th, 2019 at 03:12 AM.  
January 18th, 2019, 11:37 AM  #16  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  Quote:
I'm game if you can show me how to compute, or even define, all of these monotonically increasing sequences you speak of without using $x$ in the computation and/or definition.  
January 19th, 2019, 07:21 AM  #17 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
I think I see point of OP. Interesting. For example, what is the cut for Pi? You can't work with all members of a convergent sequence for Pi, and how do you know it is exactly Pi? The only way of knowing is if you had an independent mathematical definition for Pi. I'm guessing the Axiom of Separation* says you can pick a subset out of any set. It's only interesting for an infinite set when for example you say you can make a cut for any member of an infinite set, in which case it might apply to OP in the sense that, since the set of rational numbers is infinite, you could never create a cut for every member. Not my cup of tea. I'm perfectly happy with: if something is true for any n, it is true for all n. {just trying to get a feeling for it.} *googled EDIT Sorry about that huge icon, don't know how to limit it to an url. If you can't create a cut for Pi, that invalidates cuts as a definition of the real numbers. A cut is a real number. Prove all real numbers are cuts. But real numbers are defined by cuts, so every cut is a real number. Circular. Last edited by zylo; January 19th, 2019 at 07:31 AM. 
January 19th, 2019, 09:36 AM  #18  
Senior Member Joined: Jun 2014 From: USA Posts: 525 Thanks: 40  Quote:
 
January 19th, 2019, 10:11 AM  #19  
Senior Member Joined: Aug 2012 Posts: 2,308 Thanks: 706  Quote:
 
January 19th, 2019, 10:18 AM  #20 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
What is the cut for Pi?


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