March 6th, 2018, 04:26 AM  #21  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra 
Unfortunately you've stopped listening to what I say. You are arguing against straw men that run counter to points I have made explicitly. I'll give it one last go. Quote:
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Everthing we know of the infinite is the result of logical deduction, just like our knowledge of the cosmos and subatomic particles. In point of fact, nothing here requires the infinite. We only need the natural numbers. If you accept that they are unbounded (whether or not you accept the infinite), the rest is pure logic. Quote:
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There is no need for infinite actions because the logical steps allow us to aggregate over huge domains. It's no different to saying that $x^2 \ge 0$ for all $x$. Quote:
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Here's a question: how can you claim any knowledge of "infinite sums" if you deny that anyone can know anything about the "infinite"? Quote:
No, I don't. A limit is a limit. It's a number. A sequence is not a number. Not at all. A limit has a definition that doesn't refer to summation and doesn't even refer to the end point. A limit expressly says nothing about an "infinite sum".  
March 6th, 2018, 04:38 AM  #22 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra  Because it's just a representation of the number. The number is the abstract thing that sits behind the representations $2,\frac63,2.000\ldots$ or the calculation of the limit of a particular sequence. The representation 2.000 doesn't mean anything without having a mechanism to assign a value to it. And simply saying "we add up all the terms of the implied series" doesn't work because we can't add infinitely many terms. The best we can do is to calculate the limit of the partial sums. Therefore we arrive at the point where the limit is the number. Last edited by v8archie; March 6th, 2018 at 04:42 AM. 
March 6th, 2018, 07:15 AM  #23 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 227 Thanks: 77 Math Focus: Algebraic Number Theory, Arithmetic Geometry  I'm a little confused by this: it seems like you're defining real numbers in terms of something that itself needs to be defined. In particular, how are you defining limits of cauchy rational sequences that don't have rational limits?
Last edited by cjem; March 6th, 2018 at 07:25 AM. 
March 6th, 2018, 07:40 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra 
I don't understand how you can use the term Cauchy sequences and not know how their limits are defined?

March 6th, 2018, 08:09 AM  #25  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 227 Thanks: 77 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
We can certainly define what it means for rational sequences to have a rational limit without mentioning the reals. But then the issue is that most cauchy sequences of rationals (and all of the ones we're interested in here) don't actually have limits in this sense. Do you have a definition for the limit of a rational sequence that doesn't reference the reals and that leads to every cauchy sequence having a limit? I suspect to come up with one and make sure it's meaningful, you'd have to essentially do the hard work of any other construction of the reals. But I'd be very happy if you could show me wrong! Last edited by cjem; March 6th, 2018 at 08:28 AM.  
March 6th, 2018, 09:11 AM  #26  
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  Quote:
Let $(s_n)$ and $(t_n)$ be Cauchy sequences of rationals. Suppose that for all $\epsilon$ there exists $N$ such that for all $m, n > N$, we have $s_m  t_n < \epsilon$. This is an equivalence relation on rational Cauchy sequences, and the set of equivalence classes are ... voilĂ ! The reals. All this needs proof of course, just epsilon chasing. https://en.wikipedia.org/wiki/Constr...uchy_sequences Last edited by Maschke; March 6th, 2018 at 09:20 AM.  
March 6th, 2018, 09:16 AM  #27  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 227 Thanks: 77 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
If you notice, however, v8archie proposed an alternative definition (see what I quoted in post #23) to this  I'm simply trying to see if it holds any water. Last edited by cjem; March 6th, 2018 at 09:28 AM.  
March 6th, 2018, 09:22 AM  #28 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,403 Thanks: 2477 Math Focus: Mainly analysis and algebra 
And all sequences in the equivalence class have the same limit: a real number. Thus every limit of a Cauchy sequence of rationals is a real number.

March 6th, 2018, 09:25 AM  #29  
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  Quote:
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In other words if we have the rational Cauchy sequence $1, 1.4, 1.41, 1.414, \dots$, we would LIKE to say but we CAN NOT say "Well $\sqrt{2}$ is just that limit, can't you see??" We can't say that because there is a hole in the rationals right where we imagine there should be a real. [Again, for v8, not you]. It is a theorem that every real is the limit of a sequence of rationals, once you've defined the reals purely in terms of the rationals. Last edited by Maschke; March 6th, 2018 at 10:00 AM.  
March 6th, 2018, 09:53 AM  #30 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 227 Thanks: 77 Math Focus: Algebraic Number Theory, Arithmetic Geometry  The issue is that the usual definition of a limit makes use of real numbers (at least implicitly). So defining real numbers in terms of these limits is completely circular.


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0.999, 0.999..., 0.999...=1, arguments, fail, finitism, real numbers, usual 
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