My Math Forum Question on Dedekind Cuts

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January 19th, 2019, 10:18 AM   #21
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 Originally Posted by AplanisTophet Numbers like $\pi$ are computable, so I believe we could assert the sets $A$ and $B$ given there is a formula that the axiom schema of separation could utilize to do so. My question centers around using Dedekind cuts to ‘construct’ real numbers when in reality we’re just assuming the existence of the reals and then embedding them in $\mathcal{P}(\mathbb{Q})$. It’s the uncountably many reals that we have no way of computing, or even defining if restricting our definitions to that which could be created using only finite sentences from a finite formal language, that leads me to question the purpose of Dedekind cuts and what we mean when we generally assert that we use them to construct the reals.
It means you construct the set of all reals in one go. The set of reals is definable and constructable. Individual real numbers are not.
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
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Quote:
 Originally Posted by zylo How do you specify a cut for Pi
Why would you have to?

January 19th, 2019, 12:23 PM   #24
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 Originally Posted by Micrm@ss Why would you have to?
If you can't, cuts don't define the real numbers. They are just jargon, but then they don't have anything to do with analysis.

January 19th, 2019, 12:45 PM   #25
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I did, yes, and thank you. I’m just responding to zylo.

January 19th, 2019, 01:50 PM   #26
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 Originally Posted by AplanisTophet I'm game if you can show me how to compute, or even define, all of these monotonically increasing sequences you speak of ...
Erm, obviously it's not possible to do that, otherwise the numbers wouldn't be non-computable.

On the other hand, if you believe that non-computable 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
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Quote:
 Originally Posted by zylo 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.
Each cut you describe would be for a rational number. Jumping from having a cut for each rational number that is the sum after $n$ ‘terms’ of a nonterminating decimal to having a cut for the nonterminating decimal itself is an error in your logic, however. I’ll assume your nonterminating decimal is not computable. You could derive a cut for the nonterminating decimal if you had a function for computing the rationals you speak of, but since you lack a formula for deriving your sequence of rationals (due to the nonterminating decimal being noncomputable), you are unable to derive a cut for the nonterminating decimal.

 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
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Quote:
 Originally Posted by zylo If I can construct a cut for each n of Sn, I can construct it for all n, ie, for the endless decimal

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