My Math Forum Question on Dedekind Cuts

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January 17th, 2019, 02:52 PM   #11
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Quote:
 Originally Posted by AplanisTophet In this case, $c_n$ would have to equal $x$ (where $n$ is sufficiently large as you state). If it didn't, the sets $A$ and $B$ would not form a Dedekind cut for $x$. Rather, they would form a Dedekind cut for $c_n$.
Which $c_n$ are you referring to? My construction refers to all of them (or at least infinitely many of them).

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 $(x-c_n)$ is non-negative because the sequence $(c_n)$ is monotonically increasing with limit $x$). Thus, for every $a < x$ we can pick $\epsilon < (x-a)$ and thus find our value of $N$. "All sufficiently large $n$" then refers to "all $n > N$".

January 17th, 2019, 07:44 PM   #12
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Quote:
 Originally Posted by v8archie Which $c_n$ are you referring to? My construction refers to all of them (or at least infinitely many of them). 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 $(x-c_n)$ is non-negative because the sequence $(c_n)$ is monotonically increasing with limit $x$). Thus, for every $a < x$ we can pick $\epsilon < (x-a)$ and thus find our value of $N$. "All sufficiently large $n$" then refers to "all $n > N$".
You’re using $x$ in the definition of your sequence, which is then used to define $A = \{a \in \mathbb{Q} : \exists c_n \in \mathbb{R}(a < c_n < x)\}$. You might as well just use $x$ in the definition of $A$ in that case: $A = \{a \in \mathbb{Q} : a < x \}$.

January 17th, 2019, 08:10 PM   #13
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Quote:
 Originally Posted by v8archie I would suggest that, given that there exists a monotonically increasing sequence $(c_n)$ such that $\displaystyle \lim_{n \to \infty} c_n = x$ ...
Unless I am missing your point, I can't say I agree with or even understand this. All you have is the rationals. What is $x$ if you intend to find a cut representing an irrational?

Quote:
 Originally Posted by v8archie Given that the reals are the completion of the rationals ...
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
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Quote:
 Originally Posted by Maschke 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 $x$.
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
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Quote:
 Originally Posted by AplanisTophet You’re using $x$ in the definition of your sequence, which is then used to define $A = \{a \in \mathbb{Q} : \exists c_n \in \mathbb{R}(a < c_n < x)\}$. You might as well just use $x$ in the definition of $A$ in that case: $A = \{a \in \mathbb{Q} : a < x \}$.
No, I'm merely observing that there exists a sequence with the properties given. If you are looking to construct the reals, you follow the procedure with all such monotonically increasing sequences. You don't need to pick out any in particular.

Your cut for $\sqrt2$ used $\sqrt2$ as much as mine used $x$. You only picked your definition for $A$ and $B$ because the polinomial $x^2-2$ has a root equal to $\sqrt2$.

Quote:
 Originally Posted by Maschke The problem is that we don't yet know that [a completion of the rationals] exists.
How is that an obstacle to implementing the procedure? I don't see why you insist that one needs to prove that completion exists: surely the aim of Dedekind cuts is purely to to create a set of objects. You can then show later that they are a completion of the rationals.

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 non-computable 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
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Quote:
 Originally Posted by v8archie Your cut for $\sqrt2$ used $\sqrt2$ as much as mine used $x$. You only picked your definition for $A$ and $B$ because the polinomial $x^2-2$ has a root equal to $\sqrt2$.
Yeah, but $\sqrt{2}$ is computable whereas we agreed $x$ was not, so that's comparing apples to oranges.

Quote:
 Originally Posted by v8archie No, I'm merely observing that there exists a sequence with the properties given. If you are looking to construct the reals, you follow the procedure with all such monotonically increasing sequences. You don't need to pick out any in particular.
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: 126 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
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Quote:
 Originally Posted by zylo 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.
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.

January 19th, 2019, 10:11 AM   #19
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Quote:
 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.

 January 19th, 2019, 10:18 AM #20 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 What is the cut for Pi?

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