Question on Dedekind Cuts

Oct 2009
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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.
 
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Mar 2015
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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.
 
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v8archie

Math Team
Dec 2013
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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 [email protected] understands the concept in his post above: #21.
 
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Mar 2015
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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.
 
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Jun 2014
650
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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.
 
Mar 2015
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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.
 
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v8archie

Math Team
Dec 2013
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If I can construct a cut for each n of Sn, I can construct it for all n, ie, for the endless decimal
:giggle::bounce::giggle:

:redcard: