Cont: Deeper than primes - Continuation 2

[jsfisher]

I am not talking about terms like countable or non-countable, but about the inherent incompleteness of an inductive set.

Then why did you cite a Wikipedia reference that was talk about non-countability? And what do you believe is the consequence of the "inherent incompleteness of an inductive set"? No, let's stick to the set of natural numbers: What is the consequence of the "inherent incompleteness" of the set of natural numbers?

I am not claiming such thing, exactly because the number of infinite _i's values is greater than any given _i value.

"Infinite _i's values"? All the subscripts are natural numbers, so what is all this?

 

[doronshadmi]

This is not a matter of interpretation. This is a matter of definition. There is, by definition, no "addition" going on in the set posited by the axiom of infinity. It is defined as a set which contains all of its members' successors, not a set which instantly adds those successors to itself whenever you ask.

Your idea of "parallel addition", or whatever it is that you are calling it this time around, remains completely incoherent and useless. Even if it weren't, you still wouldn't be talking about the set defined in the axiom of infinity, because that set explicitly does not behave that way.

Since the term addition causes confusion (it is wrongly interpreted in terms of adding new elements as done in case of finite sets) I have changed this term to "there is always the next member (the successor)" as an inherent property of any inductive set.

Such inherent property can't be found among finite sets.

So from now on, the term addition "gets off stage".

Morevore, instead of using "+" that is used as the operator of addition, that add elements to finite sets, let's use the symbol "|->" in order to represent the notion of "there is always the next member (the successor) as an inherent property of any inductive set".

The notion of aleph₀ can't be used in order to to express this inherent property, since, for example, aleph₀ |-> 1 = aleph₀.

On the contrary infinite large numbers like 1,000,000,000,... can express this inherent property, since, for example,
1 |-> 1,000,000,000,... > 1,000,000,000,... by 1.

B.t.w, by using such notion in reverse we get "there is the previous member (the predecessor) as an inherent property of any infinite set, for example: 1 <-| 1,000,000,000,... < 1,000,000,000,... by 1.

 

[doronshadmi]

Then why did you cite a Wikipedia reference that was talk about non-countability?

Since there is always the next member (the successor) as an inherent property of any inductive set, one can't claim that there is a complete inductive set.

What is the consequence of the "inherent incompleteness" of the set of natural numbers?

Since the set of natural numbers is some particular case of an inductive set, there is always the next member (the successor) as an inherent property that prevents its completeness.

This inherent property is not found among finite sets.

"Infinite _i's values"? All the subscripts are natural numbers, so what is all this?

There is a set of infinitely many finite _i values, such that there is always the next member (the successor) as an inherent property of such set.

This inherent property is not found among finite sets.

 

[doronshadmi]

Let's give some concrete example of how (in the case of an inductive set like natural numbers) there are infinitely many finite values, as follows:

{1  ,  2  ,  3  ,  4  ,  ...}
 +     +     +     +
 1 |-> 1 |-> 1 |-> 1 |-> ...

Each result in the "vertical direction" is some finite value, but there are always the next values (the successors, that are represented here by infinitely many 1's at the "horizontal direction") as an inherent property that prevents the completeness of the set of natural numbers.

 

[doronshadmi]

In other words, given any finite member in the following inductive set:

{1  ,  2  ,  3  ,  4  ,  ...}

there are always the next members (by using the 1's successors) as an inherent property of such set.

Therefore the set of all natural numbers does not exist.

 

[jsfisher]

What is the consequence of the "inherent incompleteness" of the set of natural numbers?

Since the set of natural numbers is some particular case of an inductive set, there is always the next member (the successor) as an inherent property that prevents its completeness.

You didn't answer my question. Instead, you simple circled.

You have added a new term, inherent incompleteness, to doronetics. I simply wish to know that it means. So far, nothing you have said about "inherent incompleteness" differs from "the set has infinitely many members."

This inherent property is not found among finite sets.

It is not found in non-finite sets, either. This "successor" property you allege exists just happens to be an ordering relationship. Sets are not ordered.

Are you requiring that all non-finite sets be ordered sets?

 

[jsfisher]

...
Therefore the set of all natural numbers does not exist.

And therefore you reject the Axiom of Infinity. You know, that axiom that asserts the existence of the set of natural numbers? Yeah, that one.

What would you like in its place?

 

[Nonpareil]

Since the term addition causes confusion (it is wrongly interpreted in terms of adding new elements as done in case of finite sets) I have changed this term to "there is always the next member (the successor)" as an inherent property of any inductive set.

Such inherent property can't be found among finite sets.

So from now on, the term addition "gets off stage".

Morevore, instead of using "+" that is used as the operator of addition, that add elements to finite sets, let's use the symbol "|->" in order to represent the notion of "there is always the next member (the successor) as an inherent property of any inductive set".

The notion of aleph₀ can't be used in order to to express this inherent property, since, for example, aleph₀ |-> 1 = aleph₀.

On the contrary infinite large numbers like 1,000,000,000,... can express this inherent property, since, for example,
1 |-> 1,000,000,000,... > 1,000,000,000,... by 1.

So now you are attempting to use an operator that only functions on sets... on a number.

Just stop.

Besides that, a set of cardinality Aleph-0 can express the idea that an infinite set contains all of its members' successors. That is, in fact, rather what Aleph-0 means. That's what "countably infinite" means, too, in a broader and more general sense.

You just don't understand the concepts you are trying to argue against.

In other words, given any finite member in the following inductive set:

{1  ,  2  ,  3  ,  4  ,  ...}

there are always the next members (by using the 1's successors) as an inherent property of such set.

Therefore the set of all natural numbers does not exist.

Yes, so. As jsfisher said. Outright rejection of the axiom of infinity, and ignoring the entire definition of the set I.

Believe it or not, doron, we do actually understand what you are saying. It is just wrong, and rather silly.

 

[doronshadmi]

This "successor" property you allege exists just happens to be an ordering relationship. Sets are not ordered.

I talking about this part:

" ... the set formed by taking the union of x with its singleton {x} ... "

The existence of singleton {x} for each union of x with its singleton {x}, is equivalent to my use of 1's as successors.

By the standard interpretation, a successor is the union of x with its singleton {x}.

By my non-standard interpretation a successor is the existence of singleton {x} for each union, which guarantees that there are always the next members (by using the 1's successors) as an inherent property of such set.

Therefore the set of all natural numbers does not exist (given x, it always has its singleton as its successor).

 

[jsfisher]

...<snip>...

Don't care, and it doesn't matter since you've nuked the Axiom of Infinity.

What have you in its place, or does doronetics not have infinite sets at all?

 

[Nonpareil]

By my non-standard interpretation a successor is the existence of singleton {x} for each union, which guarantees that there are always the next members (by using the 1's successors) as an inherent property of such set.

Therefore the set of all natural numbers does not exist (given x, it always has its singleton as its successor).

Yes, yes. You said this. Never mind, for the moment, that it ignores that the set defined by the axiom of infinity is simply defined as containing all of its members' successors, since that has been pointed out before and you obviously do not understand or care.

As jsfisher says, you reject the axiom outright. Fine. What does this result in?

 

[doronshadmi]

Don't care, and it doesn't matter since you've nuked the Axiom of Infinity.

No, jsfisher, I simply interpret the meaning of successor differently than its standard meaning.

Therefore my interpretation of the axiom of infinity is simply different than the standard interpretation of it.

By using my non-standard interpretation of successor, one easily understands that the set of all natural numbers does not exist (given x, it always has its singleton as its successor).

EDIT:

I also wish to correct a mistake that was written in my last posts:

Any "the next member (the successor)" has to be replaced by "the next singleton (the successor)".

 

[Little 10 Toes]

Just because you can't list all the members of a set, it doesn't mean the set doesn't exist.

For example: let set X be all numbers between -1 and +1 inclusive. -1>=X>=+1. According to you doronshadmi, set X does not exist.

 

[Nonpareil]

No, jsfisher, I simply interpret the meaning of successor differently than its standard meaning.

Therefore my interpretation of the axiom of infinity is simply different than the standard interpretation of it.

By using my non-standard interpretation of successor, one easily understands that the set of all natural numbers does not exist (given x, it always has its singleton as its successor).

Which means that you have rejected the axiom of infinity.

 

[doronshadmi]

Just because you can't list all the members of a set, it doesn't mean the set doesn't exist.

I do not reject the existence of the non-finite set of natural numbers.

I do reject the existence of the non-finite set of all natural numbers.

 

[doronshadmi]

Which means that you have rejected the axiom of infinity.

Which means that I have rejected the standard interpretation of the axiom of infinity, no less, no more.

 

[jsfisher]

I do not reject the existence of the non-finite set of natural numbers.

I do reject the existence of the non-finite set of all natural numbers.

If the set of natural numbers doesn't contain all the natural numbers, then it isn't the set of natural numbers.

You have rejected the Axiom of Infinity.

The Axiom of Infinity postulates the set of natural numbers. Your feeble attempt at a semantic quibble is a failure.

 

[realpaladin]

Awesome. Still at it.

An axiom or postulate as defined in classic philosophy, is a statement (in mathematics often shown in symbolic form) that is so evident or well-established, that it is accepted without controversy or question. Thus, the axiom can be used as the premise or starting point for further reasoning or arguments, usually in logic or in mathematics. The word comes from the Greek axíōma (ἀξίωμα) 'that which is thought worthy or fit' or 'that which commends itself as evident.'

(Source: Wikipedia)

Bolding mine.

 

[doronshadmi]

This "successor" property you allege exists just happens to be an ordering relationship. Sets are not ordered.

{4  ,  2  ,  1  ,  3  ,  ...}
 +     +     +     +
 1 |-> 1 |-> 1 |-> 1 |-> ...

holds as well, so no order is involved.

As for the singleton as a successor, it is always ahead (a next element that is out of the range) of any member of the set of natural numbers, which actually causes such set to be non-finite AND inherently incomplete, for example:

{{}} u { {{}} }
{{},{{}}} u { {{},{{}}} }
{{},{{}},{{},{{}}}} u { {{},{{}},{{},{{}}}} }
...

etc. ad infinitum

Also in this representation, order is not involved, for example:

{{},{{}},{{},{{}}}} u { {{},{{}},{{},{{}}}} }
{{}} u { {{}} }
{{},{{}}} u { {{},{{}}} }
...

etc. ad infinitum

If the set of natural numbers doesn't contain all the natural numbers, then it isn't the set of natural numbers.

Because of the "next" mechanism of a singleton as a successor, the completeness of the non-finite set of natural numbers is not satisfied, and this is exactly the meaning of being an inductive set (where the non-finite set of natural numbers is some particular case of an inductive set).

The axiom of infinity is exactly this meaning, which guarantees the incompleteness of any inductive set from within (as its inherent property).

 

[jsfisher]

...snip of irrelevant Gish Gallop...

If the set of natural numbers doesn't contain all the natural numbers, then it isn't the set of natural numbers.

Because of the "next" mechanism of a singleton as a successor, the completeness of the non-finite set of natural numbers is not satisfied, and this is exactly the meaning of being an inductive set (where the non-finite set of natural numbers is some particular case of an inductive set).

You seemed to have missed this part in what I wrote: If the set of natural numbers doesn't contain all the natural numbers, then it isn't the set of natural numbers.