Cont: Deeper than primes - Continuation 2

[doronshadmi]

Which ones are missing? In the set of all natural numbers, which ones are not there?

Sine you determine, in the first place, that there is such a thing like the set of all natural numbers, your question is meaningless.

In fact, the set of natural numbers defines what numbers are the natural numbers. Were any missing (that is, not in the set of natural numbers), then they wouldn't be natural numbers, by definition.

Sine I do not determine such a thing, in the first place, some natural number in a non-finite set with 1...+1 natural numbers, is not a member of the non-finite set with 1... natural numbers, and since 1... < 1+1... by 1 there is no bijection between these two non-finite sets of natural numbers.

So only natural numbers are involved.

For further explanation please look at http://www.internationalskeptics.com/forums/showpost.php?p=11259515&postcount=1192.

 

[Nonpareil]

The infinite number 1... < 1+1... by 1, where in both cases an infinite number is involved (or as you put it, it "leaves you with infinity").

Once again, the transfinite numbers system is not my infinite numbers system.

Repeating nonsense does not make it true.

Sine you determine, in the first place, that there is such a thing like the set of all natural numbers, your question is meaningless.

Again, you have rejected the axiom of infinity. The axiom infinity says that the set of all natural numbers exists, and is infinite in size.

Sine I do not determine such a thing, in the first place, some natural number in a non-finite set with 1...+1 natural numbers, is not a member of the non-finite set with 1... natural numbers, and since 1... < 1+1... by 1 there is no bijection between these two non-finite sets of natural numbers.

So only natural numbers are involved.

For further explanation please look at http://www.internationalskeptics.com/forums/showpost.php?p=11259515&postcount=1192.

I'm impressed. That's gibberish even by your usual standards.

Still utterly wrong about everything to do with infinite sets, though.

 

[doronshadmi]

Repeating nonsense does not make it true.

You indeed have to say it to yourself, about your notions of infinite sets.

 

[doronshadmi]

The axiom infinity says that the set of all natural numbers exists, and is infinite in size.

The axiom infinity says that any inductive set is inherently incomplete, because a successor is a singleton and not some member of an inductive set.

 

[jsfisher]

The axiom infinity says that any inductive set is inherently incomplete, because a successor is a singleton and not some member of an inductive set.

Really? Perhaps you could show us where. Here is the axiom again:

I would particularly interested in the stipulation requiring members be written down before they are members.

Take your time.


ETA: Oh, drat!! That disjunction should have been a conjunction, both here and the previous post of the axiom. Now corrected here.

 

[doronshadmi]

Really? Perhaps you could show us where. Here is the axiom again:

[qimg]http://www.sciweavers.org/tex2img.php?eq=%5Cexists%20x%20%5C%2C%20%28%20%5Cemptyset%20%5Cin%20x%20%5C%2C%20%5Cwedge%20%5C%2C%20%5Cforall%20y%20%5Cin%20x%20%5C%2C%20%28%28y%20%5Ccup%20%5C%7By%5C%7D%29%20%5Cin%20x%29&bc=White&fc=Black&im=jpg&fs=12&ff=arev&edit=0[/qimg]


I would particularly interested in the stipulation requiring members be written down before they are members.

A new member of an inductive set can't be written (can be explicitly defined) unless some member is unioned with a singleton that is defined as a successor (where a singleton is not a member of the considered inductive set).

In the axiom of infinity the concept of singleton as a successor is represented by the "{y}" notation, which is not a member of the defined inductive set.

 

[jsfisher]

...irrelevant gibberish snipped...

In the axiom of infinity the concept of singleton as a successor is represented by the "{y}" notation, which is not a member of the defined inductive set.

Why are you focused on {y} (for arbitrary y, a member of the set)? The Axiom of Infinity is silent on {y}'s membership (with the notable exception of {{}}).

(And no, the axiom makes no mention of "successor", but if you wanted to adopt a meaning appropriate for, say, the natural numbers, the successor of y is not {y}, so you ranting about {y} is odd.)

 

[jsfisher]

A new member of an inductive set...

..other irrelevant gibberish snipped...

There are no new members, just members.

 

[doronshadmi]

Why are you focused on {y} (for arbitrary y, a member of the set)?

Because {y} is taken by me as a general representation of a successor, that is not a member of an inductive set.

One does not have to explicitly use the word "successor" in order to understand that {y} is exactly this concept among an inductive set.

 

[doronshadmi]

There are no new members, just members.

I accept this if only a single inductive set is considered.

EDIT:
But if one inductive set of the same kind of members has 1...+1 members and the other inductive set of the same kind of members has 1... members, there is one more member in the set of 1...+1 members, which is new and inaccessible to the rest of the members of the set with 1... members.

 

[doronshadmi]

The Axiom of Infinity is silent on {y}'s membership (with the notable exception of {{}}).

Being silent on {y}'s (non)membership (with the notable exception of {{}}, if {} is considered as a member) is the standard notion of {y} in this axiom.

I reject this silence about {y}.

 

[doronshadmi]

...the successor of y is not {y}...)

Well, this is the standard notion about {y}, and I disagree with it.

Again, in order to understand it, one first has to understand the notion of a singleton as a successor (for example http://www.internationalskeptics.com/forums/showpost.php?p=11259142&postcount=1179 , http://www.internationalskeptics.com/forums/showpost.php?p=11259373&postcount=1181 and http://www.internationalskeptics.com/forums/showpost.php?p=11259463&postcount=1185).

 

[jsfisher]

Because {y} is taken by me as a general representation of a successor, that is not a member of an inductive set.

So, you decide to label a meaningless something in a way contrary to any normal convention and then draw unsupported conclusions from it. I see.

None of that is in the axiom, so it is little more than what you want, not what actually is.

 

[jsfisher]

...if {} is considered as a member...

Is your misunderstanding of the axiom that great that you feel obligated to express this with a conditional?

 

[jsfisher]

I accept this if only a single inductive set is considered.

What you are willing to accept is irrelevant. The axiom is what it is; it postulates the existence of a set, not a time series for its construction.

The set has members. None of the members are new; none are old; they are just members.

 

[doronshadmi]

So, you decide to label a meaningless something in a way contrary to any normal convention and then draw unsupported conclusions from it. I see.

None of that is in the axiom, so it is little more than what you want, not what actually is.

{y} as a singleton successor clearly appears in the axiom of infinity.

You simply reject this notion by get y u {y} as a successor in the case of the set of natural numbers.

 

[doronshadmi]

What you are willing to accept is irrelevant.

What you are willing to not accept is irrelevant.

 

[doronshadmi]

Is your misunderstanding of the axiom that great that you feel obligated to express this with a conditional?

It depends if there is an empty member, or not.

If the inductive set is a set of natural numbers, than {} is generally not included.

 

[doronshadmi]

not a time series for its construction

Time is not involved here, maybe only in your imagination.

Also order is not involved, as clearly shown in http://www.internationalskeptics.com/forums/showpost.php?p=11259142&postcount=1179.

 

[jsfisher]

It depends if there is an empty member, or not.

Which other parts of the Axiom of Infinity do you not understand.

If the inductive set is a set of natural numbers, than {} is generally not included.

You've heard of John von Neumann, yes?


Be that as it may, we are still at the point where you claim the set of natural numbers does not exist. No Axiom of Infinity for you despite your desperate attempts to find things in the axiom that aren't there.