Cont: Deeper than primes - Continuation 1/3*9

doronshadmi said:
D is a placeholder for any given set.

∃ V ( ∅ ∈ V ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) )
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That is not a specific set. E.g., it is undetermined if {{∅}} is a member of V.

Code: 
V={
...  
}

and it does not have the largest member.
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What does "largest member" mean?

Definition 1: Cardinality is the 'size' of D iff |D| is defined by V.
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(1) You still have that bogus "iff" in there. (2) How would |D| be defined by V? Isn't that what you are trying to define?
 
jsfisher said:
That is not a specific set. E.g., it is undetermined if {{∅}} is a member of V.
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This is a member of Zermelo ordinals.

Please demonstrate how this member (or any other member of Zermelo ordinals) is also one of the members of V, which are constructed in terms of von Neumann ordinals ( for any given x ∈ V ( ( x ∪ { x } ) ∈ V )).


jsfisher said:
What does "largest member" mean?
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It means that V has the most complex member of von Neumann construction of sets.

But V does not have such member.


jsfisher said:
(1) You still have that bogus "iff" in there. (2) How would |D| be defined by V? Isn't that what you are trying to define?
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V is defined by von Neumann construction of sets.

V has at least two basic properties:

(1) Any given member is finite.

(2) It does not have the most complex member of von Neumann construction of sets.

By (1) finite cardinality is defined in terms of natural numbers.

By (2) the search for some particular finite cardinality is (potentially) non-finite since the term "all" is not satisfied (our search continues endlessly among V members, and this is exactly the reason why ω as the limit ordinal of "all" smaller ordinals (which are V members) is not established.

Since ω is not established, the difference between cardinal numbers and ordinal numbers (as done by Cantor's Transfinite system, which forces sets to be defined in terms of actual infinity (for example: "The set of all natural numbers exists")) is not established.
 
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doronshadmi said:
This is a member of Zermelo ordinals.

Please demonstrate how this member (or any other member of Zermelo ordinals) is also one of the members of V, which are constructed in terms of von Neumann ordinals ( for any given x ∈ V ( ( x ∪ { x } ) ∈ V )).
Click to expand...

You presented two requirements for your set V. (1) It contains the empty set, and (2) for every member of V, V also contains that member's "successor".

Your requirements specify things the set V must have, but they do not identify anything as something the set must not have. They neither include nor exclude {{∅}} as a member of set V.

It means that V has the most complex member of von Neumann construction of sets.
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You are offering substitute words without conveying any meaning. What does "most complex member" mean?


For that matter, what bearing does having or not having a largest member have on this discussion? If there were a set which was both infinite and did have a "largest member" (by some appropriate definition of largest member), then what? What if there were set that had no largest member but was finite, then what?
 
Little 10 Toes said:
Posted only for doronshadmi's convenience (bolds and italics added):

∃ I ( ∅ ∈ I ∧ ∀ x ∈ I ( ( x ∪ { x } ) ∈ I ) )
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I always thought playing with the spacing a bit helped, too:
∃I (∅∈I ∧ ∀x∈I ((x∪{x})∈I))​
[/pedant]
 
jsfisher said:
You presented two requirements for your set V. (1) It contains the empty set, and (2) for every member of V, V also contains that member's "successor".

Your requirements specify things the set V must have, but they do not identify anything as something the set must not have. They neither include nor exclude {{∅}} as a member of set V.
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For every member of V, V contains that member's "successor" that is defined by the union of x with {x} (written as x ∪ {x}).

Please define {{{}}} as the union of x with {x}.

jsfisher said:
You are offering substitute words without conveying any meaning. What does "most complex member" mean?
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As seen in case of V members' construction, they become more and more complex (where the least complex member is {}), yet V does not have the most complex member, by this members' construction.

jsfisher said:
For that matter, what bearing does having or not having a largest member have on this discussion? If there were a set which was both infinite and did have a "largest member" (by some appropriate definition of largest member), then what?
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Please define such set, but please remember that its members are pure sets (constructed only by "{"..."}" pairs, at the base level of sets' representations).


jsfisher said:
What if there were set that had no largest member but was finite, then what?
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Again, please think about non-empty sets that its distinct members are pure sets (constructed only by "{"..."}" pairs, at the base level of sets' representations).

jsfisher said:
The Axiom of Infinity doesn't identify any particular set; it doesn't provide a membership function. It merely states two properties the set has (i.e. that it contains the empty set and that every member of the set also has its successor as a member).

The Axiom is silent on whether, for example, {{{{ }}}} is in the set.

The Axiom must to be coupled with other axioms to conclude von Neumann's ordinal is a set in ZF.
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By ∃ V ( ∅ ∈ V ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) ) we get V members.

By ∃ X ( ∅ ∈ X ∧ for any given x ∈ X ( { x } ∈ X ) ) we don't get V members (except ∅ that is not a successor in both constructions, and it is specifically included in both sets).

You see jsfisher, you take V as a placeholder for any set, but V has a unique construction, which is different than X unique construction.

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=13187379&postcount=179 .

As you can see there, D is a placeholder for any set, but V is a unique set that is used to define cardinality, whether it is finite or (potentially) non-finite (since its largest member does not exist).
 
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doronshadmi said:
For every member of V, V contains that member's "successor" that is defined by the union of x with {x} (written as x ∪ {x}).

Please define {{{}}} as the union of x with {x}.
Click to expand...

The requirement you have cited does not rule out {{∅}} from being in V. It does guarantee that if {{∅}} be a member, then so must { {{∅}}, {{{∅}}} }, but it is silent as to whether {{∅}} is a member in the first place.

As seen in case of V members' construction....
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You use the term, construction, as if the set comes into existence by a sequence of steps. It does not.
 
jsfisher said:
I always thought playing with the spacing a bit helped, too:
∃I (∅∈I ∧ ∀x∈I ((x∪{x})∈I))​
[/pedant]
Click to expand...

I prefer the other one. It's harder to distinguish symbols like } and ) when they are right next to each other than when they are spaced a bit apart. I also find it easier to tell the structure of a proposition when it's spaced out a bit.
 
caveman1917 said:
I prefer the other one. It's harder to distinguish symbols like } and ) when they are right next to each other than when they are spaced a bit apart. I also find it easier to tell the structure of a proposition when it's spaced out a bit.
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Yeah, that is all true. The default font size this forum uses doesn't help much either.

Here it is at SIZE = 5: ∃I (∅∈I ∧ ∀x∈I ((x∪{x})∈I))

I also considered substituting brackets for parenthesis at one place to alternate the grouping symbols.

Something like this: ∃I (∅∈I ∧ ∀x∈I [(x∪{x})∈I])

The bold isn't all that helpful at SIZE = 5, but braces are more obvious.

Switching to Times New Roman: ∃I (∅∈I ∧ ∀x∈I [(x∪{x})∈I])


YMMV
 
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caveman1917 said:
I prefer the other one. It's harder to distinguish symbols like } and ) when they are right next to each other than when they are spaced a bit apart. I also find it easier to tell the structure of a proposition when it's spaced out a bit.
Click to expand...

Never thought I would ever say this about me....






The Nominate button is right down there...
vvvvvvvv
 
I notice doronshadmi dodged at least two direct questions:
jsfisher said:
You are offering substitute words without conveying any meaning. What does "most complex member" mean?
Click to expand...
and
For that matter, what bearing does having or not having a largest member have on this discussion? If there were a set which was both infinite and did have a "largest member" (by some appropriate definition of largest member), then what? What if there were set that had no largest member but was finite, then what?
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(original quote edited)

Is there a reason you can't/won't answer these questions doronshadmi?
 
jsfisher said:
The requirement you have cited does not rule out {{∅}} from being in V. It does guarantee that if {{∅}} be a member, then so must { {{∅}}, {{{∅}}} }, but it is silent as to whether {{∅}} is a member in the first place.
Click to expand...

∃ V ( ∅ ∈ V ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) ) has the following members according to its given construction:
Code: 
V={
    
       ∅, 
       {∅}, 
       { ∅, {∅} }, 
       { ∅, {∅} , {∅, {∅}} }, 
       { ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, 
... 

}

∃ X ( ∅ ∈ X ∧ for any given x ∈ X ( { x } ∈ X ) ) has the following members according to its given construction:
Code: 
X={
    
       ∅, 
       {∅}, 
       {{∅}}, 
       {{{∅}}}, 
       {{{{∅}}}}, 
... 

}


jsfisher said:
You use the term, construction, as if the set comes into existence by a sequence of steps. It does not.
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No steps (from the Platonist point of view) the unique construction of V or X sets is taken at once (or in parallel, if you wish).

From the, so called, constructionist point of view, it may be taken serially.

Some common properties of both V and X sets are:

1) No successor of x is a member of x (in order to be considered as a successor, in the first place)

2) Any given member of X or Y is a finite set.
 
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Some corrections of my previous post:

2) Any given member of V or X is a finite set.

3) No successor is the largest member of V or X.
 
jsfisher said:
Yeah, that is all true. The default font size this forum uses doesn't help much either.

Here it is at SIZE = 5: ∃I (∅∈I ∧ ∀x∈I ((x∪{x})∈I))

I also considered substituting brackets for parenthesis at one place to alternate the grouping symbols.

Something like this: ∃I (∅∈I ∧ ∀x∈I [(x∪{x})∈I])

The bold isn't all that helpful at SIZE = 5, but braces are more obvious.

Switching to Times New Roman: ∃I (∅∈I ∧ ∀x∈I [(x∪{x})∈I])


YMMV
Click to expand...

Yes if you increase the font size your version is better as it "groups" terms together. But with the default font size putting symbols right next to each other makes it less readable.
 
Little 10 Toes said:
Never thought I would ever say this about me....






The Nominate button is right down there...
vvvvvvvv
Click to expand...

Do mathematical languages count for the language awards? Like nominating your post for the most readable rendition of the existence of N without needing to increase font size?
 
doronshadmi said:
∃ V ( ∅ ∈ V ∧ for any given x ∈ V ( ( x ∪ { x } ) ∈ V ) )
Click to expand...

That would be the Axiom on Infinity. It stipulates the existence of a set (that's the ∃V part) with two properties. It does not, however, tell us specifically what that set is.

...has the following members according to its given construction
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No, there no construction, neither step-by-step nor all at once.


Doronshadmi, I have a bag of marbles. All of my blue marbles are in the bag. Is there a red marble in the bag?

Answer: From the information I've given you, you cannot tell.

Nor can you tell if {{∅}} be a member of the set stipulated by the Axiom of Infinity.
 
I notice doronshadmi dodged at least two direct questions:
jsfisher said:
You are offering substitute words without conveying any meaning. What does "most complex member" mean?
Click to expand...
and
For that matter, what bearing does having or not having a largest member have on this discussion? If there were a set which was both infinite and did have a "largest member" (by some appropriate definition of largest member), then what? What if there were set that had no largest member but was finite, then what?
Click to expand...

(original quote edited)

Is there a reason you can't/won't answer these questions doronshadmi, yet again?
 
jsfisher said:
I apologize. It seems I have introduced some confusion I did not intend by using the term "von Neumann ordinal" in a non-standard way.

The Axiom of Infinity establishes the existence of a set with certain properties. The set contains the empty set and what I will call other base elements. The set contains the successors of all of its elements (which would be its base elements, and their successors, and their successors, and ...).

A set which can be considered the minimal set satisfying the axiom's conditions is the set I will call N. It includes only the empty set as a base element. The empty set is the only element of N that is not the successor another element.

So, a set X is a non-finite set if and only if |N| <= |X|, where N is the minimal set established by the Axiom of Infinity.
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∃ N ( ∅ ∈ N ∧ for any given x ∈ N ( ( x ∪ { x } ) ∈ N ) )

Since "base elements [predecessors], and their successors" are defined by the axiom of infinity, N members are ordered, otherwise the term successor is not used, in the first place.

Moreover, x is the immediate predecessor of ( x ∪ { x } ) and ( x ∪ { x } ) is the immediate successor of x, so the way of how any given non-empty x member is defined, depends on how the successor of any given non-empty x is defined (they are its distinct foot prints that it leaves 'behind').

Definition 1: Set Y is called the immediate successor of set X iff X is the set of all the predecessors of Y AND Y is not a member of X (where the term all is valid only if X is a finite set, as will be shown in the case of set N).

So set N has ∅ as its base member, which is the predecessor of any given non-empty set that is a member of set N.

Since order is important (otherwise the terms predecessor or successor are not used, in the first place) N members (by the axiom of infinity) are as follows:

Code: 
N={
    
       ∅ (the base member), 
       [COLOR="Magenta"]∅∪{∅}[/COLOR]={∅} , 
       [COLOR="magenta"]{∅}∪{{∅}}[/COLOR]={ ∅, {∅} }, 
       [COLOR="magenta"]{∅,{∅}}∪{{∅,{∅}}}[/COLOR]={ ∅, {∅} , {∅, {∅}} }, 
       [COLOR="magenta"]{∅,{∅},{∅,{∅}}}∪{{∅,{∅},{∅,{∅}}}}[/COLOR]={ ∅, {∅} , {∅, {∅}}, {∅, {∅}, {∅, {∅}}} }, 
... 

}

So as can clearly be seen, Zermelo sets (except the first two members) are not defined by ZF axiom of infinity.

-------------

(B.T.W the largest member of a given set, is the member of that set, which has no successor.
 
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doronshadmi said:
Since "base elements [predecessors]...are defined by the axiom of infinity....
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No, they are not. The Axiom of Infinity requires the null set be in the set. It says nothing about other possibilities. The Axiom does not identify a specific set.
 
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