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

[doronshadmi]

deleted ...

 

[doronshadmi]

EDITED:

"Is a successor set" is a set property -- that's your Definition #1.
"Successor set of" is a set function -- not defined.

Lets carefully check that.

Part 1:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p})

First we check the case where p is any given finite set.

By definition 1 S can't be a successor set unless set p members AND set {p} member, are exactly the members of S, such that p (as a member of S) prevents the bijection from p to S.

Here is some example:

If

p={ ∅, {∅}, {∅, {{∅}}} }

then

S={ ∅, {∅}, {∅, {{∅}}} , { ∅, {∅}, {∅, {{∅}}} } }

where there is no bijection from p to S because of { ∅, {∅}, {∅, {{∅}}} } in S (which is set p itself, as a member of S)

So, S identity is not p identity exactly because there is no bijection from p to S.

Also be aware that no set is a member of itself, whether S or p are considered.

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

Part 2:

Now let's swim to the deep water, where p={a,b,c,...}

Also in that case S={a,b,c,... ,{a,b,c,...}} ≠ p={a,b,c,...}, but one says: "I don't care about identity, but about cardinality, such that there is bijection from p to S in case that p is a non-finite set", for example:

p={    a,         b,  c,  d,...}
       ↓          ↓   ↓   ↓
S={{a,b,c,d,...}, a,  b,  c,...}

But in order to show such mapping in the first place and also plug in universal quantification, we can't skip on a set that its members are finite sets, so we go back to the shallow water where p is any given finite set.

1) Let G be the set of any given finite set p AND any given successor set of p.

2)There can't be but injective non-surjective function form any given finite set p to the successor set of p, in set G.

3) Since injective non-surjective function is actually constant among the finite sets in set G, equality (=) is not satisfied, and set G cardinality can't be equal to any particular value.

4) Universal quantification has nothing to do with any set that its cardinality can't be equal to any particular value.

5) The bijection in the example above, is between two sets that their cardinalities are not satisfied.

 

[Little 10 Toes]

(without ignore the difference between potential infinity and actual infinity, as done by most of the, so called, classical mathematicians, as if this debate belongs to Philosophy, middle ages, history and more aviation techniques)

please quit with your personal views/attacks. just because you can't properly express yourself doesn't mean they were wrong. Please notice that this thread has been stuck in the Religion and Philosophy section.

 

[jsfisher]

Lets carefully check that.

Part 1:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p})

Yep, that defines the "successor set" property for a set. A set has the successor set property if (and only if) a certain condition is met.

First we check the case where p is any given finite set.

It can't be just any give set, finite or otherwise. And why, oh, why are you bringing in the finite qualifier out of the blue?

By definition 1 S can't be a successor set unless set p members AND set {p} member, are exactly the members of S

No. AND, in caps, in this thread has bean used to stand for logical conjunction. If you insist in writing out S = p∪{p} as a statement in English, at least use the correct words. You have also passed over the quantification aspect of p. Kinda important that.

...such that p (as a member of S) prevents the bijection from p to S.

Nothing in the definition prevents anything. Either S is or S is not a successor set. That is all it says. You'd need to prove your assertion, 'cause the definition doesn't do it for you.

Here is some example:

If

p={ ∅, {∅}, {∅, {{∅}}} }

then

S=...

Why are you working this backwards. S is the given, not p.


And you haven't defined "successor set" as a set-valued function, yet.

 

[jsfisher]

(without ignore the difference between potential infinity and actual infinity, as done by most of the, so called, classical mathematicians, as if this debate belongs to Philosophy, middle ages, history and more aviation techniques)

please quit with your personal views/attacks. just because you can't properly express yourself doesn't mean they were wrong. Please notice that this thread has been stuck in the Religion and Philosophy section.

(I will assume the unattributed quote is something Doronshadmi edited out from a recent post of his.) What Doronshadmi refuses to accept is that Mathematics doesn't ignore the difference between potential and actual infinities. Each have their places where one perspective is more useful than the other.

Set theory typically prefers the perspective Doronshadmi abhors. And, as Doronshadmi has repeatedly demonstrated, it is a subject Doronshadmi doesn't understand.

 

[Little 10 Toes]

(I will assume the unattributed quote is something Doronshadmi edited out from a recent post of his.) What Doronshadmi refuses to accept is that Mathematics doesn't ignore the difference between potential and actual infinities. Each have their places where one perspective is more useful than the other.

Set theory typically prefers the perspective Doronshadmi abhors. And, as Doronshadmi has repeatedly demonstrated, it is a subject Doronshadmi doesn't understand.

I have included a screen shot via the snipping tool of the original post showing the last time doronshadmi posted it.

Please note that the post that I am showing was edited after I edited my post. I edited my post due to misspellings.

Edit: any help to have the snip shown at regular size is appreciated.

 

[Little 10 Toes]

To clarify, yes jsfisher, doronshadmi did do the typical "edit a post after someone quoted it".

 

[doronshadmi]

Either S is or S is not a successor set. That is all it says.

S is undefined unless p (where p is not {}) is involved.

Since p is involved, there is injective non-surjective function form p to S (where p is not {}), which is the basis of S as a successor set of p, where p is any given non-empy set but its successor set, since no non-empty set is its successor set because of the different identities between being a given non-empty set, and being its successor set.

The simplest example, which is without loss of generality, is this:

p={{}}

By S = p∪{p}, S = {{},{{}}}

and we get the different identities {{}} and {{},{{}}}.

Remark: I start with the simplest non-empty set, since if we actually dealing with a non-finite set, it is, by definition, can't be empty.

Nevertheless, if p={}, then S=p∪{p}={{}}, and we get the different identities {} and {{}}.

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

No. AND, in caps, in this thread has bean used to stand for logical conjunction.

You are right, in order to avoid confusion, "AND" has to be changed to "and" in http://www.internationalskeptics.com/forums/showpost.php?p=13217932&postcount=422 .

 

[jsfisher]

S is undefined unless p (where p is not {}) is involved.

Not what your definition says.

 

[doronshadmi]

Not what your definition says.

You are right, I am going to change post http://www.internationalskeptics.com/forums/showpost.php?p=13217932&postcount=422 in order to avoid such restriction.

Thank you.

 

[doronshadmi]

EDITED 2:

First, lets correct post #428

Either S is or S is not a successor set. That is all it says.

S is undefined unless p is involved.

Since p is involved, there is no bijection form p to S, which is the basis of S as a successor set of p, where p is any given set but its successor set, since no set is its successor set because of the different identities between being a given set, and being its successor set.

The simplest example, without loss of generality, is:

p={}

S = p∪{p}, S = {{}}

and we get the different identities {} and {{}}.

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

No. AND, in caps, in this thread has bean used to stand for logical conjunction.

You are right, in order to avoid confusion, "AND" has to be changed to "and" in http://www.internationalskeptics.com/forums/showpost.php?p=13217932&postcount=422 .

 

[jsfisher]

S is undefined unless p is involved.

Still wrong. That's not what your definition says.

 

[doronshadmi]

Still wrong. That's not what your definition says.

Please try to omit p from the definition, and tell me what is left.

 

[doronshadmi]

4 EDITINGS: Today, 05:15 AM

Part 1:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p}) ∧ p≠S

First we check the case where p is any given finite set.

By definition 1 S can't be a successor set unless set p members and set {p} member are exactly the members of S, such that the member of {p} prevents the bijection from p to S.

Here is some example:

If

p={ ∅, {∅}, {∅, {{∅}}} }

then

S={ ∅, {∅}, {∅, {{∅}}} , { ∅, {∅}, {∅, {{∅}}} } }

where there is no-bijection from p to S because of { ∅, {∅}, {∅, {{∅}}} } in S (which is set p itself, as a member of S)

So, S identity is not p identity exactly because there is no bijection from p to S.

The same holds in case that p={} :

p={}

S = p∪{p}, S = {{}}

and we get the different identities {} and {{}} since there is no bijection from p to S.

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

Part 2:

Now let's swim to the deep water, where p={a,b,c,...}

Also in that case S={a,b,c,... ,{a,b,c,...}} ≠ p={a,b,c,...}, but one says: "I don't care about identity, but about cardinality, such that there is bijection from p to S in case that p is a non-finite set", for example:

p={    a,         b,  c,  d,...}
       ↓          ↓   ↓   ↓
S={{a,b,c,d,...}, a,  b,  c,...}

But in order to show such mapping in the first place and also plug in universal quantification, we can't skip on a set that its members are finite sets, so we go back to the shallow water where p is any given finite set.

1) Let G be the set of any given finite set p and any given successor set of p.

2)There is no bijection form any given finite set p to the successor set of p, in set G.

3) Since no-bijection is actually constant among the finite sets in set G, equality (=) is not satisfied, and set G cardinality can't be equal to any particular value.

4) Universal quantification has nothing to do with any set that its cardinality can't be equal to any particular value.

5) The bijection in the example above, is between two sets that their cardinalities are not satisfied.

 

[jsfisher]

Please try to omit p from the definition, and tell me what is left.

Why would I do that?

 

[jsfisher]

Part 1:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p})

First we check the case where p is any given finite set....

It is still the quantification you don't understand, isn't it?

 

[doronshadmi]

It is still the quantification you don't understand, isn't it?

S=p∪{p}

It is still the fact that there exits p as a member of S, because the member of {p} is a member of S.

It is still the fact that the members of p are the members of S, and the members of {p} prevent the bijection from p to S.

 

[jsfisher]

S=p∪{p}

It is still the fact that there exits p as a member of S, because the member of {p} is a member of S.

It is still the fact that the members of p are the members of S, and the members of {p} prevent the bijection from p to S.

Those are not facts. "S = p∪{p}" is a proposition. It is either true or false.

 

[doronshadmi]

Definition 1 have been changed to:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p}) ∧ p≠S

The details are given in http://www.internationalskeptics.com/forums/showpost.php?p=13218580&postcount=434.

 

[jsfisher]

Definition 1 have been changed to:

Definition 1: S is a successor set ⇔ ∃p∈S (S = p∪{p}) ∧ p≠S

This would be the "shotgun method" of fixing a problem. You don't understand the original. You think it needs fixing, but you have no idea how (because you don't understand any of it), so you fire a round or two from the 12-gauge in the hopes something, anything hits the target.

In addition to simply not understanding the definition, you are blinded by some compulsion to make it far more than just a definition. Some deep truth is there, and it is explicit.

Well, no it is not. It is simply a definition. Nothing more, nothing less. It succinctly describes what "is a successor set" means. Or, it did. With this latest addition, not so much.