Deeper than primes

[jsfisher]

OK, let us follow ZF where any member must be also a set.

1) This rule does not hold in the case of the empty set, because the "member" of {} is not a set.

Wrong. Every member of {} is a set, without exception.

2) Let us say that 0={} and 1={{}}, so in this case {1}={{{}}}.

Ok. You have unnecessarily complicated HatRack's example (as is your custom), but proceed.

3) Any other set (accept {{}})

(The word is "except" by the way.)

...is not a member of {{{}}},

Oddly presented, but substantially what HatRack posted. The set containing {{}} as its member has {{}} as its member and no other set (including itself) as a member.

...so also in this case a given set is used as a factor in order to determine itself as a part of its own derermination.

Well, that's HatRack's point, now wasn't it? Such a construction is always possible, so if it invalidates things, then everything, including Doronetics, is invalid. But, of course, since it is not circular reasoning (except in the twisted but fortunately invalid logic embodied in Doronetics), the world is safe. Mathematics still works.

So thank you HatRack, according to (1) and (3) we see that ZF is based on an invalid reasoning.

Well, if we overlook your failures in logic, that may be true, but only in some alternate universe.

 

[jsfisher]

Because you ignore the rest of what I wrote, you are really js.

Doron, you post idiocy. This does not help your cause. Words have meaning, and the rest of what you originally wrote, the part you presume incorrectly I ignored, doesn't change the meaning of the initial part.

It is no different than if I said, "Unicorns exist that have feelings for hermaphroditic oxen." Everything after the "unicorns exist" part is irrelevant to whether unicorns exist. They don't exist.

You said that sets and power sets share members. That was a bogus statement on your part. The additional information you provided to characterize the shared members was irrelevant since the initial part is false.

I will except colloquial expressions with their inherent imprecision, Doron, but not sloppy, illogical equivocations.

 

[HatRack]

So {} doesn't exist, but its contents do.

Exactly the opposite, HatRack, exactly the opposite.

So you're saying that the empty set {} exists now. All the way back on page 318, a multi-page argument ensued where you vehemently denied the existence of the empty set. Also, on the last page, you insisted that numbers (including 0) are not sets. So let's summarize all we know about Doronetics so far:

{} exists, but {} doesn't exist.
{} = 0, but {} ≠ 0.

Wow, what an insight! This will truly bridge the gap between logic and ethics.

 

[zooterkin]

You do not get the following:

I said that 1 (if we are not following ZF) is not itself a set, so in this case there is a difference between 1 (which is not a set) and {1} (which is a set), and HatRack's argument is not equivalent to my argument about set {} that is not a member of set {} (which is a circular reasoning because set {} is a factor that deremines set {}).

That was not in the post you originally replied to, but in a later one.

 

[doronshadmi]

Wrong. Every member of {} is a set, without exception.

In that case please show a member of {}, which is a set.

The rest of your post is based on the ability to show it.

 

[doronshadmi]

a multi-page argument ensued where you vehemently denied the existence of the empty set.

Please provide the exact qoute (and link) that supports your claim.

Also, on the last page, you insisted that numbers (including 0) are not sets.

In order to get my claim, please show that some irrational number (which is a member of R set) is a set.

 

[HatRack]

In that case please show a member of {}, which is a set.

Show us a member of {} which is not a set. jsfisher's claim that every member of {} is a set holds true unless there is a member of {} which is not a set. So, go ahead, show us.

 

[zooterkin]

Here's the post again, Doron.

No, that's not correct either. Circular reasoning is when you assume a proposition A to prove A. Every set uses itself as a factor in its definition. For example, the set {1} is defined as the set such that 1 is its member and every other set X (including {1} itself) is not a member. This type of "self-reference" is inescapable in defining sets.

You have a simple mistake, 1 is not a set, but can be a member of a set.

The mistake is yours. HatRack's subsequent explanation notwithstanding, in the quoted piece '1' is not used to refer to a set, only to a member of a set.

Now, please show, in the quoted piece to which you were responding, not a later post by HatRack, where '1' was used to refer to a set.

 

[HatRack]

From page 318:

In other words the empty set is used to define itself, which is a circular reasoning.

Here, you are clearly denying the existence of the empty set, because it uses "circular reasoning" as you claim.

 

[HatRack]

In order to get my claim, please show that some irrational number (which is a member of R set) is a set.

My goodness Doron, do you not even know Dedekind's construction of the reals? Here is an irrational number (the first one discovered in fact), which is defined as a set:

[latex]$$\sqrt{2} = \left \{ x \in Q : x^2 < 2 \lor x < 0\right \}$$[/latex]​

 

[doronshadmi]

From page 318:

In other words the empty set is used to define itself, which is a circular reasoning.

Here, you are clearly denying the existence of the empty set, because it uses "circular reasoning" as you claim.

No, I denial the validity of the reasoning that is used to determine the existence of {}, by using {} as one of the factors that determine {} existence.

It says nothing about the existence of {}, it says a lot about the invalid reasoning that is used by formal frameworks like ZF in order to establish {} existence, exactly because by ZF reasoning any member must be a set, and in that case one can't avoid circular reasoning, because sets are determined by sets.

 

[HatRack]

No, I denial the validity of the reasoning that is used to determine the existence of {}, by using {} as one of the factors that determine {} existence.

It says nothing about the existence of {}, it says a lot about the invalid reasoning that is used by formal frameworks like ZF in order to establish {} existence, exactly because by ZF reasoning any member must be a set, and in that case one can't avoid circular reasoning, because sets are determined by sets.

Give us your axiom of the empty set then.

 

[doronshadmi]

My goodness Doron, do you not even know Dedekind's construction of the reals? Here is an irrational number (the first one discovered in fact), which is defined as a set:

[latex]$$\sqrt{2} = \left \{ x \in Q : x^2 < 2 \lor x < 0\right \}$$[/latex]​

All you showed is that sqrt(2) is a member of R set, but you did not show that sqrt(2) is itself a set.

Also please show that the member of {} is itself a set.

 

[doronshadmi]

Give us your axiom of the empty set then.

There exists a set that its member is exactly Emptiness (that has no predecessor).

EDIT: By the way, by cross-contexts framework ( http://www.internationalskeptics.com/forums/showpost.php?p=6667634&postcount=13318 ) the empty set not an axiom anymore, but it is the result of an existing set that its member is Emptiness, which is not a set.

 

[doronshadmi]

Show us a member of {} which is not a set.

Emptiness.

 

[doronshadmi]

Here's the post again, Doron.

Now, please show, in the quoted piece to which you were responding, not a later post by HatRack, where '1' was used to refer to a set.

In set theory, 1 is traditionally defined to be the set {0}. Everything is a set in set theory

( http://www.internationalskeptics.com/forums/showpost.php?p=6706816&postcount=13577 )

 

[HatRack]

Here is an irrational number, which is <<<<<*****defined*****>>>>> as a set:

[latex]$$\sqrt{2} = \left \{ x \in Q : x^2 < 2 \lor x < 0\right \}$$[/latex]​

you did not show that sqrt(2) is itself a set.

 

[HatRack]

There exists a set that its member is exactly Emptiness (that has no predecessor).

Define emptiness and predecessor.

 

[doronshadmi]

http://www.internationalskeptics.com/forums/imagehosting/322964b855ca82f9b1.jpg

This is nothing but playing with notations, sqrt(2) is a member of a set that does not belong to the collection of rational numbers that are < sqrt(2) AND does not belong to the collection of rational numbers that are > sqrt(2).

It says nothing about sqrt(2) being itself a set.

In order to show it please use sqrt(2) as a set in order to show further things about sets.

 

[doronshadmi]

Define emptiness and predecessor.

http://www.internationalskeptics.com/forums/showpost.php?p=6667634&postcount=13318

EDIT: Please also look at http://www.internationalskeptics.com/forums/showpost.php?p=6713975&postcount=13634 .