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February 1st, 2016, 10:51 AM   #1
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Axiom of Regularity is wrong

AR: ZF Axiom of Regularity*: x has a member y st x$\displaystyle \cap$y=0.

a$\displaystyle \notin$a. Proof
1) Let x={a}
2) x$\displaystyle \cap$a=0, AR requirement
3) members of x: a
4) members of a: b (a has to have a member)
5) x$\displaystyle \cap$a=0 implies b$\displaystyle \neq$a
6) $\displaystyle \therefore$ a can not be a member of itself.

Problem with proof: Step 4)
You are not excluding b=a. If you exclude b=a, the proof is circular.
Basically, you begin the proof by allowing a$\displaystyle \in$a, which is impossible because "=" and "$\displaystyle \in$" can't be the same thing.

The situation is similar to division by 0 in the rational or real number system, where it is explicitly excluded. I vaguely recall a proof that 2+2=5 which allows non-transparent division by zero.
---------------
Let:
A={a,b,c}
a={d,e} d,e$\displaystyle \neq$a,b,c
b={f,g} f,g$\displaystyle \neq$a,b,c
c={h,A} h,A$\displaystyle \neq$a,b,c

Members of A: a,b,c
Members of a: d,e
Members of b: f,g
Members of c: h,A
Therefore
A$\displaystyle \cap$a=0
A$\displaystyle \cap$b=0
A$\displaystyle \cap$c=0

$\displaystyle \therefore$ A={a,b,{h,A}} is an acceptable set by AR.
WRONG. It is circular.

*
https://en.wikipedia.org/wiki/Axiom_of_regularity
https://en.wikipedia.org/wiki/Zermel...kel_set_theory
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February 1st, 2016, 10:58 AM   #2
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You are an idiot. Please stop posting nonsense about things you don't understand.
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February 1st, 2016, 12:37 PM   #3
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Quote:
zylo
Basically, you begin the proof by allowing a∈∈a, which is impossible because "=" and "∈∈" can't be the same thing.
zylo, you seem fixated on this statement or something very similar.

But you are introducing things that are not relevent when you look carefully at the logic of the dichotomy that is being presented.

Basically you take a set U and partition it into two parts.

Subset A which is all the members of U that have some property P.

All the other members of U which do not have this property form subset B

Now since no member of A is also a member of B they can never be equal.

However, depending upon how you define U, the subset A may or may not be a member of subset B.

Paradox arise when you make U too all embracing.

Zermelo's solution was to make his sets about numbers. That made sense because that is what he really wanted to discuss.
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February 1st, 2016, 12:47 PM   #4
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Quote:
Originally Posted by zylo View Post
$\displaystyle \therefore$ A={a,b,{h,A}} is an acceptable set by AR.
WRONG. It is circular.
The axiom of regularity doesn't state "a set is acceptable if...", so you are misusing the axiom. Hence what you did was wrong, rather than the axiom of regularity.

As the axiom of regularity was designed to exclude various things that you would call "circular" or involving circularity in their description, it would make more sense if you supported it. After all, you have not found any set that complies with your ideas but doesn't satisfy the axiom of regularity.
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February 1st, 2016, 01:10 PM   #5
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I think the problem is that his ego cannot cope when he finds something that he can't understand. He feels compelled to prove it false to make himself superior to all those who understand what he doesn't.

He seems quite intelligent at times, but this problem causes him to write utter nonsense constantly. I wish he'd put all this effort into understanding the existing work.
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February 2nd, 2016, 07:49 AM   #6
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Referring to another member as "an idiot" is unacceptable. Please be civil when offering criticism. If you can't, then don't post.
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February 2nd, 2016, 07:50 AM   #7
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Quote:
Originally Posted by skipjack View Post
The axiom of regularity doesn't state "a set is acceptable if...", so you are misusing the axiom. Hence what you did was wrong, rather than the axiom of regularity.

As the axiom of regularity was designed to exclude various things that you would call "circular" or involving circularity in their description, it would make more sense if you supported it. After all, you have not found any set that complies with your ideas but doesn't satisfy the axiom of regularity.
A={a,b,{h,A}} satisfies the axiom of regularity, shown very clearly and transparently in the OP, and it is obviously circular.

What is AR saying other than it is a necessary condition for a set? Why is it a necessary condition. What's the point?
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February 2nd, 2016, 08:18 AM   #8
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Originally Posted by greg1313 View Post
Referring to another member as "an idiot" is unacceptable. Please be civil when offering criticism. If you can't, then don't post.
My problem is that this other member only seems to answer polite discourse when pushed.

How is it possible to conduct a discussion on under these circumstances?
This is, after all, a discussion forum.

Last edited by studiot; February 2nd, 2016 at 08:21 AM.
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February 2nd, 2016, 08:50 AM   #9
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I've already stated that the point is that it excludes various things that you would call "circular" or that involve circularity in their description.

Last edited by skipjack; February 2nd, 2016 at 09:28 AM.
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February 2nd, 2016, 09:20 AM   #10
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Quote:
Originally Posted by greg1313 View Post
Referring to another member as "an idiot" is unacceptable.
I think it's a very good word to describe someone who pontificates on subjects of which he has no understanding, assumes that his lack of comprehension implies the untruth of that which he doesn't understand and refuses to put any effort into remedying his lack of understanding.
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