My Math Forum ZFC Axiom of regularity

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March 10th, 2018, 10:43 AM   #11
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Quote:
 Originally Posted by zylo If A= A, you can't have A$\displaystyle \epsilon$A. It's a fundamental mistake in set theory.
Which part of Brian Scott's proof do you think is wrong?

His one-line proof is simple and clear.

Anyway, there are non well-founded set theories. https://en.wikipedia.org/wiki/Non-we...ded_set_theory

Last edited by Maschke; March 10th, 2018 at 10:49 AM.

March 10th, 2018, 11:11 AM   #12
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Quote:
 Originally Posted by Maschke Which part of Brian Scott's proof do you think is wrong? His one-line proof is simple and clear. Anyway, there are non well-founded set theories. https://en.wikipedia.org/wiki/Non-we...ded_set_theory
Since A=A, you can't assume A$\displaystyle \epsilon$A. = and $\displaystyle \epsilon$ are mutually exclusive symbols.

You have to assume A$\displaystyle \epsilon$A to arrive at his final conclusion.

Or how did you arrive at the conclusion? He just says "it follows immediately."

March 10th, 2018, 11:19 AM   #13
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Quote:
 Originally Posted by zylo Since A=A, you can't assume A$\displaystyle \epsilon$A. = and $\displaystyle \epsilon$ are mutually exclusive symbols.
That "= and $\displaystyle \epsilon$ are mutually exclusive symbols" isn't something you can take for granted - in fact, it's precisely what Brian is proving here!

 March 10th, 2018, 05:42 PM #14 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 I'll attempt to explain Brian's assertion that $\displaystyle A \cap \{A\} = \emptyset$ implies that $\displaystyle A \notin A$. The key is that $\{A\}$ contains $A$. From the intersection statement, we know that $A$ contains no elements from $\{A\}$ (otherwise $A \cap \{A\} \neq \emptyset$). This directly means that $A$ cannot contain $A$ (since $A$ is an element of $\{A\}$). In symbols, this is $A \notin A$.
 March 11th, 2018, 07:07 AM #15 Senior Member   Joined: Jun 2014 From: USA Posts: 413 Thanks: 26 The axiom of regularity implies that the following are both true statements: $$A \cap \{A\} = \emptyset$$ $$A \cup \{A\} \neq A$$ On the surface it may appear that any set merely containing the empty set would pass muster with the axiom, but that is also not true. The axiom would not allow, for example, a set $A = \{\emptyset, A\}$. The axiom also would not allow a set like: $$B = \{ \emptyset,x,y,z,\dots,B \} = \{ \emptyset,x,y,z,\dots,\{\emptyset,x,y,z,\dots,B \} \} = \{\emptyset,x,y,z,\dots,\{\emptyset,x,y,z,\dots,\{ \emptyset,x,y,z,\dots,B \} \} \} = \dots$$ If we choose to express $\mathbb{N}$ in ZF as...: $$\mathbb{N} = \{ 1, 2, 3, \dots \} = \{ \{0\}, \{0, 1\}, \{0, 1, 2\}, \dots \} = \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \}, \{ \emptyset, \{ \emptyset \}, \{ \emptyset, \{ \emptyset \}\} \}, \dots \}$$ ..., then the axiom ensures that: $$\mathbb{N} \cup \{\mathbb{N}\} \neq \mathbb{N}$$ ... even though the definition of $\mathbb{N}$ also asserts that all on its own, so my clarification is redundant. Can you figure out why the axiom does all of the above zylo? Zermelo was no slouch...

 Tags axiom, regularity, zfc

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