March 10th, 2018, 10:43 AM  #11  
Senior Member Joined: Aug 2012 Posts: 2,157 Thanks: 631  Quote:
His oneline proof is simple and clear. Anyway, there are non wellfounded set theories. https://en.wikipedia.org/wiki/Nonwe...ded_set_theory Last edited by Maschke; March 10th, 2018 at 10:49 AM.  
March 10th, 2018, 11:11 AM  #12  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124  Quote:
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 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 308 Thanks: 102 Math Focus: Number Theory, Algebraic Geometry  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: 479 Thanks: 36 
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... 

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axiom, regularity, zfc 
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