# A Question about axiom of regularity

Tags:
1. Sep 13, 2016

### Ling Min Hao

It has been stated that in axiom of regularity , a set cannot be an element of itself and there is a proof for which S={S} . I can understand his proof since S is the only element and hence its method of proof is viable here . But , what if I change the question to S= {S,b} ( it is a set which contain itself with another element , b) . How can we prove from here that every set cannot be an element of itself by using the axiom of regularity?

2. Sep 13, 2016

### Ssnow

Hi, I don't know if can help but we are here also to discuss, so if you can prove (with your argument) that $S=\{S\}$ and $b=\{b\}$ separately, then $S\cup b = \{S\}\cup\{b\}=\{\{S\},\{b\}\}=\{S,b\}$ and not $S$.

3. Sep 13, 2016

### Erland

The axiom of regularity says that every non-empty set $S$ has an element $b$ such that $S\cap b =\varnothing$.
If there was a set $S$ such that $S\in S$ (which covers both cases $S=\{S\}$ and $S=\{S,b\}$), then consider the set $\{S\}$. Since $S\in S$, we have $\{S\}\cap S =\{S\}\neq\varnothing$, which contradicts the axiom.

4. Sep 13, 2016

### Ling Min Hao

Yes it does show that contradiction occurs when S= {S} , what you show is the condition when S is the only element in the set S.What if there is another element b in the set S such that S={S,b}, does it hold true that S is not an element of S ?

5. Sep 14, 2016

### Ling Min Hao

So what you meant is if S={S} is a false statement , it means that S= { S,b} is a false statement too due to S is not an element of S from S={S}?

6. Sep 14, 2016

### Erland

$S=\{S,b\}$ is not false because $S=\{S\}$ is false, but because it contradicts the axiom of regularity, which doesn't say just that $S\neq\{S\}$. See my previous post.
Perhaps it is easier to see this if we rename the sets: Assume that $A=\{A,b\}$. Put $S=\{A\}$. $S$ has only one element: $A$. But $S\cap A=\{A\}\cap\{A,b\}=\{A\}\neq \varnothing$. Hence, the nonempty set $S$ has no element which is disjoint from $S$. This contradicts the axiom of regularity. Hence $A=\{A,b\}$ is false.

7. Sep 14, 2016

### Ling Min Hao

From what you do , it still only show that
S = {A} is false . Yes, contradiction did occur on the side of S = {A} , but this is the case when A is the only element of S , but what if S = {A,c} ? Surely S won't be able to disjoint with A now but it is possible that S and c are disjoint . According to axiom of regularity , a set is true if at least one element is disjoint from the set . If S is disjoint with c , then the set S should hold the axiom of regularity right ? Now the problem is , how do we show that S and c are not disjoint so that it contradicts with axiom of regularity ?

Correct me if I am wrong

8. Sep 14, 2016

### Erland

$S$ is defined as $\{A\}$. For any set $A$, there exists a set $\{A\}$, which we here call $S$, so $S=\{A\}$ cannot be false. What is false is $A=\{A,b\}$, because then, we will get a contradiction if we apply the axiom of regularity to $S=\{A\}$. By the axiom, $S$, which is nonempty, must contain an element disjoint from $S$. Since $A$ is the only element in $S$, this means that $S\cap A=\varnothing$, but $S\cap A = \{A\}$, which is a contradiction. The only way out of this contradiction is then that the assumption $A=\{A,b\}$ is false.

Notice that the axiom of regularity is applied to $S=\{A\}$, not to $A=\{A,b\}$. Therefore, there is no need to show that $S$ and $b$ or $S$ and $c$ etc. are disjoint.

9. Sep 14, 2016

### Ling Min Hao

Okay , basically i understand the logic behind it . I just have one more question to clear it out .
Can we define a set ourselves like S={A} so that we can imply that A={A,b} is false by using the axiom of regularity? I mean , does mathematics allow us to define something , then we prove something based on what we defined? Thanks anyway for spending your time on my problem.

10. Sep 14, 2016

### Erland

Yes, in this case we can. The Pairing Axiom, one of the Zermelo-Fraenkel (ZF) axioms, says that to any sets $a$ and $b$, there exists a set $\{a,b\}$. If we apply this with $a=b=A$, we see that there exists a set $\{A,A\}=\{A\}$, which we may call $S$ if we want to.

(As the Wikipedia article shows, the pairing axiom is actually derivable from the other axioms in ZF.)

11. Sep 15, 2016

### Ling Min Hao

Finally I understand your method through axiom of pairing!!
But if I want to say given A={A} , and it is shown that A is not an element of A from axiom of regularity , can I imply that since A is not element of A , the statement A={A,b} is false as well ?

12. Sep 15, 2016

### Erland

Using the axiom of regularity, we can prove both $A\neq\{A\}$ and $A\neq\{A,b\}$ for all $A$ and $b$.

Perhaps you mean this:

Can we, without using the axiom of regularity, prove that $A\neq\{A\}$ for all $A$ implies $A\neq\{A,b\}$ for all $A$ and $b$?

I am not sure, but I don't think so. Anyone who knows?

13. Sep 15, 2016

### Ling Min Hao

I mean , if we proved that A is not an element of set A from axiom of regularity , does this result implies that A is not an element of the set A={A,b} too ?

14. Sep 15, 2016

### Erland

Well, if $A\notin A$, then we cannot have $A=\{A,b\}$, because then $A\in\{A,b\}$, that is: $A\in A$.

15. Sep 15, 2016

### Ling Min Hao

But the result of "A is not an element of A " is shown from A = { A}, can this result be applied to any other case like A = {A,b} since the result of A is not element of A was based on A ={A}

16. Sep 15, 2016

### Erland

No $A\notin A$ is showed by applying the axiom of regularity to the set $S=\{A\}$. $A \neq \{A\}$ can then be considered as a consequence of $A\notin A$, and $A\neq \{A,b\}$ is also a consequence of $A\notin A$.

17. Sep 15, 2016

### Ling Min Hao

Oh okay thanks for spending your time explaining it to me .