A Question about axiom of regularity

[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?

 

[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 $S$.

 

[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.

 

[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 ?

 

[Ling Min Hao]

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.

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}?

 

[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.

 

[Ling Min Hao]

$S=\{S,b\}$ 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.

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

 

[Erland]

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

$S$ 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.

 

[Ling Min Hao]

$S$ 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.

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.

 

[Erland]

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.

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.)

 

[Ling Min Hao]

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.)

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 ?

 

[Erland]

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 ?

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?

 

[Ling Min Hao]

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?

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 ?

 

[Erland]

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 ?

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

 

[Ling Min Hao]

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

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}

 

[Erland]

But the result of "A is not an element of A " is shown from A = { A}.

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 $A\notin A$.

 

[Ling Min Hao]

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 $A\notin A$.

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