Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Question about axiom of regularity

Tags:
  1. Sep 13, 2016 #1
    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. jcsd
  3. Sep 13, 2016 #2

    Ssnow

    User Avatar
    Gold Member

    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##.
     
  4. Sep 13, 2016 #3

    Erland

    User Avatar
    Science Advisor

    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.
     
  5. Sep 13, 2016 #4
    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 ?
     
  6. Sep 14, 2016 #5
    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}?
     
  7. Sep 14, 2016 #6

    Erland

    User Avatar
    Science Advisor

    ##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.
     
  8. Sep 14, 2016 #7
    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
     
  9. Sep 14, 2016 #8

    Erland

    User Avatar
    Science Advisor

    ##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.
     
  10. Sep 14, 2016 #9
    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.
     
  11. Sep 14, 2016 #10

    Erland

    User Avatar
    Science Advisor

    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.)
     
  12. Sep 15, 2016 #11
    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 ?
     
  13. Sep 15, 2016 #12

    Erland

    User Avatar
    Science Advisor

    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?
     
  14. Sep 15, 2016 #13
    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 ?
     
  15. Sep 15, 2016 #14

    Erland

    User Avatar
    Science Advisor

    Well, if ##A\notin A##, then we cannot have ##A=\{A,b\}##, because then ##A\in\{A,b\}##, that is: ##A\in A##.
     
  16. Sep 15, 2016 #15
    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}
     
  17. Sep 15, 2016 #16

    Erland

    User Avatar
    Science Advisor

    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##.
     
  18. Sep 15, 2016 #17
    Oh okay thanks for spending your time explaining it to me .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Question about axiom of regularity
  1. Question about axioms (Replies: 5)

Loading...