Axiom of Regularity and infinite sequences












2












$begingroup$


I'm wrestling with the elementary implications of the Axiom of Regularity.



The axiom:



$∀A(A≠∅→(∃x∈A)(A∩x=∅))$



implies that every set A either has $∅∈A$, or it has some element $x$ such that $x∩A=∅$.



My questions:
(1) If $x∈A$, and $x$ is not a subset (or, x is a subset that consists of only one element, i.e. {1}), then does $x∩A=∅$? I'm trying to understand how $x∩{1,2,3}=∅$



(2)How does this guarantee no infinitely descending sequences?



Thank you for your time.






elementary-set-theory







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I'm wrestling with the elementary implications of the Axiom of Regularity.



    The axiom:



    $∀A(A≠∅→(∃x∈A)(A∩x=∅))$



    implies that every set A either has $∅∈A$, or it has some element $x$ such that $x∩A=∅$.



    My questions:
    (1) If $x∈A$, and $x$ is not a subset (or, x is a subset that consists of only one element, i.e. {1}), then does $x∩A=∅$? I'm trying to understand how $x∩{1,2,3}=∅$



    (2)How does this guarantee no infinitely descending sequences?



    Thank you for your time.






    elementary-set-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I'm wrestling with the elementary implications of the Axiom of Regularity.



      The axiom:



      $∀A(A≠∅→(∃x∈A)(A∩x=∅))$



      implies that every set A either has $∅∈A$, or it has some element $x$ such that $x∩A=∅$.



      My questions:
      (1) If $x∈A$, and $x$ is not a subset (or, x is a subset that consists of only one element, i.e. {1}), then does $x∩A=∅$? I'm trying to understand how $x∩{1,2,3}=∅$



      (2)How does this guarantee no infinitely descending sequences?



      Thank you for your time.






      elementary-set-theory







      share|cite|improve this question











      $endgroup$




      I'm wrestling with the elementary implications of the Axiom of Regularity.



      The axiom:



      $∀A(A≠∅→(∃x∈A)(A∩x=∅))$



      implies that every set A either has $∅∈A$, or it has some element $x$ such that $x∩A=∅$.



      My questions:
      (1) If $x∈A$, and $x$ is not a subset (or, x is a subset that consists of only one element, i.e. {1}), then does $x∩A=∅$? I'm trying to understand how $x∩{1,2,3}=∅$



      (2)How does this guarantee no infinitely descending sequences?



      Thank you for your time.






      elementary-set-theory




      elementary-set-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 14 '13 at 8:27









      Sergiy Kozerenko

      21628




      21628










      asked Jul 8 '13 at 2:33









      user84815user84815

      596




      596






















          2 Answers
          2






          active

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          4












          $begingroup$

          You’ve stated the axiom of regularity correctly,



          $$forall ABig(Anevarnothingtoexists xin A(Acap x=varnothing)Big);,tag{1}$$



          but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=varnothing$, not $varnothingin A$.



          Your first question isn’t entirely clear. In the specific example, are you asking what element of $A={1,2,3}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1={0},2={0,1}$, and $3={0,1,2}$. If you’re using that definition, then $x=1$ works:



          $$xcap A=1cap{1,2,3}={0}cap{1,2,3}=varnothing;,$$



          because the only element of $1$ is $0$, and $0notin{1,2,3}$.



          For your second question, suppose that $A=langle a_n:ninBbb N}$ is a family of sets with the property that $a_{n+1}in a_n$ for each $ninBbb N$:



          $$ldotsin a_4in a_3in a_2in a_1in a_0;.$$



          Let $xin A$. Then $x=a_n$ for some $ninBbb N$, and $a_{n+1}in xcap A$, so $a_{n+1}nevarnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
            $endgroup$
            – user84815
            Jul 8 '13 at 2:56










          • $begingroup$
            @user84815: You’re welcome!
            $endgroup$
            – Brian M. Scott
            Jul 8 '13 at 3:07



















          2












          $begingroup$

          First, the alternative isn't $emptyset in A$, as you have it, but $emptyset=A$. Hence every nonempty $A$ has an element that isn't a subset.



          Suppose now that $$A={1,2,{1}, {{1}},{1,2,4},3}$$



          If we take $x={1}$, then $xcap A={1}neq emptyset$. If we take $x={1,2,4}$, then $xcap A={1,2}neqemptyset$. However if we take $x=1$, then $xcap A=emptyset$.



          Suppose now that there were an infinitely descending subsequence $$x_1ni x_2 ni x_3 ni cdots$$



          We could then take $$A={x_1,x_2,x_3,ldots}$$ which would voilate the axiom, because for each $i$, $Acap x_i$ is a set which contains $x_{i+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the help! This definitely clears things up.
            $endgroup$
            – user84815
            Jul 8 '13 at 3:00












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          2 Answers
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          4












          $begingroup$

          You’ve stated the axiom of regularity correctly,



          $$forall ABig(Anevarnothingtoexists xin A(Acap x=varnothing)Big);,tag{1}$$



          but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=varnothing$, not $varnothingin A$.



          Your first question isn’t entirely clear. In the specific example, are you asking what element of $A={1,2,3}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1={0},2={0,1}$, and $3={0,1,2}$. If you’re using that definition, then $x=1$ works:



          $$xcap A=1cap{1,2,3}={0}cap{1,2,3}=varnothing;,$$



          because the only element of $1$ is $0$, and $0notin{1,2,3}$.



          For your second question, suppose that $A=langle a_n:ninBbb N}$ is a family of sets with the property that $a_{n+1}in a_n$ for each $ninBbb N$:



          $$ldotsin a_4in a_3in a_2in a_1in a_0;.$$



          Let $xin A$. Then $x=a_n$ for some $ninBbb N$, and $a_{n+1}in xcap A$, so $a_{n+1}nevarnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
            $endgroup$
            – user84815
            Jul 8 '13 at 2:56










          • $begingroup$
            @user84815: You’re welcome!
            $endgroup$
            – Brian M. Scott
            Jul 8 '13 at 3:07
















          4












          $begingroup$

          You’ve stated the axiom of regularity correctly,



          $$forall ABig(Anevarnothingtoexists xin A(Acap x=varnothing)Big);,tag{1}$$



          but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=varnothing$, not $varnothingin A$.



          Your first question isn’t entirely clear. In the specific example, are you asking what element of $A={1,2,3}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1={0},2={0,1}$, and $3={0,1,2}$. If you’re using that definition, then $x=1$ works:



          $$xcap A=1cap{1,2,3}={0}cap{1,2,3}=varnothing;,$$



          because the only element of $1$ is $0$, and $0notin{1,2,3}$.



          For your second question, suppose that $A=langle a_n:ninBbb N}$ is a family of sets with the property that $a_{n+1}in a_n$ for each $ninBbb N$:



          $$ldotsin a_4in a_3in a_2in a_1in a_0;.$$



          Let $xin A$. Then $x=a_n$ for some $ninBbb N$, and $a_{n+1}in xcap A$, so $a_{n+1}nevarnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
            $endgroup$
            – user84815
            Jul 8 '13 at 2:56










          • $begingroup$
            @user84815: You’re welcome!
            $endgroup$
            – Brian M. Scott
            Jul 8 '13 at 3:07














          4












          4








          4





          $begingroup$

          You’ve stated the axiom of regularity correctly,



          $$forall ABig(Anevarnothingtoexists xin A(Acap x=varnothing)Big);,tag{1}$$



          but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=varnothing$, not $varnothingin A$.



          Your first question isn’t entirely clear. In the specific example, are you asking what element of $A={1,2,3}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1={0},2={0,1}$, and $3={0,1,2}$. If you’re using that definition, then $x=1$ works:



          $$xcap A=1cap{1,2,3}={0}cap{1,2,3}=varnothing;,$$



          because the only element of $1$ is $0$, and $0notin{1,2,3}$.



          For your second question, suppose that $A=langle a_n:ninBbb N}$ is a family of sets with the property that $a_{n+1}in a_n$ for each $ninBbb N$:



          $$ldotsin a_4in a_3in a_2in a_1in a_0;.$$



          Let $xin A$. Then $x=a_n$ for some $ninBbb N$, and $a_{n+1}in xcap A$, so $a_{n+1}nevarnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.






          share|cite|improve this answer









          $endgroup$



          You’ve stated the axiom of regularity correctly,



          $$forall ABig(Anevarnothingtoexists xin A(Acap x=varnothing)Big);,tag{1}$$



          but your verbal paraphrase isn’t quite right: $(1)$ says that either $A=varnothing$, or some element $x$ of $A$ is disjoint from $A$. That is, your first alternative should be $A=varnothing$, not $varnothingin A$.



          Your first question isn’t entirely clear. In the specific example, are you asking what element of $A={1,2,3}$ is disjoint from $A$? In order to answer that, you have to have some definition of $1,2$, and $3$ as sets. The usual one is that $1={0},2={0,1}$, and $3={0,1,2}$. If you’re using that definition, then $x=1$ works:



          $$xcap A=1cap{1,2,3}={0}cap{1,2,3}=varnothing;,$$



          because the only element of $1$ is $0$, and $0notin{1,2,3}$.



          For your second question, suppose that $A=langle a_n:ninBbb N}$ is a family of sets with the property that $a_{n+1}in a_n$ for each $ninBbb N$:



          $$ldotsin a_4in a_3in a_2in a_1in a_0;.$$



          Let $xin A$. Then $x=a_n$ for some $ninBbb N$, and $a_{n+1}in xcap A$, so $a_{n+1}nevarnothing$. Since $A$ is clearly non-empty, this contradicts $(1)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 8 '13 at 2:46









          Brian M. ScottBrian M. Scott

          460k40517918




          460k40517918












          • $begingroup$
            Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
            $endgroup$
            – user84815
            Jul 8 '13 at 2:56










          • $begingroup$
            @user84815: You’re welcome!
            $endgroup$
            – Brian M. Scott
            Jul 8 '13 at 3:07


















          • $begingroup$
            Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
            $endgroup$
            – user84815
            Jul 8 '13 at 2:56










          • $begingroup$
            @user84815: You’re welcome!
            $endgroup$
            – Brian M. Scott
            Jul 8 '13 at 3:07
















          $begingroup$
          Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
          $endgroup$
          – user84815
          Jul 8 '13 at 2:56




          $begingroup$
          Thanks for another helpful explanation! I was assuming that 1,2,3 were defined by the standard construction of natural numbers and I apologize for mistakenly omitting that assumption.
          $endgroup$
          – user84815
          Jul 8 '13 at 2:56












          $begingroup$
          @user84815: You’re welcome!
          $endgroup$
          – Brian M. Scott
          Jul 8 '13 at 3:07




          $begingroup$
          @user84815: You’re welcome!
          $endgroup$
          – Brian M. Scott
          Jul 8 '13 at 3:07











          2












          $begingroup$

          First, the alternative isn't $emptyset in A$, as you have it, but $emptyset=A$. Hence every nonempty $A$ has an element that isn't a subset.



          Suppose now that $$A={1,2,{1}, {{1}},{1,2,4},3}$$



          If we take $x={1}$, then $xcap A={1}neq emptyset$. If we take $x={1,2,4}$, then $xcap A={1,2}neqemptyset$. However if we take $x=1$, then $xcap A=emptyset$.



          Suppose now that there were an infinitely descending subsequence $$x_1ni x_2 ni x_3 ni cdots$$



          We could then take $$A={x_1,x_2,x_3,ldots}$$ which would voilate the axiom, because for each $i$, $Acap x_i$ is a set which contains $x_{i+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the help! This definitely clears things up.
            $endgroup$
            – user84815
            Jul 8 '13 at 3:00
















          2












          $begingroup$

          First, the alternative isn't $emptyset in A$, as you have it, but $emptyset=A$. Hence every nonempty $A$ has an element that isn't a subset.



          Suppose now that $$A={1,2,{1}, {{1}},{1,2,4},3}$$



          If we take $x={1}$, then $xcap A={1}neq emptyset$. If we take $x={1,2,4}$, then $xcap A={1,2}neqemptyset$. However if we take $x=1$, then $xcap A=emptyset$.



          Suppose now that there were an infinitely descending subsequence $$x_1ni x_2 ni x_3 ni cdots$$



          We could then take $$A={x_1,x_2,x_3,ldots}$$ which would voilate the axiom, because for each $i$, $Acap x_i$ is a set which contains $x_{i+1}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the help! This definitely clears things up.
            $endgroup$
            – user84815
            Jul 8 '13 at 3:00














          2












          2








          2





          $begingroup$

          First, the alternative isn't $emptyset in A$, as you have it, but $emptyset=A$. Hence every nonempty $A$ has an element that isn't a subset.



          Suppose now that $$A={1,2,{1}, {{1}},{1,2,4},3}$$



          If we take $x={1}$, then $xcap A={1}neq emptyset$. If we take $x={1,2,4}$, then $xcap A={1,2}neqemptyset$. However if we take $x=1$, then $xcap A=emptyset$.



          Suppose now that there were an infinitely descending subsequence $$x_1ni x_2 ni x_3 ni cdots$$



          We could then take $$A={x_1,x_2,x_3,ldots}$$ which would voilate the axiom, because for each $i$, $Acap x_i$ is a set which contains $x_{i+1}$.






          share|cite|improve this answer









          $endgroup$



          First, the alternative isn't $emptyset in A$, as you have it, but $emptyset=A$. Hence every nonempty $A$ has an element that isn't a subset.



          Suppose now that $$A={1,2,{1}, {{1}},{1,2,4},3}$$



          If we take $x={1}$, then $xcap A={1}neq emptyset$. If we take $x={1,2,4}$, then $xcap A={1,2}neqemptyset$. However if we take $x=1$, then $xcap A=emptyset$.



          Suppose now that there were an infinitely descending subsequence $$x_1ni x_2 ni x_3 ni cdots$$



          We could then take $$A={x_1,x_2,x_3,ldots}$$ which would voilate the axiom, because for each $i$, $Acap x_i$ is a set which contains $x_{i+1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 8 '13 at 2:46









          vadim123vadim123

          76.5k897191




          76.5k897191












          • $begingroup$
            Thank you for the help! This definitely clears things up.
            $endgroup$
            – user84815
            Jul 8 '13 at 3:00


















          • $begingroup$
            Thank you for the help! This definitely clears things up.
            $endgroup$
            – user84815
            Jul 8 '13 at 3:00
















          $begingroup$
          Thank you for the help! This definitely clears things up.
          $endgroup$
          – user84815
          Jul 8 '13 at 3:00




          $begingroup$
          Thank you for the help! This definitely clears things up.
          $endgroup$
          – user84815
          Jul 8 '13 at 3:00


















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