Axiom of Regularity and infinite sequences
$begingroup$
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
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
$endgroup$
add a comment |
$begingroup$
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
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
$endgroup$
add a comment |
$begingroup$
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
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
$endgroup$
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
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
edited Sep 14 '13 at 8:27
Sergiy Kozerenko
21628
21628
asked Jul 8 '13 at 2:33
user84815user84815
596
asked Jul 8 '13 at 2:33
user84815user84815
596
asked Jul 8 '13 at 2:33
user84815user84815
596
596
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $(1)$.
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $x_{i+1}$.
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
Thank you for the help! This definitely clears things up.
$endgroup$
– user84815
Jul 8 '13 at 3:00
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $(1)$.
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(1)$.
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(1)$.
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
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 $(1)$.
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
answered Jul 8 '13 at 2:46
Brian M. ScottBrian M. Scott
460k40517918
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $x_{i+1}$.
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
Thank you for the help! This definitely clears things up.
$endgroup$
– user84815
Jul 8 '13 at 3:00
add a comment |
$begingroup$
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 $x_{i+1}$.
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
$endgroup$
$begingroup$
Thank you for the help! This definitely clears things up.
$endgroup$
– user84815
Jul 8 '13 at 3:00
add a comment |
$begingroup$
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 $x_{i+1}$.
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
$endgroup$
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 $x_{i+1}$.
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
answered Jul 8 '13 at 2:46
vadim123vadim123
76.5k897191
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
add a comment |
$begingroup$
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
add a comment |
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