Sets Without a Minimal Element (Axiom of Foundation)
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I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,
i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}
every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?
elementary-set-theory
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
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add a comment |
$begingroup$
I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,
i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}
every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?
elementary-set-theory
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
$endgroup$
3
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
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– user3482749
Dec 31 '18 at 17:46
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"Proper" should be "nonempty."
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– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15
add a comment |
$begingroup$
I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,
i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}
every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?
elementary-set-theory
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
$endgroup$
I’m trying to move from an initial appreciation of Cantor’s arguments concerning infinite sets to a more rigorous, axiom-based understanding. I’m having trouble with axiom 8, the Axiom of Foundation or Regularity, in Zermelo-Fraenkel set theory, however. This axiom is explained as saying that every non-empty set has a minimal member, where “minimal” means “does not contain a proper sub-set of the set”.
The other discussions I’ve seen here focus on avoiding infinite descending chains of sets, but it seems straightforward to construct non-descending counter-examples to the axiom.
In a set where the nth member is the set of positive integers starting with value n, for example,
i.e. {{1, 2, 3, 4, …}, {2, 3, 4, 5, …}, {3, 4, 5, 6, …}, …}
every member contains the next member (and all subsequent members) as a proper subset.
Isn’t this true of every set whose members are either infinite series repeated with progressively later starting values, as above, or infinitely sub-divided sets? (E.g. the set of integers contains the set of even numbers, which contains the set of multiples of 4, and so on.)
If not, why not?
elementary-set-theory
elementary-set-theory
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
asked Dec 31 '18 at 17:40
Bill SwanBill Swan
31
31
3
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46
$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15
add a comment |
3
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46
$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15
3
3
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46
$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Regularity is focusing on $in$, not $subseteq$.
First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$
In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.
That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.
Regularity says:
For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.
Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
$endgroup$
add a comment |
$begingroup$
every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
$endgroup$
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Regularity is focusing on $in$, not $subseteq$.
First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$
In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.
That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.
Regularity says:
For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.
Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
$endgroup$
add a comment |
$begingroup$
Regularity is focusing on $in$, not $subseteq$.
First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$
In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.
That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.
Regularity says:
For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.
Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
$endgroup$
add a comment |
$begingroup$
Regularity is focusing on $in$, not $subseteq$.
First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$
In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.
That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.
Regularity says:
For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.
Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
$endgroup$
Regularity is focusing on $in$, not $subseteq$.
First, let me focus on what we're trying to do. Intuitively, what we want to prevent is the situation $$color{red}{ani bni cni dni eni ...}$$ we're perfectly happy with $$asupseteq bsupseteq csupseteq dsupseteq esupseteq ...$$
In the example you give, we don't have a descending $in$-sequence since ${2,3,4,...}notin{1,2,3,4,...}$; what you have is a descending $subseteq$-sequence, which isn't what we're worried about.
That's clear enough, but in my experience (and I think this is the situation here) the confusion arises because of the odd phrasing of the axiom of regularity.
Regularity says:
For every nonempty set $X$, there is some $min X$ such that $mcap X=emptyset$.
Note that I wrote "$mcap X=emptyset$" instead of "$m$ contains no nonempty subset of $X$" - I think the latter is more confusing (especially because it means "contains" in the sense of $supseteq$, not $ni$!). Considering your example, any of the elements of the set work as $m$: e.g. ${1,2,3,4,...}$ works, since none of the elements of this are also elements of your set.
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
answered Dec 31 '18 at 18:05
Noah SchweberNoah Schweber
126k10150288
126k10150288
add a comment |
add a comment |
$begingroup$
every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
$endgroup$
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
add a comment |
$begingroup$
every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
$endgroup$
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
add a comment |
$begingroup$
every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
$endgroup$
every member contains the next member (and all subsequent members) as a proper subset. yes, but the axiom says that there is minimal element with respect to $∈$, here you have that any element is minimal with respect to $∈$, non with respect to $⊆$ (${1,2,3,ldots}⊇{2,3,ldots}$ but ${1,2,3,ldots}∌{2,3,ldots}$)
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
answered Dec 31 '18 at 18:04
HoloHolo
5,87421131
5,87421131
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
add a comment |
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
$begingroup$
Thanks for these replies. I can’t (can’t yet?) read the more formal ones (e.g. I hadn’t encountered the sideways hook symbol with a line under it, which I don’t know how to type), but the common theme seems to be that elements of sets are not the same thing as sets of elements. So, for example, {1,2} and {1,2,3} have a non-empty intersection because they share elements 1 and 2, not because they both contain {1,2}. Is that, crudely and intuitively, about right? Or am I still missing the point?
$endgroup$
– Bill Swan
Jan 2 at 10:48
add a comment |
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3
$begingroup$
Here minimality is with respect to $in$, not to $subseteq$.
$endgroup$
– Lord Shark the Unknown
Dec 31 '18 at 17:41
$begingroup$
No element of that set contains a proper subset of the set. Or, indeed, any subset of the set: in particular, every element of that set is a set of numbers, whereas every element of any element of that set is a number, so the intersection of your set with any of its elements is empty.
$endgroup$
– user3482749
Dec 31 '18 at 17:46
$begingroup$
"Proper" should be "nonempty."
$endgroup$
– Noah Schweber
Dec 31 '18 at 18:06
$begingroup$
math.stackexchange.com/questions/214408/…
$endgroup$
– Asaf Karagila♦
Dec 31 '18 at 18:15