Mixing
Axiom of Specification as in Halmos' Naive set theory
$begingroup$
As I understand, to apply the Axiom of Specification in ZF set theory, we have to specify a set $A$ to apply it to. However, I don't quite understand why Halmos says that
In case $S(x)$ is $(xnotin x)$ or $(x=x)$, the specified $x$'s do not constitute a set.
by the end of Section 3 in his book Naive Set Theory.
I want to verify what kind of set he is using to apply the Axiom of Specification. Besides, in the previous section he already showed that we can produce a set from an arbitrary set with the sentence $x notin x$. So how come that ”the specified x’s do not constitute a set”? I would appreciate if someone could help explain this to me.
elementary-set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
$endgroup$
add a comment |
$begingroup$
As I understand, to apply the Axiom of Specification in ZF set theory, we have to specify a set $A$ to apply it to. However, I don't quite understand why Halmos says that
In case $S(x)$ is $(xnotin x)$ or $(x=x)$, the specified $x$'s do not constitute a set.
by the end of Section 3 in his book Naive Set Theory.
I want to verify what kind of set he is using to apply the Axiom of Specification. Besides, in the previous section he already showed that we can produce a set from an arbitrary set with the sentence $x notin x$. So how come that ”the specified x’s do not constitute a set”? I would appreciate if someone could help explain this to me.
elementary-set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
$endgroup$
$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
$endgroup$
– Macrophage
Jan 19 at 2:06
add a comment |
$begingroup$
As I understand, to apply the Axiom of Specification in ZF set theory, we have to specify a set $A$ to apply it to. However, I don't quite understand why Halmos says that
In case $S(x)$ is $(xnotin x)$ or $(x=x)$, the specified $x$'s do not constitute a set.
by the end of Section 3 in his book Naive Set Theory.
I want to verify what kind of set he is using to apply the Axiom of Specification. Besides, in the previous section he already showed that we can produce a set from an arbitrary set with the sentence $x notin x$. So how come that ”the specified x’s do not constitute a set”? I would appreciate if someone could help explain this to me.
elementary-set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
$endgroup$
As I understand, to apply the Axiom of Specification in ZF set theory, we have to specify a set $A$ to apply it to. However, I don't quite understand why Halmos says that
In case $S(x)$ is $(xnotin x)$ or $(x=x)$, the specified $x$'s do not constitute a set.
by the end of Section 3 in his book Naive Set Theory.
I want to verify what kind of set he is using to apply the Axiom of Specification. Besides, in the previous section he already showed that we can produce a set from an arbitrary set with the sentence $x notin x$. So how come that ”the specified x’s do not constitute a set”? I would appreciate if someone could help explain this to me.
elementary-set-theory axioms
elementary-set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.6k449115
66.6k449115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
asked Jan 18 at 9:20
MacrophageMacrophage
1,181115
1,181115
$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
$endgroup$
– Macrophage
Jan 19 at 2:06
add a comment |
$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
$endgroup$
– Macrophage
Jan 19 at 2:06
$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
$endgroup$
– Macrophage
Jan 19 at 2:06
$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
$endgroup$
– Macrophage
Jan 19 at 2:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See page 7 :
To specify a set, it is not enough to pronounce some magic words (which may form a
sentence such as "$x notin x$"); it is necessary also to have at hand a set to
whose elements the magic words apply.
We have to review the text of the Axiom of Specification :
To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.
This means that we can use the formula $x notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x in A$ that satisfy the condition.
In symbols :
$B = {x : x in A text { and } S(x) }$.
Having said that :
what kind of set he is using to apply the Axiom of Specification ?
A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.
Regarding Halmos' explanation :
in this notation, the role of $S(x)$ is now played by $x notin x$.
It follows that, whatever the set $A$ may be, if $B = {x : x in A text { and } x notin' x }$, then, for all $y$,
$y in B text { if and only if } (y in A text { and } y notin y)$.
we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").
how come that ”the specified x’s do not constitute a set”?
Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
$endgroup$
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
$endgroup$
– Mauro ALLEGRANZA
Jan 18 at 14:59
|
show 3 more comments
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$begingroup$
See page 7 :
To specify a set, it is not enough to pronounce some magic words (which may form a
sentence such as "$x notin x$"); it is necessary also to have at hand a set to
whose elements the magic words apply.
We have to review the text of the Axiom of Specification :
To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.
This means that we can use the formula $x notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x in A$ that satisfy the condition.
In symbols :
$B = {x : x in A text { and } S(x) }$.
Having said that :
what kind of set he is using to apply the Axiom of Specification ?
A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.
Regarding Halmos' explanation :
in this notation, the role of $S(x)$ is now played by $x notin x$.
It follows that, whatever the set $A$ may be, if $B = {x : x in A text { and } x notin' x }$, then, for all $y$,
$y in B text { if and only if } (y in A text { and } y notin y)$.
we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").
how come that ”the specified x’s do not constitute a set”?
Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
$endgroup$
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
$endgroup$
– Mauro ALLEGRANZA
Jan 18 at 14:59
|
show 3 more comments
$begingroup$
See page 7 :
To specify a set, it is not enough to pronounce some magic words (which may form a
sentence such as "$x notin x$"); it is necessary also to have at hand a set to
whose elements the magic words apply.
We have to review the text of the Axiom of Specification :
To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.
This means that we can use the formula $x notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x in A$ that satisfy the condition.
In symbols :
$B = {x : x in A text { and } S(x) }$.
Having said that :
what kind of set he is using to apply the Axiom of Specification ?
A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.
Regarding Halmos' explanation :
in this notation, the role of $S(x)$ is now played by $x notin x$.
It follows that, whatever the set $A$ may be, if $B = {x : x in A text { and } x notin' x }$, then, for all $y$,
$y in B text { if and only if } (y in A text { and } y notin y)$.
we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").
how come that ”the specified x’s do not constitute a set”?
Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
$endgroup$
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
$endgroup$
– Mauro ALLEGRANZA
Jan 18 at 14:59
|
show 3 more comments
$begingroup$
See page 7 :
To specify a set, it is not enough to pronounce some magic words (which may form a
sentence such as "$x notin x$"); it is necessary also to have at hand a set to
whose elements the magic words apply.
We have to review the text of the Axiom of Specification :
To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.
This means that we can use the formula $x notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x in A$ that satisfy the condition.
In symbols :
$B = {x : x in A text { and } S(x) }$.
Having said that :
what kind of set he is using to apply the Axiom of Specification ?
A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.
Regarding Halmos' explanation :
in this notation, the role of $S(x)$ is now played by $x notin x$.
It follows that, whatever the set $A$ may be, if $B = {x : x in A text { and } x notin' x }$, then, for all $y$,
$y in B text { if and only if } (y in A text { and } y notin y)$.
we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").
how come that ”the specified x’s do not constitute a set”?
Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
$endgroup$
See page 7 :
To specify a set, it is not enough to pronounce some magic words (which may form a
sentence such as "$x notin x$"); it is necessary also to have at hand a set to
whose elements the magic words apply.
We have to review the text of the Axiom of Specification :
To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.
This means that we can use the formula $x notin x$ as "condition $S(x)$" only in conjunction with and already existent $A$ to "carve out" the subset $B$ of $A$ of all and only those elements $x in A$ that satisfy the condition.
In symbols :
$B = {x : x in A text { and } S(x) }$.
Having said that :
what kind of set he is using to apply the Axiom of Specification ?
A set $A$ whatever, provided that we already know that it exists. This is the gist of the axiom : it does not produce new sets out of nothing, but allows use to manufacture them only as subsets of alredy existsing sets.
Regarding Halmos' explanation :
in this notation, the role of $S(x)$ is now played by $x notin x$.
It follows that, whatever the set $A$ may be, if $B = {x : x in A text { and } x notin' x }$, then, for all $y$,
$y in B text { if and only if } (y in A text { and } y notin y)$.
we have only to note that the "condition" $S(x)$ is expressed by a formula whatever of the language of set theory, with a free variable $x$ (a "parameter").
how come that ”the specified x’s do not constitute a set”?
Because in ordet to specify a set we have to consider an already existsing set $A$ and a "specifying condition" $S(x)$.
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
edited Jan 20 at 14:15
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
answered Jan 18 at 10:44
Mauro ALLEGRANZAMauro ALLEGRANZA
66.6k449115
66.6k449115
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
$endgroup$
– Mauro ALLEGRANZA
Jan 18 at 14:59
|
show 3 more comments
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
$endgroup$
– Mauro ALLEGRANZA
Jan 18 at 14:59
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
This I understand. However, when Halmos uses notations like ${x:S(x)}$, he's not explicitly stating the set $A$. Does he only tacitly do this when there's no confusion that possibly arise (such as when $S(x)$ stands for $xneq x$)?
$endgroup$
– Macrophage
Jan 18 at 10:48
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
Thank you for adding extra explanations about formulas. However, I'm not sure if that address my problem. With $A(x)=x notin x$ we can still find a set B for any set A...so how does Halmos say what I quoted. still didn't get it.
$endgroup$
– Macrophage
Jan 18 at 14:33
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
But what I read is that in the section for the Axiom of Specification he used that sentence on any arbitrary set A to prove there's a set B that does not belong to A? So from any arbitrary set A, this B should be an existent set that can be induced with $S(x)=x notin x$?
$endgroup$
– Macrophage
Jan 18 at 14:45
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
$endgroup$
– Macrophage
Jan 18 at 14:51
$begingroup$
I'm not sure what is the set he specifies for the Axiom in the section for unordered pairs where I quoted the sentence in my question, though.
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– Macrophage
Jan 18 at 14:51
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@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
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– Mauro ALLEGRANZA
Jan 18 at 14:59
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@Macrophage - now the issue is about Axiom of Pairing ? This is a new one: it asserts that, given two sets $a$ and $b$, it exists a third set - call it $A$ - that contain them. Again, the following comment is a little bit subtle... Given Pairing and using Spec, Halmos proves that exists a set with only $a$ and $b$. Pairing asserts that $A$ contains $a$ and $b$ but not asserts that it contains only them...
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– Mauro ALLEGRANZA
Jan 18 at 14:59
|
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$begingroup$
Potential duplicate: math.stackexchange.com/questions/1962070/…
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– Macrophage
Jan 19 at 2:06