Mixing
Alternative axioms for NBG or MK
$begingroup$
While I was thinking about NBG and MK I had the idea for two alternative axioms. As usual $V$ is the class of sets.
The first one:
For a boolean function $f : {T,F}^n to {T,F}$ let $varphi_f(x_1,dots,x_n)$ be a formal representation of $f$. That means for $a_1,dots, a_n in {T,F}$ we have $f(a_1,dots,a_n) = T Leftrightarrow modelsvarphi_f(a_1,dots,a_n)$. (I'm identifying T, F with $top$, $bot$). So for example if $f$ is the AND-function we have $varphi_f = (x_1 wedge x_n)$.
Axiom 1 (scheme):
For all boolean functions $f : {T,F}^n to {T,F}$ and all $R_1, dots, R_n in {=,in,subseteq}$:
For all $b_1,dots, b_n in V$ we have
${x; varphi_f(x R_1 b_1, dots, x R_n b_n)} in V quad Leftrightarrow quad negvarphi_f(bot, dots, bot)$ $ $ (or semantically $f(F,dots,F) = F$)
This axiom implies
- EmptySet $quad$ (choose $n=0$ and $f(langlerangle) = F$; $ langlerangle in {T,F}^0$ is the empty sequence)
- Pairing $quad$ (choose $n=2$, "$f = AND$" and $R_1,R_2 =; =$)
- Powerset $quad$ (choose $n=1$, "$f = $ identity" and $R_1 =; subseteq$)
- SmallUnion $quad$ (choose $n=2$, "$f = OR$" and $R_1, R_2 =; in$)
and others... (Let $a in^2 b :Leftrightarrow exists c (a in c wedge c in b)$. If we allow the $R_i$ to be $in^2$ we have Union too. )
Further: the axiom states that many classes are proper (without the help of other axioms).
And I'm quite sure that this axiom follows from NBG/MK.
If we choose Extensionality, (Foundation), Class Comprehension, Limitation of Size, Infinity and our Axiom 1 we have a version of NBG resp. MK which is easy to remember. What do you think?
The second:
Axiom 2:
If $X$ is a class of non-empty disjoint sets, then $X$ is a set iff there is a choice set for $X$.
(I think the formalisation is clear)
This axiom is a fusion of choice and (at least) a part of replacement. So for example if we have a class function $f: A to B$ and $A$ is a set that contains no pairs, we could build the class $X = { {x, langle x, f(x)rangle}; x in A}$ and use our axiom to conclude that $X$ is a set (since $A$ is obviously a choice set of $X$). With Union and Separation we get that the image of $f$ is a set too.
My first question:
Is Axiom 2 equivalent to choice and replacement (modulo other standard axioms)?
My second question:
Are similar axioms studied somewhere?
set-theory axioms foundations
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
$endgroup$
add a comment |
$begingroup$
While I was thinking about NBG and MK I had the idea for two alternative axioms. As usual $V$ is the class of sets.
The first one:
For a boolean function $f : {T,F}^n to {T,F}$ let $varphi_f(x_1,dots,x_n)$ be a formal representation of $f$. That means for $a_1,dots, a_n in {T,F}$ we have $f(a_1,dots,a_n) = T Leftrightarrow modelsvarphi_f(a_1,dots,a_n)$. (I'm identifying T, F with $top$, $bot$). So for example if $f$ is the AND-function we have $varphi_f = (x_1 wedge x_n)$.
Axiom 1 (scheme):
For all boolean functions $f : {T,F}^n to {T,F}$ and all $R_1, dots, R_n in {=,in,subseteq}$:
For all $b_1,dots, b_n in V$ we have
${x; varphi_f(x R_1 b_1, dots, x R_n b_n)} in V quad Leftrightarrow quad negvarphi_f(bot, dots, bot)$ $ $ (or semantically $f(F,dots,F) = F$)
This axiom implies
- EmptySet $quad$ (choose $n=0$ and $f(langlerangle) = F$; $ langlerangle in {T,F}^0$ is the empty sequence)
- Pairing $quad$ (choose $n=2$, "$f = AND$" and $R_1,R_2 =; =$)
- Powerset $quad$ (choose $n=1$, "$f = $ identity" and $R_1 =; subseteq$)
- SmallUnion $quad$ (choose $n=2$, "$f = OR$" and $R_1, R_2 =; in$)
and others... (Let $a in^2 b :Leftrightarrow exists c (a in c wedge c in b)$. If we allow the $R_i$ to be $in^2$ we have Union too. )
Further: the axiom states that many classes are proper (without the help of other axioms).
And I'm quite sure that this axiom follows from NBG/MK.
If we choose Extensionality, (Foundation), Class Comprehension, Limitation of Size, Infinity and our Axiom 1 we have a version of NBG resp. MK which is easy to remember. What do you think?
The second:
Axiom 2:
If $X$ is a class of non-empty disjoint sets, then $X$ is a set iff there is a choice set for $X$.
(I think the formalisation is clear)
This axiom is a fusion of choice and (at least) a part of replacement. So for example if we have a class function $f: A to B$ and $A$ is a set that contains no pairs, we could build the class $X = { {x, langle x, f(x)rangle}; x in A}$ and use our axiom to conclude that $X$ is a set (since $A$ is obviously a choice set of $X$). With Union and Separation we get that the image of $f$ is a set too.
My first question:
Is Axiom 2 equivalent to choice and replacement (modulo other standard axioms)?
My second question:
Are similar axioms studied somewhere?
set-theory axioms foundations
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
$endgroup$
$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25
add a comment |
$begingroup$
While I was thinking about NBG and MK I had the idea for two alternative axioms. As usual $V$ is the class of sets.
The first one:
For a boolean function $f : {T,F}^n to {T,F}$ let $varphi_f(x_1,dots,x_n)$ be a formal representation of $f$. That means for $a_1,dots, a_n in {T,F}$ we have $f(a_1,dots,a_n) = T Leftrightarrow modelsvarphi_f(a_1,dots,a_n)$. (I'm identifying T, F with $top$, $bot$). So for example if $f$ is the AND-function we have $varphi_f = (x_1 wedge x_n)$.
Axiom 1 (scheme):
For all boolean functions $f : {T,F}^n to {T,F}$ and all $R_1, dots, R_n in {=,in,subseteq}$:
For all $b_1,dots, b_n in V$ we have
${x; varphi_f(x R_1 b_1, dots, x R_n b_n)} in V quad Leftrightarrow quad negvarphi_f(bot, dots, bot)$ $ $ (or semantically $f(F,dots,F) = F$)
This axiom implies
- EmptySet $quad$ (choose $n=0$ and $f(langlerangle) = F$; $ langlerangle in {T,F}^0$ is the empty sequence)
- Pairing $quad$ (choose $n=2$, "$f = AND$" and $R_1,R_2 =; =$)
- Powerset $quad$ (choose $n=1$, "$f = $ identity" and $R_1 =; subseteq$)
- SmallUnion $quad$ (choose $n=2$, "$f = OR$" and $R_1, R_2 =; in$)
and others... (Let $a in^2 b :Leftrightarrow exists c (a in c wedge c in b)$. If we allow the $R_i$ to be $in^2$ we have Union too. )
Further: the axiom states that many classes are proper (without the help of other axioms).
And I'm quite sure that this axiom follows from NBG/MK.
If we choose Extensionality, (Foundation), Class Comprehension, Limitation of Size, Infinity and our Axiom 1 we have a version of NBG resp. MK which is easy to remember. What do you think?
The second:
Axiom 2:
If $X$ is a class of non-empty disjoint sets, then $X$ is a set iff there is a choice set for $X$.
(I think the formalisation is clear)
This axiom is a fusion of choice and (at least) a part of replacement. So for example if we have a class function $f: A to B$ and $A$ is a set that contains no pairs, we could build the class $X = { {x, langle x, f(x)rangle}; x in A}$ and use our axiom to conclude that $X$ is a set (since $A$ is obviously a choice set of $X$). With Union and Separation we get that the image of $f$ is a set too.
My first question:
Is Axiom 2 equivalent to choice and replacement (modulo other standard axioms)?
My second question:
Are similar axioms studied somewhere?
set-theory axioms foundations
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
$endgroup$
While I was thinking about NBG and MK I had the idea for two alternative axioms. As usual $V$ is the class of sets.
The first one:
For a boolean function $f : {T,F}^n to {T,F}$ let $varphi_f(x_1,dots,x_n)$ be a formal representation of $f$. That means for $a_1,dots, a_n in {T,F}$ we have $f(a_1,dots,a_n) = T Leftrightarrow modelsvarphi_f(a_1,dots,a_n)$. (I'm identifying T, F with $top$, $bot$). So for example if $f$ is the AND-function we have $varphi_f = (x_1 wedge x_n)$.
Axiom 1 (scheme):
For all boolean functions $f : {T,F}^n to {T,F}$ and all $R_1, dots, R_n in {=,in,subseteq}$:
For all $b_1,dots, b_n in V$ we have
${x; varphi_f(x R_1 b_1, dots, x R_n b_n)} in V quad Leftrightarrow quad negvarphi_f(bot, dots, bot)$ $ $ (or semantically $f(F,dots,F) = F$)
This axiom implies
- EmptySet $quad$ (choose $n=0$ and $f(langlerangle) = F$; $ langlerangle in {T,F}^0$ is the empty sequence)
- Pairing $quad$ (choose $n=2$, "$f = AND$" and $R_1,R_2 =; =$)
- Powerset $quad$ (choose $n=1$, "$f = $ identity" and $R_1 =; subseteq$)
- SmallUnion $quad$ (choose $n=2$, "$f = OR$" and $R_1, R_2 =; in$)
and others... (Let $a in^2 b :Leftrightarrow exists c (a in c wedge c in b)$. If we allow the $R_i$ to be $in^2$ we have Union too. )
Further: the axiom states that many classes are proper (without the help of other axioms).
And I'm quite sure that this axiom follows from NBG/MK.
If we choose Extensionality, (Foundation), Class Comprehension, Limitation of Size, Infinity and our Axiom 1 we have a version of NBG resp. MK which is easy to remember. What do you think?
The second:
Axiom 2:
If $X$ is a class of non-empty disjoint sets, then $X$ is a set iff there is a choice set for $X$.
(I think the formalisation is clear)
This axiom is a fusion of choice and (at least) a part of replacement. So for example if we have a class function $f: A to B$ and $A$ is a set that contains no pairs, we could build the class $X = { {x, langle x, f(x)rangle}; x in A}$ and use our axiom to conclude that $X$ is a set (since $A$ is obviously a choice set of $X$). With Union and Separation we get that the image of $f$ is a set too.
My first question:
Is Axiom 2 equivalent to choice and replacement (modulo other standard axioms)?
My second question:
Are similar axioms studied somewhere?
set-theory axioms foundations
set-theory axioms foundations
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
edited Jan 16 at 11:56
Popov Florino
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
asked Nov 29 '18 at 11:18
Popov FlorinoPopov Florino
1467
1467
$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25
add a comment |
$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25
$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25
add a comment |
1 Answer
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$begingroup$
I found an answer to the first question: Yes
Let $f : A to B$ be a class function and $A$ a set
(i) If $f$ is injective we define $C := {a in A; f(a) notin A}$. With separation it is a set. Now we build the class $X = { {a,f(a)}; a in C }$. Since $C$ is a choice set of $X$ (and $X$ is a class of non-empty disjoint sets) we get with Axiom 2 that $X$ is a set. If we have small union, union and separation this implies, that $operatorname{im} f = f(C) cup (A cap operatorname{im} f)$ is a set.
(ii) Now consider an arbitrary class function $f$. $f$ defines an equivalence relation $a sim b :Leftrightarrow f(a) = f(b)$ on $A$. With powerset and separation we get, that the class of equivalence classes $A/{sim} := { [a]_sim; a in A}$ is a set. Since $f_{sim}: A/{sim} to B, [a]_{sim} mapsto f(a)$ is injective, we can use (i) to show that $operatorname{im} f = operatorname{im} f_{sim}$ is a set.
So Axiom 2 implies replacement.
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
$endgroup$
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
add a comment |
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$begingroup$
I found an answer to the first question: Yes
Let $f : A to B$ be a class function and $A$ a set
(i) If $f$ is injective we define $C := {a in A; f(a) notin A}$. With separation it is a set. Now we build the class $X = { {a,f(a)}; a in C }$. Since $C$ is a choice set of $X$ (and $X$ is a class of non-empty disjoint sets) we get with Axiom 2 that $X$ is a set. If we have small union, union and separation this implies, that $operatorname{im} f = f(C) cup (A cap operatorname{im} f)$ is a set.
(ii) Now consider an arbitrary class function $f$. $f$ defines an equivalence relation $a sim b :Leftrightarrow f(a) = f(b)$ on $A$. With powerset and separation we get, that the class of equivalence classes $A/{sim} := { [a]_sim; a in A}$ is a set. Since $f_{sim}: A/{sim} to B, [a]_{sim} mapsto f(a)$ is injective, we can use (i) to show that $operatorname{im} f = operatorname{im} f_{sim}$ is a set.
So Axiom 2 implies replacement.
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
$endgroup$
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
add a comment |
$begingroup$
I found an answer to the first question: Yes
Let $f : A to B$ be a class function and $A$ a set
(i) If $f$ is injective we define $C := {a in A; f(a) notin A}$. With separation it is a set. Now we build the class $X = { {a,f(a)}; a in C }$. Since $C$ is a choice set of $X$ (and $X$ is a class of non-empty disjoint sets) we get with Axiom 2 that $X$ is a set. If we have small union, union and separation this implies, that $operatorname{im} f = f(C) cup (A cap operatorname{im} f)$ is a set.
(ii) Now consider an arbitrary class function $f$. $f$ defines an equivalence relation $a sim b :Leftrightarrow f(a) = f(b)$ on $A$. With powerset and separation we get, that the class of equivalence classes $A/{sim} := { [a]_sim; a in A}$ is a set. Since $f_{sim}: A/{sim} to B, [a]_{sim} mapsto f(a)$ is injective, we can use (i) to show that $operatorname{im} f = operatorname{im} f_{sim}$ is a set.
So Axiom 2 implies replacement.
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
$endgroup$
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
add a comment |
$begingroup$
I found an answer to the first question: Yes
Let $f : A to B$ be a class function and $A$ a set
(i) If $f$ is injective we define $C := {a in A; f(a) notin A}$. With separation it is a set. Now we build the class $X = { {a,f(a)}; a in C }$. Since $C$ is a choice set of $X$ (and $X$ is a class of non-empty disjoint sets) we get with Axiom 2 that $X$ is a set. If we have small union, union and separation this implies, that $operatorname{im} f = f(C) cup (A cap operatorname{im} f)$ is a set.
(ii) Now consider an arbitrary class function $f$. $f$ defines an equivalence relation $a sim b :Leftrightarrow f(a) = f(b)$ on $A$. With powerset and separation we get, that the class of equivalence classes $A/{sim} := { [a]_sim; a in A}$ is a set. Since $f_{sim}: A/{sim} to B, [a]_{sim} mapsto f(a)$ is injective, we can use (i) to show that $operatorname{im} f = operatorname{im} f_{sim}$ is a set.
So Axiom 2 implies replacement.
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
$endgroup$
I found an answer to the first question: Yes
Let $f : A to B$ be a class function and $A$ a set
(i) If $f$ is injective we define $C := {a in A; f(a) notin A}$. With separation it is a set. Now we build the class $X = { {a,f(a)}; a in C }$. Since $C$ is a choice set of $X$ (and $X$ is a class of non-empty disjoint sets) we get with Axiom 2 that $X$ is a set. If we have small union, union and separation this implies, that $operatorname{im} f = f(C) cup (A cap operatorname{im} f)$ is a set.
(ii) Now consider an arbitrary class function $f$. $f$ defines an equivalence relation $a sim b :Leftrightarrow f(a) = f(b)$ on $A$. With powerset and separation we get, that the class of equivalence classes $A/{sim} := { [a]_sim; a in A}$ is a set. Since $f_{sim}: A/{sim} to B, [a]_{sim} mapsto f(a)$ is injective, we can use (i) to show that $operatorname{im} f = operatorname{im} f_{sim}$ is a set.
So Axiom 2 implies replacement.
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
edited Dec 14 '18 at 17:10
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
answered Nov 29 '18 at 12:15
Popov FlorinoPopov Florino
1467
1467
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
add a comment |
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
$begingroup$
I hope it is correct now!
$endgroup$
– Popov Florino
Nov 29 '18 at 13:16
add a comment |
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$begingroup$
With "choice set of X" I mean a set $A$, so that $A cap x$ is a singleton for all $x in X$.
$endgroup$
– Popov Florino
Nov 29 '18 at 12:08
$begingroup$
I see, thanks...
$endgroup$
– Carl Mummert
Nov 29 '18 at 12:25