From an axiomatic set theoric approach why can we take uncountable unions?
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From an axiomatic set theoric approach why can we take uncountable unions in the first place? Sure if $A$ and $B$ are sets then the thing denoted by $Acup B$ exists and is a set, but if we want to extend this to an uncountable union shouldn't we assume some form of transfinite induction (and hence $sf AC$)?
elementary-set-theory
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
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add a comment |
$begingroup$
From an axiomatic set theoric approach why can we take uncountable unions in the first place? Sure if $A$ and $B$ are sets then the thing denoted by $Acup B$ exists and is a set, but if we want to extend this to an uncountable union shouldn't we assume some form of transfinite induction (and hence $sf AC$)?
elementary-set-theory
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
$endgroup$
add a comment |
$begingroup$
From an axiomatic set theoric approach why can we take uncountable unions in the first place? Sure if $A$ and $B$ are sets then the thing denoted by $Acup B$ exists and is a set, but if we want to extend this to an uncountable union shouldn't we assume some form of transfinite induction (and hence $sf AC$)?
elementary-set-theory
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
$endgroup$
From an axiomatic set theoric approach why can we take uncountable unions in the first place? Sure if $A$ and $B$ are sets then the thing denoted by $Acup B$ exists and is a set, but if we want to extend this to an uncountable union shouldn't we assume some form of transfinite induction (and hence $sf AC$)?
elementary-set-theory
elementary-set-theory
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
asked Mar 10 at 8:54
Stupid Questions IncStupid Questions Inc
12811
12811
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then ${amidexists xin X: ain x}$ is a set, and we denote that set as $bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X={A,B}$ we normally write $Acup B$ and not $bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $bigcapvarnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
add a comment |
$begingroup$
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that
$$ forall xcolon (xin Uleftrightarrow exists yin Acolon xin y).$$
edited Mar 10 at 18:38
Tanner Swett
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
add a comment |
$begingroup$
No, one does not need the axiom of choice. The axiom of replacement (the image of any set under any definable mapping is also a set) together with the axiom of union allows one to form countable unions, and, indeed, unions indexed by sets of any already defined cardinality. This is why the scale of alephs can be constructed in ZF without using choice, it is only needed to prove that every set is bijective to an aleph. It is interesting that Fraenkel added replacement to Zermelo's original axioms when it was realized that without it one can not go beyond aleph omega.
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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active
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$begingroup$
The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then ${amidexists xin X: ain x}$ is a set, and we denote that set as $bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X={A,B}$ we normally write $Acup B$ and not $bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $bigcapvarnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
add a comment |
$begingroup$
The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then ${amidexists xin X: ain x}$ is a set, and we denote that set as $bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X={A,B}$ we normally write $Acup B$ and not $bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $bigcapvarnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
add a comment |
$begingroup$
The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then ${amidexists xin X: ain x}$ is a set, and we denote that set as $bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X={A,B}$ we normally write $Acup B$ and not $bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $bigcapvarnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
$endgroup$
The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then ${amidexists xin X: ain x}$ is a set, and we denote that set as $bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X={A,B}$ we normally write $Acup B$ and not $bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $bigcapvarnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
answered Mar 10 at 9:05
Asaf Karagila♦Asaf Karagila
308k33441775
308k33441775
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
add a comment |
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
1
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
Mar 10 at 11:18
add a comment |
$begingroup$
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that
$$ forall xcolon (xin Uleftrightarrow exists yin Acolon xin y).$$
edited Mar 10 at 18:38
Tanner Swett
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
add a comment |
$begingroup$
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that
$$ forall xcolon (xin Uleftrightarrow exists yin Acolon xin y).$$
edited Mar 10 at 18:38
Tanner Swett
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
add a comment |
$begingroup$
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that
$$ forall xcolon (xin Uleftrightarrow exists yin Acolon xin y).$$
edited Mar 10 at 18:38
Tanner Swett
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that
$$ forall xcolon (xin Uleftrightarrow exists yin Acolon xin y).$$
edited Mar 10 at 18:38
Tanner Swett
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
edited Mar 10 at 18:38
Tanner Swett
4,3291739
edited Mar 10 at 18:38
Tanner Swett
4,3291739
edited Mar 10 at 18:38
Tanner Swett
4,3291739
4,3291739
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
answered Mar 10 at 9:07
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
add a comment |
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
4
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
Mar 10 at 13:01
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
$begingroup$
I've taken the liberty of changing "$y in U$" to "$y in A$", since it seems clear that that's what you meant.
$endgroup$
– Tanner Swett
Mar 10 at 18:40
add a comment |
$begingroup$
No, one does not need the axiom of choice. The axiom of replacement (the image of any set under any definable mapping is also a set) together with the axiom of union allows one to form countable unions, and, indeed, unions indexed by sets of any already defined cardinality. This is why the scale of alephs can be constructed in ZF without using choice, it is only needed to prove that every set is bijective to an aleph. It is interesting that Fraenkel added replacement to Zermelo's original axioms when it was realized that without it one can not go beyond aleph omega.
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
$endgroup$
add a comment |
$begingroup$
No, one does not need the axiom of choice. The axiom of replacement (the image of any set under any definable mapping is also a set) together with the axiom of union allows one to form countable unions, and, indeed, unions indexed by sets of any already defined cardinality. This is why the scale of alephs can be constructed in ZF without using choice, it is only needed to prove that every set is bijective to an aleph. It is interesting that Fraenkel added replacement to Zermelo's original axioms when it was realized that without it one can not go beyond aleph omega.
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
$endgroup$
add a comment |
$begingroup$
No, one does not need the axiom of choice. The axiom of replacement (the image of any set under any definable mapping is also a set) together with the axiom of union allows one to form countable unions, and, indeed, unions indexed by sets of any already defined cardinality. This is why the scale of alephs can be constructed in ZF without using choice, it is only needed to prove that every set is bijective to an aleph. It is interesting that Fraenkel added replacement to Zermelo's original axioms when it was realized that without it one can not go beyond aleph omega.
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
$endgroup$
No, one does not need the axiom of choice. The axiom of replacement (the image of any set under any definable mapping is also a set) together with the axiom of union allows one to form countable unions, and, indeed, unions indexed by sets of any already defined cardinality. This is why the scale of alephs can be constructed in ZF without using choice, it is only needed to prove that every set is bijective to an aleph. It is interesting that Fraenkel added replacement to Zermelo's original axioms when it was realized that without it one can not go beyond aleph omega.
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
answered Mar 10 at 9:11
ConifoldConifold
7,06521744
7,06521744
add a comment |
add a comment |
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