From an axiomatic set theoric approach why can we take uncountable unions?A question regarding the validity of the Union AxiomIf $B$ is an uncountable set and $A$ is a countable set, then prove that $B$ is similar to $B-A$.its about induction and i got a contradictionWhy doesn't this definition of natural numbers hold up in axiomatic set theory?Extract countable set from uncountable setUncountable neighborhoods of an uncountable setProof explanation for the statement that $Bbb R$ can be partitioned into a union of uncountable sets where the index set is also uncountableWhy Cantor set is uncountable despite each element is rational.Given an infinite set, can we give an infinite chain of finite subsets of that set, with the chain ordered by the proper subset operator?The minimal uncountable well-ordered set has no largest element
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From an axiomatic set theoric approach why can we take uncountable unions?
A question regarding the validity of the Union AxiomIf $B$ is an uncountable set and $A$ is a countable set, then prove that $B$ is similar to $B-A$.its about induction and i got a contradictionWhy doesn't this definition of natural numbers hold up in axiomatic set theory?Extract countable set from uncountable setUncountable neighborhoods of an uncountable setProof explanation for the statement that $Bbb R$ can be partitioned into a union of uncountable sets where the index set is also uncountableWhy Cantor set is uncountable despite each element is rational.Given an infinite set, can we give an infinite chain of finite subsets of that set, with the chain ordered by the proper subset operator?The minimal uncountable well-ordered set has no largest element
$begingroup$
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 2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
Stupid Questions IncStupid Questions Inc
12811
asked 2 days ago
Stupid Questions IncStupid Questions Inc
12811
asked 2 days ago
Stupid Questions IncStupid Questions Inc
12811
asked 2 days ago
Stupid Questions IncStupid Questions Inc
12811
asked 2 days ago
Stupid Questions IncStupid Questions Inc
12811
12811
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
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 2 days ago
Tanner Swett
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
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 2 days ago
ConifoldConifold
7,02521744
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
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active
oldest
votes
$begingroup$
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 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
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 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
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 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
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 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
answered 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
answered 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
answered 2 days ago
Asaf Karagila♦Asaf Karagila
306k33438769
306k33438769
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
add a comment |
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
1
1
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
$begingroup$
thanks a ton our $sf AC$ master
$endgroup$
– Stupid Questions Inc
2 days ago
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 2 days ago
Tanner Swett
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
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 2 days ago
Tanner Swett
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
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 2 days ago
Tanner Swett
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
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 2 days ago
Tanner Swett
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
edited 2 days ago
Tanner Swett
4,2581639
edited 2 days ago
Tanner Swett
4,2581639
edited 2 days ago
Tanner Swett
4,2581639
4,2581639
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 2 days ago
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
add a comment |
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
4
4
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
You mean "$yin A$".
$endgroup$
– freakish
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
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 2 days ago
ConifoldConifold
7,02521744
$endgroup$
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 2 days ago
ConifoldConifold
7,02521744
$endgroup$
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 2 days ago
ConifoldConifold
7,02521744
$endgroup$
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 2 days ago
ConifoldConifold
7,02521744
answered 2 days ago
ConifoldConifold
7,02521744
answered 2 days ago
ConifoldConifold
7,02521744
answered 2 days ago
ConifoldConifold
7,02521744
7,02521744
add a comment |
add a comment |
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