Mixing
Banach–Tarski paradox fails to iterate over all points
Clash Royale CLAN TAG#URR8PPP
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The Banach–Tarski Paradox explanation goes along these lines
(1) Take a point on a sphere
(2) Use 2pi/3 rotations to construct a subset of points
(3) After infinite number of rotations you've selected [an infinite] countable set of points on the sphere
(4) For each point not on the set repeat step (2)
(5) Repeat step (4) while there are unprocessed points
Done. [This paradox uses an axiom of choice]
Here is my concern. I am fine with an axiom of choice in the form "for any non-empty set you can always select a member of a set". However, Banach–Tarski Paradox seems to try to do something very different.
It tries to keep selecting points from a continuum set until no more points remain. This is NOT possible. Infinite while loop can only select Countable number of points, while continuum must be selected to iterate over all points on a line or a sphere.
If this selection is possible then we can enumerate all points on (0,1) interval. Is there something in the proof that I am missing?
set-theory
edited 2 days ago
bof
45.8k348110
asked Aug 4 at 2:22
Stepan
430311
add a comment |Â
up vote
2
down vote
favorite
The Banach–Tarski Paradox explanation goes along these lines
(1) Take a point on a sphere
(2) Use 2pi/3 rotations to construct a subset of points
(3) After infinite number of rotations you've selected [an infinite] countable set of points on the sphere
(4) For each point not on the set repeat step (2)
(5) Repeat step (4) while there are unprocessed points
Done. [This paradox uses an axiom of choice]
Here is my concern. I am fine with an axiom of choice in the form "for any non-empty set you can always select a member of a set". However, Banach–Tarski Paradox seems to try to do something very different.
It tries to keep selecting points from a continuum set until no more points remain. This is NOT possible. Infinite while loop can only select Countable number of points, while continuum must be selected to iterate over all points on a line or a sphere.
If this selection is possible then we can enumerate all points on (0,1) interval. Is there something in the proof that I am missing?
set-theory
edited 2 days ago
bof
45.8k348110
asked Aug 4 at 2:22
Stepan
430311
5
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
5
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Banach–Tarski Paradox explanation goes along these lines
(1) Take a point on a sphere
(2) Use 2pi/3 rotations to construct a subset of points
(3) After infinite number of rotations you've selected [an infinite] countable set of points on the sphere
(4) For each point not on the set repeat step (2)
(5) Repeat step (4) while there are unprocessed points
Done. [This paradox uses an axiom of choice]
Here is my concern. I am fine with an axiom of choice in the form "for any non-empty set you can always select a member of a set". However, Banach–Tarski Paradox seems to try to do something very different.
It tries to keep selecting points from a continuum set until no more points remain. This is NOT possible. Infinite while loop can only select Countable number of points, while continuum must be selected to iterate over all points on a line or a sphere.
If this selection is possible then we can enumerate all points on (0,1) interval. Is there something in the proof that I am missing?
set-theory
edited 2 days ago
bof
45.8k348110
asked Aug 4 at 2:22
Stepan
430311
The Banach–Tarski Paradox explanation goes along these lines
(1) Take a point on a sphere
(2) Use 2pi/3 rotations to construct a subset of points
(3) After infinite number of rotations you've selected [an infinite] countable set of points on the sphere
(4) For each point not on the set repeat step (2)
(5) Repeat step (4) while there are unprocessed points
Done. [This paradox uses an axiom of choice]
Here is my concern. I am fine with an axiom of choice in the form "for any non-empty set you can always select a member of a set". However, Banach–Tarski Paradox seems to try to do something very different.
It tries to keep selecting points from a continuum set until no more points remain. This is NOT possible. Infinite while loop can only select Countable number of points, while continuum must be selected to iterate over all points on a line or a sphere.
If this selection is possible then we can enumerate all points on (0,1) interval. Is there something in the proof that I am missing?
set-theory
edited 2 days ago
bof
45.8k348110
asked Aug 4 at 2:22
Stepan
430311
edited 2 days ago
bof
45.8k348110
edited 2 days ago
bof
45.8k348110
edited 2 days ago
bof
45.8k348110
45.8k348110
asked Aug 4 at 2:22
Stepan
430311
asked Aug 4 at 2:22
Stepan
430311
asked Aug 4 at 2:22
Stepan
430311
430311
5
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
5
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56
add a comment |Â
5
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
5
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56
5
5
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
5
5
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
There are indeed transfinite sequences — more precisely, sequences indexed by ordinal numbers — that enumerate the entire interval $(0,1)$.
Such a sequence can be 'constructed' via transfinite recursion by the method of transfinitely iterating the operation of "pick an element that hasn't been chosen yet, or stop if there are none".
This applies to any set at all; this fact is basically the content of the well-ordering theorem.
(Note: I'm responding to your final paragraphs. Your description of B-T is somewhat incoherent as well, so I'm not entirely sure you understand what it's doing or that your final questions actually pertain to B-T)
answered Aug 4 at 2:46
Hurkyl
107k9112252
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There are indeed transfinite sequences — more precisely, sequences indexed by ordinal numbers — that enumerate the entire interval $(0,1)$.
Such a sequence can be 'constructed' via transfinite recursion by the method of transfinitely iterating the operation of "pick an element that hasn't been chosen yet, or stop if there are none".
This applies to any set at all; this fact is basically the content of the well-ordering theorem.
(Note: I'm responding to your final paragraphs. Your description of B-T is somewhat incoherent as well, so I'm not entirely sure you understand what it's doing or that your final questions actually pertain to B-T)
answered Aug 4 at 2:46
Hurkyl
107k9112252
add a comment |Â
up vote
2
down vote
accepted
There are indeed transfinite sequences — more precisely, sequences indexed by ordinal numbers — that enumerate the entire interval $(0,1)$.
Such a sequence can be 'constructed' via transfinite recursion by the method of transfinitely iterating the operation of "pick an element that hasn't been chosen yet, or stop if there are none".
This applies to any set at all; this fact is basically the content of the well-ordering theorem.
(Note: I'm responding to your final paragraphs. Your description of B-T is somewhat incoherent as well, so I'm not entirely sure you understand what it's doing or that your final questions actually pertain to B-T)
answered Aug 4 at 2:46
Hurkyl
107k9112252
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There are indeed transfinite sequences — more precisely, sequences indexed by ordinal numbers — that enumerate the entire interval $(0,1)$.
Such a sequence can be 'constructed' via transfinite recursion by the method of transfinitely iterating the operation of "pick an element that hasn't been chosen yet, or stop if there are none".
This applies to any set at all; this fact is basically the content of the well-ordering theorem.
(Note: I'm responding to your final paragraphs. Your description of B-T is somewhat incoherent as well, so I'm not entirely sure you understand what it's doing or that your final questions actually pertain to B-T)
answered Aug 4 at 2:46
Hurkyl
107k9112252
There are indeed transfinite sequences — more precisely, sequences indexed by ordinal numbers — that enumerate the entire interval $(0,1)$.
Such a sequence can be 'constructed' via transfinite recursion by the method of transfinitely iterating the operation of "pick an element that hasn't been chosen yet, or stop if there are none".
This applies to any set at all; this fact is basically the content of the well-ordering theorem.
(Note: I'm responding to your final paragraphs. Your description of B-T is somewhat incoherent as well, so I'm not entirely sure you understand what it's doing or that your final questions actually pertain to B-T)
answered Aug 4 at 2:46
Hurkyl
107k9112252
answered Aug 4 at 2:46
Hurkyl
107k9112252
answered Aug 4 at 2:46
Hurkyl
107k9112252
answered Aug 4 at 2:46
Hurkyl
107k9112252
107k9112252
add a comment |Â
add a comment |Â
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5
Your statement "for any non-empty set you can always select a member of a set" has nothing to do with the axiom of choice. However, the statement that you so heavily reject has everything to do with the axiom of choice. So it seems your problem is indeed accepting the axiom of choice. Note that this axiom does not claim anything about "computability" of the choice set, so we do not have to make one choice after the other in an "infinite while loop", but we can make all choices at once.
– M. Winter
Aug 4 at 2:32
5
Are you familiar with transfinite induction/recursion? (My terminology pet peeve: when you build something it's "recursion," when you verify a property it's "induction.") Roughly, we can in many circumstances "iterate" a process uncountably many times - or rather, along any well ordering (and this is where the axiom of choice comes in: by well-ordering whatever set we care about, we get a well-ordering along which we can perform the desired transfinite recursion).
– Noah Schweber
Aug 4 at 2:56