Set membership and a box of bananas
$begingroup$
Suppose that I have two boxes $A$ and $B$, each of which contains some number of identical (indistinguishable)* bananas.
If I treat $A$ and $B$ as multisets whose elements are bananas, it follows that
$$ain Aimplies a=banana$$
$$bin Bimplies b=banana$$
As $A$ and $B$ are physically seperate boxes, it also follows that $A$ and $B$ are disjoint:
$$Acap B=emptyset$$
However, by way of transitivity:
$$a=banana=bimplies a=b$$
$$therefore ain Aimplies ain B$$
Which leads to the contradiction $ain Aimplies ainemptyset$
How do I resolve this?
*Indistinguishable means that a particular banana, while it is in a box, carries no information that would allow it to be differentiated from any other banana in either box. If you swap any two bananas, the result is the same as doing nothing at all.
Guesses (in order of descending ridiculousness)
Potential option 1:
The relationship between a banana and its box is not adequately described by set membership and equality, devise a more exclusive relation to prevent a banana in $A$ from being in $B$ at the same time.
Potential option 2:
The bananas in box $A$ exist in an undecidable superposition with those of box $B$. Thus neither $ain A$ nor $ain B$ is either true or false, but half-true and half-false, until I open the box.
Potential option 3:
The intersection of $A$ and $B$ is not empty. There exists a "virtual" set $C$ of "potential bananas" such that $ain Aimplies ain Bimplies ain C$ and $Acap B=C$
Potentail option 4:
Options $1-3$ represent different expressions of a fundamental truth regarding set membership, and are simultaneously true (somehow)
elementary-set-theory soft-question relations multisets
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
$endgroup$
add a comment |
$begingroup$
Suppose that I have two boxes $A$ and $B$, each of which contains some number of identical (indistinguishable)* bananas.
If I treat $A$ and $B$ as multisets whose elements are bananas, it follows that
$$ain Aimplies a=banana$$
$$bin Bimplies b=banana$$
As $A$ and $B$ are physically seperate boxes, it also follows that $A$ and $B$ are disjoint:
$$Acap B=emptyset$$
However, by way of transitivity:
$$a=banana=bimplies a=b$$
$$therefore ain Aimplies ain B$$
Which leads to the contradiction $ain Aimplies ainemptyset$
How do I resolve this?
*Indistinguishable means that a particular banana, while it is in a box, carries no information that would allow it to be differentiated from any other banana in either box. If you swap any two bananas, the result is the same as doing nothing at all.
Guesses (in order of descending ridiculousness)
Potential option 1:
The relationship between a banana and its box is not adequately described by set membership and equality, devise a more exclusive relation to prevent a banana in $A$ from being in $B$ at the same time.
Potential option 2:
The bananas in box $A$ exist in an undecidable superposition with those of box $B$. Thus neither $ain A$ nor $ain B$ is either true or false, but half-true and half-false, until I open the box.
Potential option 3:
The intersection of $A$ and $B$ is not empty. There exists a "virtual" set $C$ of "potential bananas" such that $ain Aimplies ain Bimplies ain C$ and $Acap B=C$
Potentail option 4:
Options $1-3$ represent different expressions of a fundamental truth regarding set membership, and are simultaneously true (somehow)
elementary-set-theory soft-question relations multisets
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
$endgroup$
$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
$endgroup$
– R. Burton
Dec 20 '18 at 15:49
add a comment |
$begingroup$
Suppose that I have two boxes $A$ and $B$, each of which contains some number of identical (indistinguishable)* bananas.
If I treat $A$ and $B$ as multisets whose elements are bananas, it follows that
$$ain Aimplies a=banana$$
$$bin Bimplies b=banana$$
As $A$ and $B$ are physically seperate boxes, it also follows that $A$ and $B$ are disjoint:
$$Acap B=emptyset$$
However, by way of transitivity:
$$a=banana=bimplies a=b$$
$$therefore ain Aimplies ain B$$
Which leads to the contradiction $ain Aimplies ainemptyset$
How do I resolve this?
*Indistinguishable means that a particular banana, while it is in a box, carries no information that would allow it to be differentiated from any other banana in either box. If you swap any two bananas, the result is the same as doing nothing at all.
Guesses (in order of descending ridiculousness)
Potential option 1:
The relationship between a banana and its box is not adequately described by set membership and equality, devise a more exclusive relation to prevent a banana in $A$ from being in $B$ at the same time.
Potential option 2:
The bananas in box $A$ exist in an undecidable superposition with those of box $B$. Thus neither $ain A$ nor $ain B$ is either true or false, but half-true and half-false, until I open the box.
Potential option 3:
The intersection of $A$ and $B$ is not empty. There exists a "virtual" set $C$ of "potential bananas" such that $ain Aimplies ain Bimplies ain C$ and $Acap B=C$
Potentail option 4:
Options $1-3$ represent different expressions of a fundamental truth regarding set membership, and are simultaneously true (somehow)
elementary-set-theory soft-question relations multisets
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
$endgroup$
Suppose that I have two boxes $A$ and $B$, each of which contains some number of identical (indistinguishable)* bananas.
If I treat $A$ and $B$ as multisets whose elements are bananas, it follows that
$$ain Aimplies a=banana$$
$$bin Bimplies b=banana$$
As $A$ and $B$ are physically seperate boxes, it also follows that $A$ and $B$ are disjoint:
$$Acap B=emptyset$$
However, by way of transitivity:
$$a=banana=bimplies a=b$$
$$therefore ain Aimplies ain B$$
Which leads to the contradiction $ain Aimplies ainemptyset$
How do I resolve this?
*Indistinguishable means that a particular banana, while it is in a box, carries no information that would allow it to be differentiated from any other banana in either box. If you swap any two bananas, the result is the same as doing nothing at all.
Guesses (in order of descending ridiculousness)
Potential option 1:
The relationship between a banana and its box is not adequately described by set membership and equality, devise a more exclusive relation to prevent a banana in $A$ from being in $B$ at the same time.
Potential option 2:
The bananas in box $A$ exist in an undecidable superposition with those of box $B$. Thus neither $ain A$ nor $ain B$ is either true or false, but half-true and half-false, until I open the box.
Potential option 3:
The intersection of $A$ and $B$ is not empty. There exists a "virtual" set $C$ of "potential bananas" such that $ain Aimplies ain Bimplies ain C$ and $Acap B=C$
Potentail option 4:
Options $1-3$ represent different expressions of a fundamental truth regarding set membership, and are simultaneously true (somehow)
elementary-set-theory soft-question relations multisets
elementary-set-theory soft-question relations multisets
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
edited Dec 20 '18 at 16:06
R. Burton
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
asked Dec 20 '18 at 2:44
R. BurtonR. Burton
670110
670110
$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
$endgroup$
– R. Burton
Dec 20 '18 at 15:49
add a comment |
$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
$endgroup$
– R. Burton
Dec 20 '18 at 15:49
$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
$endgroup$
– R. Burton
Dec 20 '18 at 15:49
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
$endgroup$
– R. Burton
Dec 20 '18 at 15:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You're using "=" as a equivalence relation, but then using different sets so it breaks down.
answered Dec 20 '18 at 2:50
kminikmini
913
$endgroup$
add a comment |
$begingroup$
The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
$endgroup$
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You're using "=" as a equivalence relation, but then using different sets so it breaks down.
answered Dec 20 '18 at 2:50
kminikmini
913
$endgroup$
add a comment |
$begingroup$
You're using "=" as a equivalence relation, but then using different sets so it breaks down.
answered Dec 20 '18 at 2:50
kminikmini
913
$endgroup$
add a comment |
$begingroup$
You're using "=" as a equivalence relation, but then using different sets so it breaks down.
answered Dec 20 '18 at 2:50
kminikmini
913
$endgroup$
You're using "=" as a equivalence relation, but then using different sets so it breaks down.
answered Dec 20 '18 at 2:50
kminikmini
913
answered Dec 20 '18 at 2:50
kminikmini
913
answered Dec 20 '18 at 2:50
kminikmini
913
answered Dec 20 '18 at 2:50
kminikmini
913
913
add a comment |
add a comment |
$begingroup$
The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
$endgroup$
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
add a comment |
$begingroup$
The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
$endgroup$
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
add a comment |
$begingroup$
The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
$endgroup$
The problem is that each banana is different so the fact that $a$ is a banana and $b$ is a banana doesn't allow us to say $a$ = $b$.
If $a$ and $b$ came from a set of fruits, then the banana would be unique and the fact that both sets contained a banana would indeed say they have non-empty intersection. In your case each box represents a set of physical objects so, although two bananas may be indistinguishable, they are unique.
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
answered Dec 20 '18 at 3:40
John DoumaJohn Douma
5,62711420
5,62711420
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
add a comment |
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
That makes sense. Is there a formal way to denote that two elements are unique but indistinguishable? As kmini noted the relation between any two bananas is not an equivalence relation, and as Eric Wolfsey noted, the bananas can be implicitly distinguished by which box they are in; but since this is not an intrinsic property of a given banana $a$ and $b$ are explicitly "equal". (Obviously each banana is different, but there's no way to prove this without altering the system - i.e. opening the box)
$endgroup$
– R. Burton
Dec 20 '18 at 16:03
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
@R.Burton This has nothing to do with cats and uncertainty about their health. If I removed a banana from each box and asked you to tell me which box each came from, you cannot tell because they are identical. You know there are two of them so clearly you can see they are distinct. Otherwise, when you went to count the second one, you would say "I already counted this one." and conclude that there is only one banana.
$endgroup$
– John Douma
Dec 20 '18 at 16:30
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
$begingroup$
Actually, the question did originate as a macroscopic simplification of a problem in statistical mechanics, so Schrödinger's cat isn't too-bad an analogy. Anyway, the issue is explicitly stating the relation between two bananas, and each banana to each box, while they are still in the boxes, and without introducing contradictions. Reason being, that this is necessary in order for any sort of rigorous mathematics to be conducted.
$endgroup$
– R. Burton
Dec 20 '18 at 16:42
add a comment |
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$begingroup$
How can bananas truly be "indistinguishable" if they are in different boxes? That distinguishes them!
$endgroup$
– Eric Wofsey
Dec 20 '18 at 3:46
$begingroup$
@EricWofsey they are indistinguishable in the sense that if you were to shuffle them, then select one at random, there would be no way to know which banana you had or which box it originally came from; the system is invariant to the swapping of any two bananas.
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– R. Burton
Dec 20 '18 at 15:49