Ordered pair - why so complicated?
Clash Royale CLAN TAG#URR8PPP
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In wikipedia is written that Kuratowski definition of ordered pair is now-accepted:
$(a,b) := a, a,b $
My question is why people not use below simpler definition instead:
$(a,b) := a, b, emptyset $
?
elementary-set-theory
asked 4 hours ago
Kamil Kiełczewski
1063
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Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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show 6 more comments
up vote
1
down vote
favorite
1
In wikipedia is written that Kuratowski definition of ordered pair is now-accepted:
$(a,b) := a, a,b $
My question is why people not use below simpler definition instead:
$(a,b) := a, b, emptyset $
?
elementary-set-theory
asked 4 hours ago
Kamil Kiełczewski
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
1
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
2
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
1
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago
 |Â
show 6 more comments
up vote
1
down vote
favorite
1
up vote
1
down vote
favorite
1
1
In wikipedia is written that Kuratowski definition of ordered pair is now-accepted:
$(a,b) := a, a,b $
My question is why people not use below simpler definition instead:
$(a,b) := a, b, emptyset $
?
elementary-set-theory
asked 4 hours ago
Kamil Kiełczewski
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In wikipedia is written that Kuratowski definition of ordered pair is now-accepted:
$(a,b) := a, a,b $
My question is why people not use below simpler definition instead:
$(a,b) := a, b, emptyset $
?
elementary-set-theory
elementary-set-theory
asked 4 hours ago
Kamil Kiełczewski
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Kamil Kiełczewski
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Kamil Kiełczewski
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Kamil Kiełczewski
1063
asked 4 hours ago
Kamil Kiełczewski
1063
1063
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kamil Kiełczewski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
1
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
2
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
1
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago
 |Â
show 6 more comments
If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
1
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
2
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
1
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago
If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
1
1
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
2
2
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
1
1
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
There are many definitions for ordered pairs. The point is that we don't actually care which definition is used, exception in very rare cases. What is important is that there is a definition that works.
Kuratowski's definition is easy to understand: it does not rely on additional objects except $a$ and $b$, and it really codes the "order" of $a<b$ by its initial segments: $a$ and then $a,b$.
But you are right that $(a,b)=a,b,varnothing$ is also a valid definition that works.
answered 4 hours ago
Asaf Karagila♦
297k32414740
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
There are many definitions for ordered pairs. The point is that we don't actually care which definition is used, exception in very rare cases. What is important is that there is a definition that works.
Kuratowski's definition is easy to understand: it does not rely on additional objects except $a$ and $b$, and it really codes the "order" of $a<b$ by its initial segments: $a$ and then $a,b$.
But you are right that $(a,b)=a,b,varnothing$ is also a valid definition that works.
answered 4 hours ago
Asaf Karagila♦
297k32414740
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
 |Â
show 4 more comments
up vote
5
down vote
There are many definitions for ordered pairs. The point is that we don't actually care which definition is used, exception in very rare cases. What is important is that there is a definition that works.
Kuratowski's definition is easy to understand: it does not rely on additional objects except $a$ and $b$, and it really codes the "order" of $a<b$ by its initial segments: $a$ and then $a,b$.
But you are right that $(a,b)=a,b,varnothing$ is also a valid definition that works.
answered 4 hours ago
Asaf Karagila♦
297k32414740
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
 |Â
show 4 more comments
up vote
5
down vote
up vote
5
down vote
There are many definitions for ordered pairs. The point is that we don't actually care which definition is used, exception in very rare cases. What is important is that there is a definition that works.
Kuratowski's definition is easy to understand: it does not rely on additional objects except $a$ and $b$, and it really codes the "order" of $a<b$ by its initial segments: $a$ and then $a,b$.
But you are right that $(a,b)=a,b,varnothing$ is also a valid definition that works.
answered 4 hours ago
Asaf Karagila♦
297k32414740
There are many definitions for ordered pairs. The point is that we don't actually care which definition is used, exception in very rare cases. What is important is that there is a definition that works.
Kuratowski's definition is easy to understand: it does not rely on additional objects except $a$ and $b$, and it really codes the "order" of $a<b$ by its initial segments: $a$ and then $a,b$.
But you are right that $(a,b)=a,b,varnothing$ is also a valid definition that works.
answered 4 hours ago
Asaf Karagila♦
297k32414740
answered 4 hours ago
Asaf Karagila♦
297k32414740
answered 4 hours ago
Asaf Karagila♦
297k32414740
answered 4 hours ago
Asaf Karagila♦
297k32414740
297k32414740
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
 |Â
show 4 more comments
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
I disagree. We do care about definitions. We pick one that is most advantageous. The reason Kuratowski is better is for example because of the easy "being the first/second coordinate" definition (as can be read at wiki). This wouldn't be easy for OPs definition.
– freakish
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
Since we don't spend any time writing these formulas in proofs, no, we don't care beyond the proof that the definition satisfies the ordered pair property.
– Asaf Karagila♦
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
ah, but we do. Every time you use a projection you do.
– freakish
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
Really? I always wrote $pi_0$ or $pi_1$ to denote the projection, or some other ad hoc notation. I never, ever in my life, wrote the explicit formula for the projections. Have you?
– Asaf Karagila♦
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing.
– freakish
4 hours ago
 |Â
show 4 more comments
Kamil Kiełczewski is a new contributor. Be nice, and check out our Code of Conduct.
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Kamil Kiełczewski is a new contributor. Be nice, and check out our Code of Conduct.
Kamil Kiełczewski is a new contributor. Be nice, and check out our Code of Conduct.
Kamil Kiełczewski is a new contributor. Be nice, and check out our Code of Conduct.
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If $a=b=emptyset$ then this will not work.
– SmileyCraft
4 hours ago
1
@SmileyCraft why not? If $a=b=emptyset $ then $(emptyset,emptyset)=emptyset$ which is the same result in Kuratowski formula
– Kamil Kiełczewski
4 hours ago
2
Good point! But there is still an inconvenience if $b=emptyset$ and $aneqemptyset$. You would like to be able to define the first and second elements of a tuple. The usual definition lets us define the first element as the element of the unique set of cardinality 1. This breaks if $(emptyset,emptyset)=emptyset,emptyset$.
– SmileyCraft
4 hours ago
1
In what sense $ a, b, emptyset $ should be simpler than $ a, a,b $?
– Paul Frost
4 hours ago
@PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P
– Kamil Kiełczewski
4 hours ago