Ordered pair - why so complicated?

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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







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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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  • 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















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







share|cite|improve this question







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













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







share|cite|improve this question







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






share|cite|improve this question







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.











share|cite|improve this question







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.









share|cite|improve this question




share|cite|improve this question






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

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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

















  • 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











1 Answer
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5
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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.






share|cite|improve this answer




















  • 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










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1 Answer
1






active

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1 Answer
1






active

oldest

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active

oldest

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oldest

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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.






share|cite|improve this answer




















  • 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














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.






share|cite|improve this answer




















  • 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












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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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










Kamil Kiełczewski is a new contributor. Be nice, and check out our Code of Conduct.









 

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