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In wikipedia is written that Kuratowski definition of ordered pair is now-accepted (I use down-right index "K" to mark Kuratowski formula):

$p_K = (a,b)_K := \{ \{a\}, \{a,b\} \}$

My question is why people not use below simpler definition instead:

$p = (a,b) := \{ \{a\}, \{b, \varnothing \} \}$

?

UPDATE

After discussion on comments we get following

Advantages of $(a,b)$:

  • for paris $(a,\varnothing)$ the result of $(a,\varnothing) = \{ \{a\}, \{ \varnothing \} \}$ is simpler than $(a,\varnothing)_K = \{ \{a\}, \{ a, \varnothing \} \}$ (we don't need to duplicate $a$)
  • for case when $a=b$ formula $(a,a)=\{ \{a\}, \{a,\varnothing \} \}$ save property that we have two elements(pairs) in set, which is loose by formula $(a,a)_K=\{\{a\}\}$

Disatvantages of $(a,b)$:

  • for case $(\{\varnothing\},\varnothing) = \{ \{\{\varnothing\}\}, \{ \varnothing \} \}$ we loose property that when $a\neq b$ the first element of pair cardinality is $=1$ and second element cardinality is $=2$ which is saved for Kuratowski formula: $(\{\varnothing\},\varnothing)_K = \{ \{\{\varnothing\}\}, \{ \{\varnothing\}, \varnothing \} \}$. In this case both elements has cardinality $=1$.
  • for case $a=b$ formula $(a,a)_K=\{\{a\}\}$ is simpler than $(a,a)=\{ \{a\}, \{a,\varnothing \} \}$

  • for Kuratowski formula is easy to extract pair elements using union/intersection (which is not possible/easy for my formulat):

    $\pi_1( p_K ) = \bigcap (a,b)_K = \{a\}\cap \{a,b\} = \{a\}$

    $\pi_2( p_K ) = \bigcup (a,b)_K = \{a\}\cup \{a,b\} = \{a, b\}$

I also realized that my definition is similar (but not the same) to Winner's definition

Kamil Kiełczewski
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  • If $a=b=\emptyset$ then this will not work. – SmileyCraft Oct 16 '18 at 21:18
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    @SmileyCraft why not? If $a=b=\emptyset $ then $(\emptyset,\emptyset)={{\emptyset}}$ which is the same result in Kuratowski formula – Kamil Kiełczewski Oct 16 '18 at 21:23
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    Good point! But there is still an inconvenience if $b=\emptyset$ and $a\neq\emptyset$. 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 Oct 16 '18 at 21:28
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    In what sense ${ {a}, {b, \emptyset } }$ should be simpler than ${ {a}, {a,b} }$? – Paul Frost Oct 16 '18 at 21:36
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    @PaulFrost if a is "large" (and complicated) object then we not need to write it twice :P – Kamil Kiełczewski Oct 16 '18 at 21:39
  • @Kamil: When was the last time you actually wrote the ordered pairs explicitly outside an exercise about the definition of ordered pairs (or some closely related notion)? – Asaf Karagila Oct 16 '18 at 21:43
  • currently - when I explain definition to my younger friend :P (and have idea to simplify it) – Kamil Kiełczewski Oct 16 '18 at 21:44
  • I rest my case. – Asaf Karagila Oct 16 '18 at 21:46
  • If $b=\emptyset$ and $a$ is some big complicated expression, using Kuratowski you write "${{a},{a,\emptyset}}$ where $a = (big\ complicated\ expression).$" Then you still only have to write the complicated expression once. – David K Oct 16 '18 at 22:35
  • @DavidK using your case, $b=\emptyset$ my formula is simpler than Kuratowski: $(a,\emptyset)={ {a}, {\emptyset}}$ – Kamil Kiełczewski Oct 16 '18 at 22:55
  • Shorter and simpler are not the same thing. Anyway, your formula is sometimes shorter, sometimes longer. For example, consider $a=b={\emptyset}.$ – David K Oct 17 '18 at 00:12
  • @DavidK yes but you provide single example, I provide (using your argument :P ) "family of examples" where my formula is simpler (because not duplicate "a") – Kamil Kiełczewski Oct 17 '18 at 13:10
  • Your complaint about "duplicate $a$" only occurs if you insist on using inefficient notation. Even using your notation I would still want to use the "where $a =$" notation if $a$ is complicated, because it makes it easier to see the pair. In the end, "fixing" the "problem" with Kuratowski notation isn't worth the effort we've already spent on it already. If Kuratowski had used your notation instead, I'd be just as happy to argue against changing it. :-) – David K Oct 17 '18 at 14:35

1 Answers1

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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.

Asaf Karagila
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  • 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 Oct 16 '18 at 21:40
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    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 Oct 16 '18 at 21:42
  • ah, but we do. Every time you use a projection you do. – freakish Oct 16 '18 at 21:42
  • 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 Oct 16 '18 at 21:43
  • is that a reason to not care about how it is defined? Are you suggesting we should do naive maths? Amusing. – freakish Oct 16 '18 at 21:44
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    Yes. That is precisely the reason not to care about how something is defined. Once you have a definition, you can move on. Even if you write your proofs into a proof assistant you don't actually care. Because it's a function that you wrote once. Probably in a code library that someone else wrote once, and you wrote never. If you think that all mathematicians write everything they do down to the ZFC syntax, that's very naive of you. And amusing. – Asaf Karagila Oct 16 '18 at 21:46
  • @freakish: If you have a better answer, you're more than welcomed to write it yourself. – Asaf Karagila Oct 16 '18 at 21:48
  • first of all: don't try to put responsibility on me for your answer. I've already -1 it cause it doesn't answer the question. Secondly, coders make lots of mistakes with their naive "this should work" approach. No, maths is not a science of "I don't care". And neither does OP, he asked the question after all. – freakish Oct 16 '18 at 21:50
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    I am not making you responsible for my answer. I didn't ask you to edit it in any kind of way. I asked you to take your argument, that you so proudly flaunt here in the comments, and put it into an answer. Now, since this discussion has run its course, I bid you a good day sir. – Asaf Karagila Oct 16 '18 at 21:51