Ar

Ar
“Você aqui até paga o ar que respira, minha senhora!”
Ela já o sabia. Já o sabia há muito tempo. Ainda antes de emigrar, já o sabia. A melancolia inundava-a. Tinha saudades de casa, dos amigos, dos conhecidos. E das paisagens, dos cheiros e dos ruídos familiares, com que tinha crescido. Mas agora…
Que raio de decisão malfadada a tinha levado até à colónia mineira lunar, onde quem não busca o Hélio 3, vive daqueles que o procuram.

The empty set and the ordered pair

[From Wiki:
An ordered pair is a collection of two not necessarily distinct objects, one of which is distinguished as the first coordinate (or first entry or left projection) and the other as the second coordinate (second entry, right projection). The common notation for an ordered pair with first coordinate a and second coordinate b is (a, b).
Let (a1, b1) and (a2, b2) be two ordered pairs. Then the characteristic or defining property of ordered pairs is:(a1, b1) = (a2, b2) <=> (a1 = a2 & b1 = b2).
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The ordered pair (a,b) is usually defined as the Kuratowski pair:(a,b)K:= {{a}, {a,b}}
The above definition of an ordered pair is "adequate", in the sense that it satisfies the characteristic property that an ordered pair must have (namely: if (a,b)=(x,y), then a=x and b=y), but also arbitrary, as there are many other definitions which are no more complicated and would also be adequate, as:
(a,b)short:= { a, {a,b} }
End Wiki]
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As usual in set theory:
Let {} represent the empty set.
An unordered pair is any set with two elements, {a,b}= {b,a}.
Commas are just a graphical aid: {a, b}= {a b}; { , b}= {b}.
Repeated elements should be treated as only one, {a,a,a,b}= { a a a b}= {a,b}={a b},and {, , ,}={ }={}.
For the purpose of the above ordered pair definitions, the empty set can be a valid object, because none excludes it.

Also, define:
(a,b)tiny: {a, {b}}


Let us consider how to evaluate ordered pairs, when one of the sets involved is the empty set. To do so, we will need to look at four cases, C0 to C3:
C0: {a}= {}, {b}= {}
C1: {a}= {}, {b}= {x}
C2: {a}= {x}, {b}= {}
C3: {a}= {x}, {b}= {x}.

Using (a,b)K, we get:
C0: {{},{,}}= {{},{}}= {{}}
C1: {{},{,x}}= {{},{x}}
C2: {{x},{x,}}={{x},{x}}= {{x}}
C3: {{x},{x}}= {{x}}
Note: C2 and C3 are equal, and shouldn’t be, by the characteristic property.

Using (a,b)short, we get:
C0: {,{,}}= {,{}}= {{}}
C1: {,{,x}}= {{,x}}= {{x}}
C2: {x,{x,}}={x,{x}}= {x,{x}}
C3: {x,{x,x}}= {x,{x}}
Note: Again, C2 and C3 are equal, and shouldn’t be, by the characteristic property.

Using (a,b)tiny, we get:
C0: {,{}}= { {}}= {{}}
C1: {,{x}}= { {x}}= {{x}}
C2: {x,{ }}= {x,{ }}
C3: {x,{x}}= {x,{x}}
Now we have four different results, as it should be by the characteristic property.

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This tiny definition looks simpler, more elegant and discriminates between the C2 and C3 cases. Why isn’t the standard definition?
One flaw is:
({a},b)tiny= {{a},{b}}.
({b},a)tiny= {{b},{a}}= {{a},{b}}
So all definitions are flawed.

They all are unequal. They all have different flaws.
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Is there any error in the above deduction? It looks lean and clean.
For me, the most difficult part was to realize that {, , ,}= {, , }= {,}= {}.
Maybe this is wrong, but {,}~={} looks even worst to me.
But there must be an error, somewhere.
Otherwise, the Kuratowsky definition would not be the one with widespread usage.
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Any comments will be welcome.