Questions about proving the characteristic property of ordered pairs.

Dale10101

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A series of questions about the following proof.

Proving that definitions satisfy the characteristic property



Prove: (a, b) = (c, d) if and only if a = c and b = d.



1) In elementary algebra the ordered pair (a, b) delivers the location of a point on the Cartesian plane and of course (a,b) is not equal to (b,a) in general ... is that the meaning of (a,b) in the current context or ...


2) Does referring to (a, b) as an ordered pair mean that "a" is a proper subset of set "b" in a hierarchy of nested sets in which case, again, (a,b) is not equal to (b,a) in general?


The two possibilities cannot be equivalent since, if we are talking about integers for example, "a" must be less then "b" in the second case but is not required to be so in the first case.


No point in going further until I understand this most basic point. Thank you.






Proving that definitions satisfy the characteristic property


Prove: (a, b) = (c, d) if and only if a = c and b = d.Kuratowski:
If. If a = c and b = d, then {{ a}, {a, b }} = {{c}, { c, d}}. Thus (a, b) K = (c, d )K.
Only if. Two cases: a = b, and ab .If a = b:(a, b) K = {{a}, { a, b}} = {{a }, {a, a}} = {{ a}}.(c, d) K = {{c}, { c, d}} = {{a }}.Thus {c} = { c, d} = {a}, which implies a = c and a = d. By hypothesis, a = b. Hence b = d .If ab, then (a, b )K = ( c, d)K implies {{ a}, {a, b }} = {{c}, { c, d}}.Suppose {c, d} = { a}. Then c = d = a , and so {{c}, { c, d}} = {{a }, {a, a}} = {{ a}, {a}} = {{ a}}. But then {{a }, {a, b}} would also equal {{a}}, so that b = a which contradicts ab.Suppose {c} = { a, b}. Then a = b = c, which also contradicts ab .Therefore {c} = { a}, so that c = a and {c, d} = { a, b}.If d = a were true, then { c, d} = {a, a } = {a} ≠ { a, b}, a contradiction. Thus d = b is the case, so that a = c and b = d.Reverse:
(a, b)reverse = {{b}, { a, b}} = {{b}, { b, a}} = (b, a )K.
If. If (a, b )reverse = ( c, d)reverse , (b, a) K = (d, c) K. Therefore b = d and a = c.Only if. If a = c and b = d, then {{ b}, {a, b }} = {{d}, { c, d}}. Thus (a, b) reverse = (c, d )reverse.
 
You seem to be asking "what is the definition of 'ordered pair'". There are several equivalent ways but since you mention Kuratowski, his definition is "The ordered pair, (a, b), is the set {a, {ab}}. That's closest to your (2) but does NOT mean "a is a subset of b". "a" and "b" theselves are not necessarily sets at all.
 
Follow up question

You seem to be asking "what is the definition of 'ordered pair'". There are several equivalent ways but since you mention Kuratowski, his definition is "The ordered pair, (a, b), is the set {a, {ab}}. That's closest to your (2) but does NOT mean "a is a subset of b". "a" and "b" theselves are not necessarily sets at all.

Thank you. I am really struggling to get a grasp on something that seems heretofore elusive. I realize that I am not making my confusion clear. After several hours of head pounding it comes to this … does an “ordered pair” give order to the position of the two elements within the set, or does it give order to the relationships BETWEEN the elements of the set THEMSELVES.

In the first case (a,b) = {{a}, {ab}} simply identifies a first and second element, that is, for example, if you give me the ordered pair (1,2) I can plot it on the Cartesian plane because I know the first element refers to the x-axis and the second element to the y-axis, but if you give me the unordered pair {1,2} to plot, I have no response. Is that the sole content of the definition or an ordered pair, to define a first and second element within the set. I am now thinking that is it.

Or, and this is what I was thinking before, does (a,b) with respect to set theory define an order relationship between “a” and “b” themselves, like a < b, or a is a proper subset of b.

I was led to this last conception (or misconception) by the definition of the natural numbers as a successive nesting of sets,

The integer 0 is a set containing no elements

The integer 1 is a set containing one element {0}

The integer 2 is a set containing two elements {0,1}, that is, the integer 2 is the set containing the set of elements inherited from its predecessor plus (in union with) one additional element, i.e. 2 = {{0},1)}

The integer 3 is a set containing three elements {0,1,2}, that is, the integer 3 is the set containing the set of elements inherited from its predecessor plus one additional element, i.e. 3 = {{1},2)}

And so on.

I now think, because I had been studying the meaning and notation used in defining the natural numbers, I began to think of {a,{ab}} as meaning the type of 'nested hierarchical order' used to define the natural numbers, that is, first the set “a” then, in union, with “b”, the set {ab}, hence {a,{ab}}.

I now think that the first case is correct, an ordered pair is simply, in concept and notation, a distinguishing of a first and second element in a set, HOWEVER, at this point, I know full well that I can cleverly confuse myself and may well be entirely wrong. Thanks.
 
Writing ordered pair, (a, b), as the set {a, {a,b}} is not really very deep. The only difference between a set containing two things and an ordered pair is (of course) the order. The set {a, {a,b}} tells us that we are talking about two things, a and b, and they are NOT "equivalent" because one, a, occurs both as itself and in the set {a, b}. The fact that the "a" in {a, {a,b}} is first in the pair is purely "convention". We could as easily have interpreted {a, {a,b}} as defining the pair (b, a).
 
OK

Writing ordered pair, (a, b), as the set {a, {a,b}} is not really very deep. The only difference between a set containing two things and an ordered pair is (of course) the order. The set {a, {a,b}} tells us that we are talking about two things, a and b, and they are NOT "equivalent" because one, a, occurs both as itself and in the set {a, b}. The fact that the "a" in {a, {a,b}} is first in the pair is purely "convention". We could as easily have interpreted {a, {a,b}} as defining the pair (b, a).

Thank you. I see that I have been making a mountain out of a mosquito bite. :(
 
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