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 a ≠ b .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 a ≠ b, 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 a ≠ b.Suppose {c} = { a, b}. Then a = b = c, which also contradicts a ≠ b .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.
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 a ≠ b .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 a ≠ b, 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 a ≠ b.Suppose {c} = { a, b}. Then a = b = c, which also contradicts a ≠ b .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.