경문사 SET THEORY 연습문제풀이

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경문사 SET THEORY 연습문제풀이에 대한 보고서 자료입니다.

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{ { -} atop {f } }^{-1 }
defines an injection of B onto the subset
{ -} atop {A }
of A. The proof is now complete. ■
Theorem 5 (Well-ordering principle)
Every set can be well ordered.
pf) Let A be an arbitrarily given set which is to be well ordered. Consider the set
{ A}^{* }
of all well-ordered sets (
{ A}_{0 }
,
{ <= }_{ 0}
), where
{ A}_{0 }
SUBSETEQ
A. We partially order
{ A}^{* }
by writing (
{ A}_{0 }
,
{ <= }_{ 0}
)
<=
(
{ A}_{1}
,
{ <= }_{ 1}
) if and only if
(ⅰ)
{ A}_{0 }
SUBSETEQ
A
(ⅱ) x, y
IN
{ A}_{0 }
and x
{ <= }_{ 0}
y imply x
{ <= }_{ 1}
y
(ⅲ) x
IN
{ A}_{1}
-
{ A}_{0 }
imply y
{ <= }_{ 1}
x for all y
IN
{ A}_{0 }
.
This relation
{ <= }^{* }
is a partial order relation on
{ A}^{* }
(
{ A}_{0 }
,
{ <= }_{ 0}
)
{ <= }^{* }
(
{ A}_{1}
,
{ <= }_{ 1}
)⇔ 1)
{ A}_{0 }
SUBSETEQ
{ A}_{1}
2) x, y,
IN
{ A}_{0 }
, x
{ <= }_{ 0}
y⇒ x
{ <= }_{ 1}
y
3) x
IN
{ A}_{1}
-
{ A}_{0 }
⇒ y
{ <= }_{ 1}
x, ∀y
IN
{ A}_{0 }
1) reflexive : ∀(
{ A}_{0 }
,
{ <= }_{ 0}
)
IN
{ A}^{* }
⇒(
{ A}_{0 }
,
{ <= }_{ 0}
)
{ <= }^{* }
(
{ A}_{0 }
,
{ <= }_{ 0}
) : satisfy
ⅰ)
{ A}_{0 }
SUBSETEQ
{ A}_{0 }
ⅱ) x, y
IN
{ A}_{0 }
and x
{ <= }_{ 0}
y → x
{ <= }_{ 0}
y
ⅲ)
{ A}_{0 }
-
{ A}_{0 }
=0
∴ reflexive.
2) transitive : ∀(
{ A}_{0 }
,
{ <= }_{ 0}
), (
{ A}_{1}
,
{ <= }_{ 1}
), (
{ A}_{2 }, { <= }_{2 }
)
IN
{ A}^{* }
(
{ A}_{0 }
,
{ <= }_{ 0}
)
{ <= }^{* }
(
{ A}_{1}
,
{ <= }_{ 1}
)∧(
{ A}_{1}
,
{ <= }_{ 1}
)
{ <= }^{* }
(
{ A}_{2 }, { <= }_{2 }
)
⇔ ⅰ)
{ A}_{0 }
SUBSETEQ
{ A}_{1}
SUBSETEQ
{ A}_{2 }
ⅱ) x, y
IN
{ A}_{0 }
∧ x
{ <= }_{ 0}
y → x
{ <= }_{ 1}
y
z, w
IN
{ A}_{1}
∧ z
{ <= }_{ 1}
w → z
{ <= }_{ 2}
w
ⅲ) x
IN
{ A}_{1}
-
{ A}_{0 }
→ y
{ <= }_{ 1}
x for ∀y
IN
{ A}_{0 }
z
IN
{ A}_{2}
-
{ A}_{1 }
→ w
{ <= }_{ 2}
z for ∀w
IN
{ A}_{1 }
⇒ ⅰ)
{ A}_{0 }
SUBSETEQ
{ A}_{2}
ⅱ) x, y
IN
{ A}_{0 }
∧ x
{ <= }_{ 0}
y → x
{ <= }_{ 2}
y
ⅲ) x
IN
{ A}_{2}
-
{ A}_{0 }
→ y
{ <= }_{ 2}
x for ∀y
IN
{ A}_{0 }
⇔(
{ A}_{0 }
,
{ <= }_{ 0}
)
{ <= }^{* }
(
{ A}_{2 }, { <= }_{2 }
).
∴transitive.
3) antisymmetric : (
{ A}_{0 }
,
{ <= }_{ 0}
)
{ <= }^{* }
(
{ A}_{1}
,
{ <= }_{ 1}
)∧(
{ A}_{1}
,
{ <= }_{ 1}
)
{ <= }^{* }
(
{ A}_{0 }
,
{ <= }_{ 0}
)
⇔ ⅰ)
{ A}_{0 }
<=
{ A}_{1}
∧
{ A}_{1}
<=
{ A}_{0 }
⇒
{ A}_{0 }
=
{ A}_{1}
ⅱ) x, y
IN
{ A}_{0 }
∧ x
{ <= }_{ 0}
y → x
{ <= }_{ 1}
y
z, w
IN
{ A}_{1}
∧ z
{ <= }_{ 1}
w → z
{ <= }_{ 0}
w
By ⅰ), ⅱ)
{ <= }_{ 0}
=
{ <= }_{ 1}
∧
{ A}_{0 }
=
{ A}_{1}
⇒ (
{ A}_{0 }
,
{ <= }_{ 0}
)=(
{ A}_{1}
,
{ <= }_{ 1}
)
∴ antisymmetric.
∴
{ <= }^{* }
: partial order relation.
In order to apply Zorn's lemma, we show that any totally ordered subset R of
(
{ A}^{* }
,
{ <= }^{* }
) has an upper bound.
The natural candidate for this upper bound is (
{ <= }^{'}
{ BIGCUP } atop {A IN R}A
,), where x
{ <= }^{'}
y if both x and y belong to some
{ A}_{0 }
such that (
{ A}_{0 }
,
{ <= }_{ 0}
)
IN
R and x
{ <= }_{ 0}
y.■

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