MA3205
Set Theory
AY 2003/2004 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Teo Wei Hao
MA3205 Set Theory
AY 2003/2004 Sem 1
SECTION A
Question 1
S
i.e. there exists γ ∈ α such that β ∈ γ. Since α is transitive, and so β ∈ α.
(a) Let β ∈ α, S
This give us α ⊆ α.
(b) Let us be given that α is not a limit ordinal.
Then ∃β ∈ Ord such that S(β) = α, i.e. β ∈ α, and for all γ ∈ α, we have γ ∈ β or γ = β.
Since (α, ∈) is S
a linearly ordered S
set, there does not exists γ ∈ α such that β ∈ γ.
Therefore β 6∈ α. This give us α 6= α.
S
Instead let us be given that α 6=Sα.
S
From (1a.), we conclude that α 6⊆ α, i.e. there exists β ∈ α such that β 6∈ α.
If γ ∈ α, then we have β 6∈ γ, and since (α, ∈) is a linearly ordered set, we have γ ∈ β or γ = β.
Thus α ⊆ β ∪ {β}.
Instead if γ ∈ β or γ = β, then since β ∈ α, and α is transitive, we get γ ∈ α. Thus β ∪ {β} ⊆ α.
Therefore α = β ∪ {β} = S(β), i.e. α is not a limit ordinal.
(c) Let β ∈ P(α). Since α is transitive, for all γ ∈ β, we get γ ∈ α, and from that we get γ ⊆ α.
Thus γ ∈ P(α), and we conclude that P(α) is transitive.
Question 2
(a) Assume on the contrary that there exists α ∈ Ord such that α ∪ {α} = S(α) = ω.
Then since ω is infinite and {α} is finite, we have |ω| = |α ∪ {α}| = |α|.
This give us α to be infinite, contradicting the fact that ω is the first infinite ordinal.
Thus ω is a limit ordinal.
(b) By Axiom of Foundation, there exists β ∈ ω such that for all γ ∈ ω, we have γ 6∈ β.
Since ω is transitive, we have β = ∅, and so ∅ ∈ ω.
Let α ∈ ω. Since ω is a limit ordinal, we have S(α) 6= ω. Also, we cannot have ω 6∈ S(α), or else
it give us ω ∈ α or ω = α, a contradiction to the Axiom of Foundation.
Since ordinals are comparable, we conclude that S(α) ∈ ω.
Thus ω is an inductive set.
(c) Assume on the contrary that ω 6⊆ A. Then X = ω − A is non-empty. Since (ω, ∈) is a well-ordered
set, there exists α ∈ X which is the least element of X. As A is inductive, ∅ ∈ A and so α 6= ∅.
Since all finite ordinals except ∅ are successor ordinals, there exists β ∈ ω such that S(β) = α.
By our condition on α, we have β ∈ A. However A is inductive, which give us α = S(β) ∈ A, a
contradiction that α ∈ X. Therefore ω ⊆ A.
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MA3205
Set Theory
AY 2003/2004 Sem 1
Question 3
(a) Since A is countable, there exists a bijection g : A → ω.
Let f : P(A) → P(ω) be such that f (X) = g[X].
Since g is a bijection, g[X] = g[Y ] implies X = Y , i.e. f is injective.
For all Y ⊆ ω, we have f (g −1 [Y ]) = Y , and so f is surjective.
Therefore f is bijective, i.e. |P(ω)| = |P(A)|.
(b) Since Q is countable, and R is uncountable, we have R − Q to be uncountable.
(c) Let f : (0, 1) → (1, 17) be such that f (x) = 16x + 1. Since f is linear, it is injective. Also for all
y−1
y ∈ (1, 17), we have y−1
16 ∈ (0, 1) with f ( 16 ) = y, and so f is surjective.
Therefore f is bijective, i.e. |(0, 1)| = |(1, 17)|.
Let g : [0, 1] → (0, 1) be a well-defined injective function such that

1

,
x = 0;


2 + x
x
1
g(x) =
, x = , k ∈ Z+ ;


k
 1 + 2x

x,
otherwise.
y
y
If y = k1 for some k ∈ N − {0, 1, 2}, then 1−2y
∈ [0, 1] such that g( 1−2y
) = y.
1
If y = 2 , then g(0) = y.
Otherwise, g(y) = y. Thus g is surjective.
Therefore g is bijective, i.e. |(0, 1)| = |[0, 1]|.
Question 4
(a) If |A| < |B| and |B| ≤ |C|, then there exists injective functions f : A → B and g : B → C.
This give us gf : A → C to be injective, i.e. |A| ≤ |C|.
Assume on the contrary that there exists an injective function h : C → A.
This give us hg : B → A to be an injective function, a contradiction to |A| < |B|.
Therefore |A| < |C|.
(b) If |A| ≤ |B| and |B| < |C|, then there exists injective functions f : A → B and g : B → C.
This give us gf : A → C to be injective, i.e. |A| ≤ |C|.
Assume on the contrary that there exists an injective function h : C → A.
This give us f h : C → B to be an injective function, a contradiction to |B| < |C|.
Therefore |A| < |C|.
(c) Let f : A × B → B × A be such that f (a, b) = (b, a). f is bijective, and so |A × B| = |B × A|.
SECTION B
Question 5
(a) If ω1 6∈ Card, then there exists α ∈ ω1 such that α ∈ Card and α = |ω1 |. However, this would
implies that α is uncountable, a contradiction to the condition that ω1 is the first uncountable
ordinal.
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MA3205
Set Theory
AY 2003/2004 Sem 1
(b) Let f : ω1 → S be such that f (α) = α + 1. By definition of S, f is a bijection, and so |S| = |ω1 |.
Since T ⊆ ω1 , |T | ≤ |ω1 |.
For all α ∈ Ord, we have ω · α 6∈ S, since it is a limit ordinal. Now if α ∈ ω1 , then α is countable.
Thus ω · α is a countable ordinal (since |ω · α| = |ω × α| = ω), and so ω · α ∈ ω1 .
Therefore we have the well-defined function g : ω1 → T such that g(α) = ω · α.
Let α, β ∈ ω1 such that WLOG, α < β. Then there exists γ ∈ Ord such that α + γ = β.
This give us ω · α < ω · α + ω · γ = ω · (α + γ) = ω · β, and so g is injective. Thus |ω1 | ≤ |T |.
Therefore by Cantor-Bernstein Theorem, |ω1 | = |T |.
(c) Since we have |ω| < |P(ω)|, P(ω) is uncountable.
By consequence of Axiom of Choice, there exists uncountable α ∈ Card such that α = |P(ω)|.
Since ω1 is the first uncountable cardinal, |ω1 | = ω1 ≤ α = |P(ω)|.
Question 6
Let us be given that there exists κ ∈ Card such that |κ| = |R|.
Then this give us a bijection f : R → κ.
Let us define the relation @ on R, such that for r1 , r2 ∈ R, we have r1 @ r2 iff f (r1 ) ∈ f (r2 ).
This resulted in f being the isomorphism between (R, @) and (κ, ∈).
Thus (R, @) is a well-ordered set, i.e. R is well orderable.
Instead let us be given that R is well orderable, i.e.S(R, @) is a well-order.
Let define g : R → V recursively, such that g(x) = {S(g(y)) | y @ x}.
Notice that for all x, y ∈ R, if y @ x, then g(y) ∈ g(x), thus g is a order-preserving function.
This implies that g is injective, and so (g[R], ∈) is isomorphic to (R, @).
Let α ∈ g[R], then there exists x ∈ R such that g(x) = α.
Let β ∈ α, i.e. β ∈ S(g(y)) for some y @ x.
Then the set X = {y ∈ R | y @ x ∧ β ∈ S(g(y))} is non-empty.
Since (R, @) is a well-ordered set, there exists b ∈ X such that b is the least element of X.
Assume on the contrary that β ∈ g(b). Then β ∈ g(c) for some c @ b, a contradiction to the
condition on b. Therefore from β ∈ S(g(b)) = g(b) ∪ {g(b)}, we conclude that g(b) = β, i.e.
β ∈ g[R]. Axiom of Replacement give us g[R] to be a set, and so we can conclude that it is
transitive. Thus g[R] ∈ Ord. By definition of cardinals, there exists κ ∈ Card such that κ = |g[R]|.
Therefore we have |κ| = |R|.
Question 7
(a) Since A is countable, there exists an bijection between A and N.
This bijection give us an enumeration of A, i.e. we can write A = {a0 , a1 , a2 , . . .}.
Now we define a function f : A → Q ∩ (0, 1) recursively, such that,
f (ak ) =
sup({0} ∪ {f (ai ) | ai <A ak , i ∈ k}) + inf({1} ∪ {f (ai ) | ai >A ak , i ∈ k})
,
2
k ∈ N.
Notice that in this definition above, 0 < ak < 1 for any k ∈ N, also f (ak ) > f (ai ) for any ak >A ai ,
and f (ak ) < f (ai ) for any ak <A ai , i ∈ k. Thus f is a well-defined, order-preserving function.
Since <A is a linear order, we have f to be injective.
So by letting B = f [A] ⊆ Q ∩ (0, 1), and g : A → B such that g(a) = f (a) for all a ∈ A, we get g
to be an isomorphism between (A, <A ) and (B, <), i.e. (A, <A ) ∼
= (B, <).
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MA3205
Set Theory
AY 2003/2004 Sem 1
(b) Since α < ω1 , we have α to be countable, and so (α, ∈α ) is a countable linearly ordered set.
Thus using result of (7a.), we get a set Aα ∈ Q ∩ (0, 1) such that (Aα , <) ∼
= (α, ∈α ).
Let f : Aα → α be an isomorphism, and let ∅ 6= X ⊆ Aα .
Then since α is well-ordered, and ∅ =
6 f [X] ⊆ α, there exists a least element of f [X], say z. This
implies that there exists y ∈ X such that f (y) = z. Since f is an isomorphism, y is the least
element of X, and so Aα is well-ordered by <.
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