Set theory - Chapter 6: The Axiom of Regularity

6.1

≤ $\forall z \in x (rank (z) < rank (x))$ , $\forall z \in x (rank (z) + 1 \leq rank (x))$ . So $sup \{(rank (z) + 1)| z \in x\} \leq rank (x)$
≥ For all $z \in x$ , $z \in V_{rank (z) + 1} \subseteq V_{sup \{(rank (z) + 1)| z \in x\}}$ . Hence $x \in V_{sup \{(rank (z) + 1)| z \in x+ + 1}$ i.e $rank (x) \leqsup ((rank (z) + 1), z \in x)$ .

6.2

By induction on $n < ω$ , we show that $∣ V_{n} ∣$ is finite (using 3.1 (iii)). Hence $∣ V_{ω} ∣ \leq \sum_{n < ω} ∣ V_{n} ∣ = ℵ_{0} sup_{n < ω} ∣ V_{n} ∣ \leq {ℵ_{0}}^{2} = ℵ_{0}$ . (Note that we can apply Lemma 5.8, since only the first term is equal to zero). Moreover, $∣ V_{ω} ∣ \geq ℵ_{0}$ since $ω \subseteq V_{ω}$ . Finally $∣ V_{ω} ∣ = ℵ_{0}$ .

Now we shall show by induction on $α$ that $∣ V_{ω + α} ∣ = ℶ_{α}$ . We see below that it is true for $α = 0$ . If it is the case for an ordinal $α$ , then $∣ V_{ω + α + 1} ∣ = ∣ ℘ (V_{ω + α}) ∣ = 2^{ℶ_{α}} = ℶ_{α + 1}$ . Finally, consider a limit ordinal $λ$ such that the property holds below it. Then $∣ V_{ω + λ} ∣ = ∣ ⋃_{α < λ} V_{ω + α} ∣ \leq \sum_{α < λ} ∣ V_{ω + α} ∣ = λ sup_{α < λ} ℶ_{α} = λ ℶ_{λ}$ . But Beth is a normal function, so $ℶ_{λ} \geq λ$ and $∣ V_{ω + λ} ∣ = ℶ_{λ}$ .

6.3

For all $κ$ we have $κ \subseteq V_{κ}$ so $κ \leq ∣ V_{κ} ∣$ .

Suppose $κ$ is inaccessible. Then by induction on $α < κ$ , we have $ℶ_{α} < κ$ . For $α = 0$ , it is true because $κ$ is uncountable. If the property holds for an $α$ then it also holds for $α + 1$ since $κ$ is strong limit. The regularity of $κ$ allows to prove the property at limit step. Finally we have :

$∣ V_{κ} ∣ \leq \sum_{α < κ} ∣ V_{ω + α} ∣ = κ sup_{n < κ} ℶ_{α} \leq κ^{2} = κ$

6.4

Let $x, y$ be two sets of rank less than $α$ . Then by 6.1, $rank ({x; y}) = sup {rank (x) + 1; rank (y) + 1} \leq α + 1$ . $rank < x, y > = rank ({{x}; {x; y}})$ . But we've just seen that the rank ${x}$ and ${x; y}$ is $\leq α + 1$ . So $rank < x, y > \leq α + 2$ . If $t \in x \cup y$ then $rank (t) < \inf {rank (x); rank (y)} \leq α$ , so $rank (t) + 1 \leq α$ and $rank ({x; y}) \leq α$ . If $t \in ⋃ x$ then there is $z \in x$ such that $t \in x$ . So $rank (t) < rank (z)$ , $rank (t) + 1 \leq rank (z) < rank (x) \leq α$ . So $rank (⋃ x) < α$ . If $z \in ℘ (x)$ then for all $t \in z$ , $t \in x$ , $rank (t) < rank (x)$ , $rank (t) + 1 \leq α$ . Hence $rank (z) \leq α$ . Finally $rank (℘ (z)) \leq α + 1$ . $x^{y} \subseteq y \times x = \{< a, b >| a \in y \land a \in x\}$ . For $< a, b > \in x^{y}$ , $rank (a) + 1 \leq α$ and $rank (b) + 1 \leq α$ so $rank (< a, b >) = rank ({{a}; {a; b}}) \leq sup {rank (a) + 2; rank (b) + 2} \leq α + 1$ . Finally, $rank (x^{y}) \leq α + 2$ .

6.5

We can write $ℤ = \frac{ℕ^{2}}{R_{1}}$ , $ℚ = \frac{ℤ \times ℤ^{*}}{R_{2}}$ and $ℝ = \frac{ℚ^{ℕ}}{I}$ where $R_{1}, R_{2}$ are relations of equivalence on $ℕ^{2}$ and $ℤ \times ℤ^{*}$ and $I$ the ideal $\{u \in ℚ^{ℕ}| (u_{n} \underset{n \to + \infty}{\to} 0)\}$ . Hence we have :

$rank (ℕ) = rank (ω) = ω$ .
$rank (ℤ) \leq rank (℘ (℘ (ℕ^{2}))) \leq (ω + 2) + 2 = ω + 4$ .
$rank (ℚ) \leq rank (℘ (℘ (ℤ \times ℤ^{*}))) \leq (ω + 2 + 2) + 2 = ω + 6$ .
$rank (ℝ) \leq rank (℘ (℘ (ℚ^{ℕ}))) \leq (ω + 6 + 2) + 2 = ω + 10$

6.6

Let $A$ be a class and $B = \{x \in A| x \subseteq B\}$ the class of sets that are hereditarily in the class $A$ .

⇒ Let x ∈ B y ∈ TC ( x ) ⇔ y ∈ ⋃ n ∈ ℕ S n ( x ) .
- $S_{0} (x) = x \subseteq B$
- If $S_{n} (x) \subseteq B$ , then $S_{n + 1} (x) = ⋃ S_{n} (x) \subseteq B$ because $B$ is transitive. Finally, $TC (x) \subseteq B \subseteq A$
⇐
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