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Proposition 1.37. Let \( X \) be well-ordered, and let \( \Lambda \) be the set of limits in \( X \) .\n\n(a) \( X = \left\{ {y \in X : \exists n\exists x \in \Lambda, y = {S}^{n}\left( x\right) }\right\} \) . | Proof. (a) Suppose this fails. Let \( C = \left\{ {y \in X : \forall n\forall x \in {\Lambda y} \neq {S}^{n}\left( x\right) }\right\} \neq \varnothing \) . Let \( y \) be minimum in \( C.y \notin \Lambda \), so there is \( z < y \) with \( y = S\left( z\right) \) . Since \( z \notin C \) there is \( x \in \Lambda \) an... | No |
Theorem 1.40. If \( X \) is finite, so is \( \mathcal{P}\left( X\right) \) . In fact, if \( X \) has size \( n \), for some \( n \in \mathbb{N} \), then \( \mathcal{P}\left( X\right) \) has size \( {2}^{n} \) . | Proof. . We start with the base, \( n = 0 \) : If \( X \) has no elements, \( X = \varnothing \), and \( \mathcal{P}\left( X\right) = \{ \varnothing \} \), i.e., it has \( 1 = {2}^{0} \) elements.\n\nOur induction hypothesis is: every set of size \( n \) has a power set of size \( {2}^{n} \) . From this we must prove t... | Yes |
Proposition 1.42. If \( X \) is well-ordered, then so is each \( {L}_{n}\left( X\right) \) . | Proof. \( {L}_{1} \) is isomorphic to \( X \), so it is well-ordered. Suppose \( {L}_{n}\left( X\right) \) is well-ordered. We need to show that \( {L}_{n + 1} \) is well-ordered.\n\nSuppose \( C = \left\{ {{\overrightarrow{x}}_{m} : m \in \mathbb{N}}\right\} \) is an infinitely descending chain in \( {L}_{n + 1}\left(... | Yes |
Theorem 1.43. Let \( \\left\\{ {{A}_{n} : n \\in \\mathbb{N}}\\right\\} \) be a sequence of infinite subsets of \( \\mathbb{N} \) so that each \( {A}_{n} \\supseteq {A}_{n + 1} \). Then there is an infinite set \( A \\subseteq \\mathbb{N} \) so that \( A \\smallsetminus {A}_{n} \) is finite for each \( n \), i.e., for ... | Proof. We construct \( A = \\left\\{ {{k}_{n} : n \\in \\mathbb{N}}\\right\\} \) recursively. Suppose at stage \( n \), we have constructed \( \\left\\{ {{k}_{m} : m \\leq n}\\right\\} \) and suppose \( {k}_{m} \\in {A}_{m} \) for all \( m \\leq n.{}^{8} \) Since \( {A}_{n + 1} \) is infinite, it has some element \( k ... | Yes |
Example 1.46. Let \( x \in \mathbb{R} \) and let \( \mathcal{B} \) be the collection of open intervals \( \left( {x - r, x + r}\right) \) where \( r > 0 \). | \( \mathcal{B} \) is a filterbase on \( \mathbb{R}.{\mathcal{B}}^{ \supseteq } \) is a proper, nonprincipal filter on \( \mathbb{R} \) and not an ultrafilter (neither \( \{ x\} \) nor \( \mathbb{R} \smallsetminus \{ x\} \) are in \( {\mathcal{B}}^{ \supseteq } \) ). | Yes |
Example 1.47. Let \( X \) be an infinite set, and let \( \mathcal{F} = \{ F \subseteq X : X \smallsetminus F \) is finite \( \} \). | \( \mathcal{F} \) is a proper nonprincipal filter and not an ultrafilter (since every infinite set splits into two disjoint infinite subsets). | Yes |
Example 1.48. Let \( X \) be any nonempty set, choose \( x \in X \), and let \( \mathcal{F} = \{ Y \subseteq X : x \in Y\} \). | \( \mathcal{F} \) is a proper, principal ultrafilter on \( X \) (since for every \( Y \subseteq X \) either \( x \in Y \) or \( x \in X \smallsetminus Y \) ). | "Yes" |
Proposition 1.49. Suppose \( \mathcal{F} \) is an ultrafilter on a set \( X \). (a) If \( F \in \mathcal{F} \) and \( G \subseteq F \) then either \( G \in \mathcal{F} \) or \( F \smallsetminus G \in \mathcal{F} \). (b) If \( X = {A}_{1} \cup \ldots \cup {A}_{n} \) then some \( {A}_{n} \in \mathcal{F} \). | Proof. We prove (a) and (b). For (a): If \( G \notin \mathcal{F} \) then \( X \smallsetminus G \in \mathcal{F} \), so \( F \cap \left( {X \smallsetminus G}\right) = F \smallsetminus G \in \mathcal{F} \). For (b): If not, then, since no \( {A}_{i} \in \mathcal{F},\mathcal{F} \) is proper, and each \( X \smallsetminus {A... | Yes |
Example 2.3. Suppose \( X \) is the set of permutations on a set \( S \) where \( S \) has at least three elements, and we interpret \( \circ \) by composition, \( e \) by the identity permutation. Call this model \( {\mathcal{P}}_{S} \). Then \( {\mathcal{P}}_{S} \vDash \mathrm{G}1 \land \mathrm{G}2 \land \mathrm{G}3 ... | I.e., \( {\mathcal{P}}_{S} \) is not a commutative group. Since \( \mathbb{R} \) with + and 0 is a commutative group, the statement \( \forall x\forall {yx} \circ y = y \circ x \) is independent of TG. | No |
Theorem 2.4. (Gödel’s first incompleteness theorem) Let \( \mathfrak{N} \) be the structure \( \mathbb{N} \) with the unary predicate \( S \), i.e., successor under the usual interpretation. For every consistent axiomatizable theory which extends \( {PA} \) and is true in \( \mathfrak{N} \) there is a sentence \( \varp... | More precisely, Gödel’s first incompleteness theorem says: If \( T \) is a consistent axiomatizable theory which extends PA and holds in \( \mathfrak{N} \), then \( {\varphi }_{T} \) is independent of \( T \) . By the way \( {\varphi }_{\mathrm{{PA}}} \) is constructed, \( \mathfrak{N} \vDash {\varphi }_{\mathrm{{PA}}}... | Yes |
Theorem 2.5. (Gödel's second incompleteness theorem). A consistent axiomatizable theory which interprets \( {PA} \) and whose interpretation of \( {PA} \) holds in \( \mathfrak{N} \) cannot prove its own consistency. | I.e., PA can't prove it's own consistency. We think it's consistent because we think it holds in \( \mathbb{N} \). But we have to reach outside PA to make this statement. | No |
Theorem 3.2. \( \forall x\forall {yx} = y \) iff \( \left( {x \subseteq y\text{and}y \subseteq x}\right) \) . | Proof. \( x = y \) iff \( \forall z\left( {z \in x\text{iff}z \in y}\right) \) iff \( \forall z\left( {\left( {z \in x \rightarrow z \in y}\right) \text{and}\left( {z \in y \rightarrow z \in x}\right) }\right) \) iff \( (\forall z(z \in x \rightarrow \) \( z \in y) \) and \( \forall z\left( {z \in y \rightarrow z \in x... | Yes |
Theorem 3.7. \( X \vDash \) extensionality iff, for all distinct \( x, y \in X, X \cap \left( {\left( {x \smallsetminus y}\right) \cup \left( {y \smallsetminus x}\right) }\right) \neq \varnothing \) . | Proof. Suppose \( X \vDash \) extensionality. Then for each distinct pair of elements \( x, y \in X \) there is \( z \in X \) with either \( z \in x \smallsetminus y \) or \( z \in y \smallsetminus x \) . I.e., \( X \cap \left( {\left( {x \smallsetminus y}\right) \cup \left( {y \smallsetminus x}\right) }\right) \neq \v... | Yes |
Proposition 3.9. \( \left( {x, y}\right) = \left( {z, w}\right) \) iff \( x = z \) and \( y = w \) . | Proof. By extensionality, if \( x = z \) and \( y = w \), then \( \left( {x, y}\right) = \left( {z, w}\right) \) . So suppose \( \left( {x, y}\right) = \left( {z, w}\right) \) . By extensionality, either \( \{ x\} = \{ z\} \) or \( \{ x\} = \{ z, w\} \) . If \( \{ x\} = \{ z, w\} \) , then, by extensionality, \( x = w ... | Yes |
Theorem 3.13. \( X \vDash \) pairing iff \( \forall x \in X\forall y \in X\exists z \in {Xz} \cap X = \{ x, y\} \) . | Proof. Suppose \( X \vDash \) pairing, and let \( x, y \in X \) . Then \( \exists z \in {XX} \vDash (x, y \in z \) and if \( w \in z \) then \( w = x \) or \( w = y \) ). I.e., \( \exists z \in {Xz} \cap X = \{ x, y\} \) . For the other direction, suppose \( \forall x\forall y \in X\exists z \in {Xz} \cap X = \{ x, y\}... | Yes |
Theorem 3.21. \( {Ax}\forall y\left( {y \in x}\right) \} \) | Proof. Suppose not. Let \( x \) be the universal set (i.e., \( \forall {yy} \in x \) ) and let \( \varphi \) be the formula \( z \notin z \) . Let \( y = \{ z \in x : z \notin z\} \) . By separation, \( y \in x \) . Hence \( y \in y \) iff \( y \notin y \), a contradiction. | Yes |
Theorem 3.22. (a) \( x \neq \varnothing \) iff \( \exists {yy} = \bigcap x \) . | Proof. We’ve essentially already done (a); | No |
Theorem 3.23. \( X \vDash \) union iff \( \forall x \in X\exists z \in Xz \cap X = \bigcup \{ y \cap X : y \in x \cap X\} \) . | Proof. \( X \vDash \) union iff \( \forall x \in X\exists z \in X\left( {u \in z \cap X\text{iff}\exists y \in x \cap X\;z \in y}\right) \) iff \( \forall x \in X\exists z \in Xz \cap X = \) \( \bigcup \{ y \cap X : y \in x \cap X\} \) . | Yes |
Corollary 3.25. If \( \forall x \in X\mathcal{P}\left( x\right) \subseteq X \) then \( X \vDash \) separation. | Proof. Suppose \( \varphi \) is a formula with free variables \( {x}_{0},\ldots {x}_{n} \) . Let \( {y}_{0},\ldots {y}_{n - 1} \in X \) . Whatever \( \{ y \in \) \( \left. {x \cap X : X \vDash \varphi \left( {{y}_{0},\ldots {y}_{n - 1}, y}\right) }\right\} \) is, it’s in \( X \) . | No |
Proposition 3.28. \( \exists {xx} = \varnothing \) . | Proof. Let \( x \) be inductive. \( {}^{40}\varnothing = \{ y \in x : y \neq y\} \) . | No |
Proposition 3.29. Let \( x \) be inductive. Then every finite ordinal is an element of \( x \) . | Proof. Suppose there is a finite ordinal \( n \notin x \) . Then \( n + 1 \) is well ordered and nonempty, so \( \{ k \in n + 1 : k \notin x\} \) has a least element \( {n}^{ * } \) . Since \( 0 \in x,\exists m{n}^{ * } = m + 1 \) . By definition of \( {n}^{ * }, m \in x \) . But then, since \( x \) is inductive, \( {n... | Yes |
Corollary 3.30. \( \exists z\forall {yy} \in z \) iff \( y \) is a finite ordinal. | Proof. Let \( x \) be an inductive set. By the axiom of separation, \( z = \{ y \in x : y \) is a finite ordinal \( \} \). | No |
Theorem 3.32. Let \( X \) be transitive. \( X \) is a model of infinity iff \( \exists x \in X\varnothing \in x \cap X \) and \( \forall y \in \) \( x \cap {XS}\left( y\right) \in x \cap X{.}^{42} \) | Proof. Let \( X \) be transitive. \( \forall x \in X(\left( {\varnothing \in x \cap X\text{and}\forall y \in x \cap X\;S\left( y\right) \in x \cap X}\right) \) iff \( X \vDash (\varnothing \in x \) and \( \left. {\forall y \in {xS}\left( y\right) \in x}\right) ) \) | No |
Given a set \( \left\{ {{x}_{i} : i \in I}\right\} \), there is a set \( y = {\Pi }_{i \in I}{x}_{i} \) . | First, note that \( \left( {x, y}\right) = \{ \{ x\} ,\{ x, y\} \} \), so if \( x, y \in z \) then \( \left( {x, y}\right) \subseteq \mathcal{P}\left( {\mathcal{P}\left( z\right) }\right) \), hence\n\n\[ \left( {x, y}\right) \in \mathcal{P}\left( {\mathcal{P}\left( {\mathcal{P}\left( z\right) }\right) }\right) \]\n\nSe... | Yes |
Proposition 3.34. Let \( n \in \omega \) . If, for each \( i \in n,{x}_{i} \neq \varnothing \), then \( {\Pi }_{i \in n}{x}_{i} \neq \varnothing \) . | Proof. For each \( i < n \) pick \( {y}_{i} \in {x}_{i} \) . Then \( \left\{ {\left( {i,{y}_{i}}\right) : i \in n}\right\} \in {\Pi }_{i \in n}{x}_{i} \) . | Yes |
Theorem 3.35. Let \( X \) be a transitive set. \( X \vDash \) power set iff \( \forall x \in X\exists y \in Xy \cap X = X \cap \mathcal{P}\left( x\right) \) . | Proof. Let \( X \) be transitive, and suppose \( x \in X \) . \( \exists y \in Xy \cap X = X \cap \mathcal{P}\left( x\right) \) iff \( \exists y \in {Xz} \in y \cap X \) iff \( z \in X \cap \mathcal{P}\left( x\right) \) iff \( \exists y \in X\forall z \subseteq x \) if \( z \in X \) then \( z \in y \) iff \( X \vDash \... | No |
Proposition 3.38. Let \( \varphi \) be a formula with \( n + 2 \) free variables \( {x}_{0},\ldots {x}_{n - 1}, x, y.\;\forall {x}_{0},\ldots \forall {x}_{n - 1} \) if \( \forall x\forall y\forall {z\varphi }\left( {{x}_{0},\ldots {x}_{n - 1}, x, y}\right) \) and \( \left. {\varphi \left( {{x}_{0},\ldots {x}_{n - 1}, x... | Proof. Let \( z \) be as in replacement. Define \( f = \{ \left( {xy}\right) \in w \times z : \varphi \left( {xy}\right) \} \) . | No |
Define \( \omega + \omega = \omega \cup \left\{ {{S}^{n}\left( \omega \right) : n \in \omega }\right\} \) . | We show that \( \omega + \omega \) exists. Let \( \varphi \left( {x, y}\right) \) be defined as \( y = {S}^{n}\left( \omega \right) \) if \( x = n \) for some \( n \in \omega ;y = 0 \) otherwise. \( \varphi \) is a functional By replacement, \( \left\{ {{S}^{n}\left( \omega \right) : n \in \omega }\right\} \) exists. B... | No |
Define \( {V}_{0} = \varnothing ;{V}_{n + 1} = \mathcal{P}\left( {V}_{n}\right) ;{V}_{\omega } = \mathop{\bigcup }\limits_{{n < \omega }}{V}_{n} \). We show that \( {V}_{\omega } \) exists. | Define \( \varphi \left( {n, y}\right) \) iff \( \exists \left( {{x}_{0},\ldots .{x}_{n}}\right) {x}_{0} = \varnothing, y = {x}_{n} \) and for each \( i < n{x}_{i + 1} = \) \( \mathcal{P}\left( {x}_{i}\right) .\varphi \) is a functional. Let \( z = \{ y : \exists n \in \omega \varphi \left( {n, y}\right) \} \) . By rep... | Yes |
Let \( \left\{ {{a}_{n} : n < \omega }\right\} \) be a family of infinite subsets of \( \omega \) so each \( {a}_{n} \supseteq {a}_{n + 1} \). There is an infinite set \( a \) so that \( a{ \subseteq }_{ae}{a}_{n} \) for each \( n \). | We start with \( \left\{ {{a}_{n} : n \in \omega }\right\} \) and, by proposition 3.38, a function \( f \) defined by \( f\left( n\right) = {a}_{n} \). As before, we recursively define \( {k}_{n} \) to be some element in \( {a}_{n} \) with \( {k}_{n} \neq {k}_{i} \) for each \( i < n \). But not just any element. We de... | No |
Proposition 3.43. Let \( {V}_{\omega } \) be as in example 3.40. Then \( {V}_{\omega } \vDash \) extensionality, pairing, union, separation, power set, and replacement. | Proof. We will prove pairing, and replacement, leaving the rest to the exercises.\n\nFirst we prove, by induction, that each \( {V}_{n} \subseteq {V}_{n + 1} \) . Clearly \( {V}_{0} \subseteq {V}_{1} \) . Suppose we know that \( {V}_{0} \subseteq {V}_{1} \subseteq \ldots {V}_{n}, n \geq 1 \) . Let \( y \in {V}_{n} \in ... | No |
Proposition 4.2. The axiom of regularity is equivalent to: there is no infinite descending \( \in \) -chain. | Proof. If there is a set \( \left\{ {{x}_{n} : n \in \omega }\right\} \) so each \( {x}_{n + 1} \in {x}_{n} \), then regularity would fail.\n\nFor the other direction, suppose regularity fails. Let \( {x}_{0} \) be a set with no \( \in \) -minimal element. Let \( {x}_{1} \in {x}_{0} \) . Since \( {x}_{1} \) is not \( \... | Yes |
Proposition 4.3. Let \( X \) be transitive. Then \( X \vDash \) extensionality. | Proof. Suppose \( X \) is transitive. If \( x, y \in X \) and \( x \neq y \) then there is some \( z \in \left( {x \smallsetminus y}\right) \cup \left( {y \smallsetminus x}\right) \) . Since \( X \) is transitive, \( z \in X \) . | No |
Proposition 4.4. Let \( X \) be transitive.\n\n(a) if \( y \in X \) and \( y \cap X = \varnothing \) then \( y = \varnothing \) .\n\n(b) If \( x, y, z \in X \) and \( z \cap X = \{ x, y\} \) then \( z = \{ x, y\} \) .\n\n(c) If \( z, x, y \in X \) and \( z \cap X = \left( {x, y}\right) \) then \( z = \left( {x, y}\righ... | Proof. All parts of this lemma are corollaries of the following fact: if \( X \) is transitive and \( z \in X \) then \( z \cap X = z \) . | No |
Proposition 4.5. \( X \) is transitive iff \( \forall x \subseteq X \cup x \subseteq X \) . | Proof. If \( X \) is transitive, \( x \subseteq X \) and \( y \in \bigcup x \) then there is \( z \in x \) with \( y \in z.z \in X \) and \( X \) is transitive, so \( y \in X \) . Hence \( \bigcup x \subseteq X \) .\n\nSuppose \( \forall x \subseteq X \cup x \subseteq X \) . Let \( y \in X \) . Then \( \{ y\} \subseteq... | Yes |
Theorem 4.7. \( \forall {xTC}\left( x\right) \) is transitive, and if \( x \subseteq y \) and \( y \) is transitive, then \( {TC}\left( x\right) \subseteq y \) . | Proof. Suppose \( w \in {TC}\left( x\right) \) and \( z \in w \) . Then \( w \in T{C}_{n}\left( x\right) \) for some \( n \in \omega \) . Hence \( z \in T{C}_{n + 1}\left( x\right) \subseteq \) \( {TC}\left( x\right) \) .\n\nSuppose \( x \subseteq y \) and \( y \) is transitive. By induction, each \( T{C}_{n}\left( x\r... | Yes |
Corollary 4.8. \( x \) is transitive iff \( x = {TC}\left( x\right) \) . | Proof. \( {TC}\left( x\right) \) is transitive, and \( x \subseteq {TC}\left( x\right) \), so we need to prove that if \( x \) is transitive, \( {TC}\left( x\right) \subseteq x \) . Since \( x \) is transitive, each \( T{C}_{n}\left( x\right) \subseteq x \), so \( {TC}\left( x\right) \subseteq x \) . | Yes |
Proposition 4.12. Let \( \alpha \) be an ordinal.\n\n(a) If \( x \in \alpha \) then \( x \) is an ordinal.\n\n(b) \( S\left( \alpha \right) \) is an ordinal.\n\n(c) \( \alpha \notin \alpha {.}^{48} \)\n\n(d) If \( \beta \) is an ordinal then \( \beta \subset \alpha \) iff \( \beta \in \alpha \) . | Proof. For (a): If \( x \in \alpha \) then, by transitivity, \( x \subseteq \alpha \), hence, since well-ordering is hereditary, \( x \) is strictly well-ordered by \( \in \) . For \( x \) transitive, suppose \( z \in y \in x \) . Since \( \alpha \) is transitive, \( y \in \alpha \), hence \( z \in \alpha \) . Since \(... | Yes |
Proposition 4.14. If \( A \) is an initial segment of an ordinal \( \alpha \) then either \( A \in \alpha \) or \( A = \alpha \) . | Proof. \( A \) is well-ordered and transitive, so it is an ordinal. Since it is an initial segment of \( \alpha \) , \( A \subseteq \alpha \) . | No |
Proposition 4.15. If \( \alpha ,\beta \) are ordinals, then either \( \alpha < \beta \) or \( \alpha > \beta \) or \( \alpha = \beta \) . | Proof. Let \( A = \alpha \cap \beta \) . \( A \) is an initial segment of both \( \alpha \) and \( \beta \), so \( A \) is an ordinal and \( A \in \alpha \cap \beta \) . But then \( A \in A \), contradicting proposition 4.12(c). | No |
Proposition 4.16. Let \( x \) be a set of ordinals.\n\n(a) \( \bigcup x \) is an ordinal.\n\n(b) \( \exists \alpha \) an ordinal with \( x \subseteq \alpha \) . | Proof. For (a): Since every element of a ordinal is an ordinal, \( \bigcup x \) is a set of ordinals, hence strictly well-ordered. For transitive: If \( \delta \in \gamma \in \bigcup x \) then there is \( \beta \in x \) with \( \gamma \in \beta \), hence \( \delta \in \beta \), so \( \delta \in \bigcup x \) .\n\nFor (b... | Yes |
Theorem 4.17. Induction on ordinals I. Let \( \alpha \) be an ordinal, \( \varphi \) a formula with one free variable, and suppose we know that \( \forall \beta \in \alpha \) if \( \forall \gamma < {\beta \varphi }\left( \gamma \right) \), then \( \varphi \left( \beta \right) \) . Then \( \forall \beta \in {\alpha \var... | Proof. Suppose not. Then \( \{ \beta \in \alpha : \neg \varphi \left( \beta \right) \} \neq \varnothing \) . Let \( \beta \) be its least element. Then \( \forall \gamma < {\beta \varphi }\left( \gamma \right) \) . So, by hypothesis, \( \varphi \left( \beta \right) \), a contradiction. | Yes |
Theorem 4.18. Induction on ordinals II. Let \( \varphi \) be a formula, and suppose for all ordinals \( \beta \) if \( \forall \gamma < {\beta \varphi }\left( \gamma \right) \) then \( \varphi \left( \beta \right) \) . Then for all ordinals \( \beta ,\varphi \left( \beta \right) \) . | Proof. If not, let \( \beta \) be an ordinal with \( \neg \varphi \left( \beta \right) \) . Let \( \alpha = \beta + 1 \) . Proceed as above. | No |
Theorem 4.19. ON is not a set. | Proof. If ON were a set it would be well-ordered and transitive, so it would be an ordinal \( \alpha \) with \( \alpha \in \alpha \), contradicting proposition 4.12(c). | Yes |
Proposition 4.23. (a) rank \( x \) is not a limit.\n\n(b) If \( y \in x \) and rank \( x = \alpha \) then rank \( y < \alpha \) . | Proof. For (a): If \( x \in {V}_{\alpha } \) and \( \alpha \) is a limit, then, by definition, \( x \in {V}_{\beta } \) for some \( \beta < \alpha \).\n\nFor (b): Suppose rank \( x = \beta + 1 \) . Then \( x \subseteq {V}_{\beta } \) . So, for all \( y \in x \), rank \( y \leq \beta \) . | Yes |
Theorem 4.25. If \( V = \mathop{\bigcup }\limits_{{\alpha \in \mathrm{{ON}}}}{V}_{\alpha } \) then every set has an \( \in \) -minimal element. | Proof. \( {}^{50} \) Suppose \( V = \mathop{\bigcup }\limits_{{\alpha \in \text{ ON }}}{V}_{\alpha } \) . Fix \( x \neq \varnothing \) . Let \( R = \{ \operatorname{rank}y : y \in x\} \) . Let \( \beta \) be the minimum element in \( R \), and let \( y \in x \) with rank \( y = \beta \) . If \( z \in y \) then rank \( ... | No |
Theorem 4.31. Every linear order has a well-ordered cofinal subset. | Proof. Method I, from WO: Let \( X \) be linearly ordered by \( { \leq }_{X} \) . By WO, for some ordinal \( \alpha \) , \( X = \left\{ {{x}_{\beta } : \beta < \alpha }\right\} \) . Let \( C = \left\{ {{x}_{\gamma } : \forall \delta < \gamma {x}_{\delta }{ < }_{X}{x}_{\gamma }}\right\} .{x}_{0} \in C \), so \( C \) is ... | No |
Theorem 4.32. If \( F \) is a proper filter on a set \( x \) then it extends to an ultrafilter on \( x \) . | Proof. Method I, from WO: Let \( F \) be a proper filter on \( x \) . By WO, \( \mathcal{P}\left( x\right) = \left\{ {{y}_{\beta } : \beta < \alpha }\right\} \) for some ordinal \( \alpha \) . We construct a sequence \( \left\{ {{F}_{\beta } : \beta < \alpha }\right\} \) of subsets of \( \mathcal{P}\left( x\right) \) s... | No |
Theorem 4.33. If there are no infinite descending \( \in \) -chains, then every set has an \( \in \) -minimal element. | Proof. Suppose \( x \) has no \( \in \) -minimal element. Let \( b = \{ y \in x : y \cap x \) has no \( \in \) -minimal element \( \} \) . Note that \( b \neq \varnothing \) . Let \( \mathcal{C} \) be the set of finite descending \( \in \) -chains in \( b \) where the order is \( C \leq D \) iff \( D \) is an end-exten... | Yes |
Theorem 4.34. \( \forall {XX} \vDash \) regularity. | Proof. Given \( x \in X \) with \( x \cap X \neq \varnothing \), let \( y \in x \cap X \) be minimal with respect to \( \in \) . Then \( X \vDash \forall z \in \) \( {xz} \notin y \), i.e., \( X \vDash x \) has an element minimal with respect to \( \in \) . | No |
Theorem 4.36. Let \( X \) be transitive and closed under power set, union, and finite product (i.e., if \( x \in X \) then \( \mathcal{P}\left( x\right) \in X \) and \( \bigcup x \in X \), and if \( x, y \in X \) then \( x \times y \in X \) ). Then \( X \vDash \) choice. | Proof. Suppose \( I \in X,\left\{ {{x}_{i} : i \in I}\right\} \in X \) . Then \( I \times \bigcup \left\{ {{x}_{i} : i \in I}\right\} \in X \) . By transitivity, \( \bigcup \left\{ {{x}_{i} : i \in }\right. \) \( I{\} }^{I} \subseteq X \) . By AC, \( \exists f \in {\left\{ {x}_{i} : i \in I\right\} }^{I} \) with \( f\l... | Yes |
Theorem 4.39. A set of Dedekind cuts with an upper bound has a least upper bound. | Proof. Let \( B \) be a set of Dedekind cuts with an upper bound. \( \bigcup B \) is a Dedekind cut, and \( \bigcup B \neq \mathbb{Q} \) . So \( \bigcup B \) is also an upper bound for \( B \) . We show \( \bigcup B \) is a least upper bound: suppose \( x \) is also an upper bound for \( B \) . I.e., \( x \supseteq y \... | Yes |
Proposition 5.3. (a) If \( \left| x\right| \leq \left| y\right|, a \in x \), and \( b \in y \), then \( \left| {x\smallsetminus \{ a\} }\right| \leq \left| {y\smallsetminus \{ b\} }\right| \) . | Proof. For (a): Suppose \( f : x \rightarrow y \) is 1-1. Let \( c = {f}^{ \leftarrow }\left( b\right) \) and \( d = f\left( a\right) \) . Define \( {f}^{ * } : x \smallsetminus \{ a\} \rightarrow y \smallsetminus \{ b\} \) as follows: If \( z \neq c \) then \( {f}^{ * }\left( z\right) = f\left( z\right) \) ; if \( c \... | No |
Proposition 5.5. Assume AC. \( x \) is finite, iff \( \left| x\right| \neq \left| y\right| \) for any \( y \subset x \) . | Proof. For necessity, it suffices to show that if \( m < n \) then \( \left| m\right| < \left| n\right| \) . This proof is by induction and does not use AC. So suppose we know that \( \forall i < j \leq n\left| i\right| < \left| j\right| \) . Let \( x = n \smallsetminus \{ 0\}, y = n + 1 \smallsetminus \{ 0\} \) . \( \... | Yes |
Theorem 5.7. \( \mathbb{Q} \) is countable. | Proof. We give two proofs, the first quick, and the second (due to Cantor) insightful.\n\nProof I. To each \( q \in \mathbb{Q} \) we associate the unique pair \( \left( {{n}_{q},{m}_{q}}\right) q = \frac{{n}_{q}}{{m}_{q}},{m}_{q} > 0 \), and \( {m}_{q} \) is the smallest possible such denominator. Let \( r, s, t \) be ... | Yes |
Corollary 5.8. A countable union of countable sets is countable. | Proof. \( \forall n < \omega \) let \( {x}_{n} \) be countable. We may assume that the \( {x}_{n} \) ’s are pairwise disjoint (if necessary, by replacing \( {x}_{n} \) by \( \left. {{x}_{n} \smallsetminus \mathop{\bigcup }\limits_{{i < n}}{x}_{i}}\right) \) . Let \( {f}_{n} : {x}_{n} \rightarrow \left\{ {\frac{m}{n + 1... | No |
Theorem 5.9. \( \forall x\left| x\right| < \left| {\mathcal{P}\left( x\right) }\right| \) . | Proof. It’s trivial to show that \( \left| x\right| \leq \left| {\mathcal{P}\left( x\right) }\right| \) : just let \( f\left( y\right) = \{ y\} \) . We need to show that \( \left| x\right| \neq \left| {\mathcal{P}\left( x\right) }\right| \) .\n\nSuppose \( f : x \rightarrow \mathcal{P}\left( x\right) \) . We must show ... | Yes |
Theorem 5.10. \( \left| \mathbb{R}\right| = \left| {\mathcal{P}\left( \omega \right) }\right| \) . | Proof. To show that \( \left| \mathbb{R}\right| \geq \left| {\mathcal{P}\left( \omega \right) }\right| \), let \( C = \{ x \in \mathbb{R} : 0 \leq x < 1 \) and the ternary expansion of \( x \) contains no 2’s \( \} .{}^{68} \) Define \( f : C \rightarrow \mathcal{P}\left( \omega \right) \) as follows: \( f\left( x\righ... | No |
Proposition 5.12. \( \forall x\left| {\mathcal{P}\left( x\right) }\right| = \left| {2}^{x}\right| \) . | Proof. We define \( \chi : \mathcal{P}\left( x\right) \rightarrow {2}^{x} \) as follows: for \( y \subseteq x \), define \( {\chi }_{y} : x \rightarrow 2 \) by: \( {\chi }_{y}\left( z\right) = 0 \) iff \( z \in y \) . ( \( {\chi }_{y} \) is called the characteristic function of \( y \) .) Define \( \chi \left( y\right)... | Yes |
Theorem 5.15. AC iff \( \forall x\exists {\kappa \kappa } \) is an initial ordinal and \( \left| x\right| = \left| \kappa \right| \) . | Proof. Assume AC. Then \( \exists \alpha \) an ordinal, \( \left| x\right| = \left| \alpha \right| \) . Let \( \kappa \) be the least ordinal with \( \left| \kappa \right| = \left| \alpha \right| \) . Then \( \kappa \) is an initial ordinal and \( \left| x\right| = \left| \kappa \right| \) .\n\nAssume AC fails. Then th... | Yes |
Theorem 5.27. Let \( \alpha ,\beta ,\gamma \) be ordinals.\n\n(a) \( \left( {\alpha + \beta }\right) + \gamma = \alpha + \left( {\beta + \gamma }\right) \) . | Proof. We prove (a); (b) and (c) are left to the reader.\n\nThe proof of (a) is by induction on \( \gamma \) . The induction hypothesis is that, for all \( \delta < \gamma ,\left( {\alpha + \beta }\right) + \delta = \) \( \alpha + \left( {\beta + \delta }\right) \) .\n\nCase 1. \( \gamma = \delta + 1 \) for some \( \de... | No |
Fix ordinals \( \alpha ,\beta \) . Under the lexicographical ordering, the following sets have the following order types:\n\n(a) \( \left( {\{ 0\} \times \alpha }\right) \cup \left( {\{ 1\} \times \beta }\right) \) has type \( \alpha + \beta \) . | The assertions in example 5.30 are easily proved by induction. | No |
(b) \( \left\{ {{x}_{n, m} : n, m \in \omega }\right\} \) has order type \( \omega \times \omega \) . | By example 5.30 it suffices to show that \( \left\{ {{x}_{n, m} : n, m \in \omega }\right\} \) is order-isomorphic to \( \omega \times \omega \) under the lexicographic order. So suppose \( \left( {n, m}\right) { \leq }_{L}\left( {k, s}\right) \) . If \( \left( {n, m}\right) = \left( {k, s}\right) \) we are done. If no... | Yes |
Theorem 5.32. Every well-ordered set is order-isomorphic to an ordinal. | Proof. Let \( x \) be well-ordered by \( \leq \) . We define \( f \) a function whose range is \( x \) by induction. \( f\left( 0\right) \) is the minimal element of \( x \) . For each \( \alpha \), if \( f\left\lbrack \alpha \right\rbrack = \{ f\left( \beta \right) : \beta < \alpha \} \neq x \) then \( f\left( \alpha ... | No |
Proposition 5.33. Every non-zero countable limit ordinal is some \( \mathop{\bigcup }\limits_{{n < \omega }}{\beta }_{n} \) where each \( {\beta }_{n} < {\beta }_{n + 1} \) . | Proof. Suppose \( \alpha \) is a countable limit ordinal. Since \( \alpha \) is countable there is a 1-1 onto function \( h : \omega \rightarrow \alpha \) . Define an increasing function \( f : \omega \rightarrow \alpha \) as follows: \( f\left( 0\right) = h\left( 0\right) \) . Give \( {\left. f\right| }_{n + 1} \), le... | Yes |
An ordinal \( \alpha \) is countable iff some subset of \( \mathbb{R} \) has order type \( \alpha \) . | Proof. Suppose \( A \subseteq \mathbb{R} \) has order type \( \alpha \) . Then \( A = \left\{ {{x}_{\beta } : \beta < \alpha }\right\} \) where if \( \beta < \gamma < \alpha \) then \( {x}_{\beta } < {x}_{\gamma } \) . Consider the open intervals \( {I}_{\beta } = \left( {{x}_{\beta },{x}_{\beta + 1}}\right) \subset \m... | Yes |
Theorem 5.36. Let \( 0 < \alpha \leq \beta \) be ordinals. Then there are ordinals \( \delta ,\gamma \) with \( \gamma < \alpha \) and \( \beta = \alpha \cdot \delta + \gamma \) . | Proof. Let \( \delta = \sup \{ \rho : \alpha \cdot \rho \leq \beta \} \) . By definition of \( \delta ,\alpha \cdot \delta \leq \beta \) and \( \alpha \cdot \left( {\delta + 1}\right) > \beta \) . Let \( \gamma \) be the order type of \( \beta \smallsetminus \alpha \cdot \delta \) . Then \( \alpha \cdot \delta + \gamma... | Yes |
Theorem 5.37. For all ordinals \( \alpha ,\beta ,\alpha + \beta = \beta \) iff \( \beta \geq \alpha \cdot \omega \) . | Proof. If \( \alpha + \beta = \beta \) then \( \beta \geq \alpha \) so there are \( \delta ,\gamma \) with \( \beta = \alpha \cdot \delta + \gamma ,\gamma < \alpha \) . \n\n\[ \n\alpha + \beta = \alpha + \left( {\alpha \cdot \delta }\right) + \gamma = \alpha \left( {1 + \delta }\right) + \gamma . \n\] \n\nNote that \( ... | Yes |
Theorem 5.39. A continuous strictly increasing class function defined on ON has arbitrarily high fixed points, i.e., for every \( \alpha \) there is some \( \beta > \alpha \) with \( f\left( \beta \right) = \beta \) . | Proof. Note that, by strictly increasing, each \( f\left( \alpha \right) \geq \alpha \) . Let \( {\beta }_{0} = f\left( \alpha \right) \) and define \( {\beta }_{n + 1} = f\left( {\beta }_{n}\right) \) for all finite \( n \) . Let \( \beta = \sup \left\{ {{\beta }_{n} : n < \omega }\right\} \) . By continuity, \( f\lef... | Yes |
Corollary 5.40. For each ordinal \( \alpha \) there are arbitrarily high limit ordinals with \( \alpha + \beta = \beta = \alpha \cdot \beta \) . | Proof. Let \( \beta \) be a fixed point for the function \( f\left( \beta \right) = \alpha \cdot \beta \) where \( \beta \geq \alpha \cdot \omega \) . By theorem 5.39, \( \alpha + \beta = \beta \) . | Yes |
Proposition 5.45. Let \( \kappa \) be an infinite cardinal. \( \kappa = \left| {\{ \alpha < \kappa : \alpha \text{even}\} }\right| = \left| {\{ \alpha < \kappa : \alpha \text{odd}\} }\right| \) . | Proof. Define \( f\left( {\alpha + {2n}}\right) = \alpha + n;g\left( {\alpha + {2n}}\right) = \alpha + {2n} + 1 \) . \( f \) shows that \( \kappa = \left| {\{ \alpha < \kappa : \alpha \text{even}\} }\right| \) and \( g \) shows that \( \left| {\{ \alpha < \kappa : \alpha \text{even}\} }\right| = \left| {\{ \alpha < \ka... | No |
Theorem 5.46. Let \( \kappa ,\lambda \) be infinite cardinals. Then\n\n(a) \( \kappa + \kappa = \kappa = \kappa \cdot \kappa \)\n\n(b) \( \kappa + \lambda = \sup \{ \kappa ,\lambda \} = \kappa \cdot \lambda \) | Proof. For (a): By proposition 5.45, \( \kappa + \kappa = \kappa \) . We need to show that \( \kappa = \left| {\kappa \times \kappa }\right| \) .\n\nDefine the following order \( \leq \) on \( \kappa \times \kappa \) : for \( \left( {\alpha ,\beta }\right) ,\left( {\gamma ,\delta }\right) \in \kappa \times \kappa \), w... | Yes |
Theorem 5.47. (a) For \( \kappa \) infinite, \( 2 \leq \lambda \leq \kappa ,{\lambda }^{\kappa } = {\kappa }^{\kappa } \) . | Proof. For (a) Let \( \varphi \) be a 1-1 function from \( \kappa \times \kappa \) onto \( \kappa \) . A function \( f : \kappa \rightarrow \kappa \) is a subset of \( \kappa \times \kappa \) . So we define \( \psi \left( f\right) = \{ \varphi \left( {\alpha, f\left( \alpha \right) }\right) : \alpha < \kappa \} \) . Th... | Yes |
Theorem 5.50. (a) An infinite ordinal of cardinality \( \kappa \) has cofinality at most \( \kappa \) | Proof. For (a): Suppose \( \left| \alpha \right| = \kappa \) and \( \rho = \operatorname{cf}\left( \alpha \right) \) . Let \( f : \kappa \rightarrow \alpha \) be 1-1 and onto, and define \( g \) with dom \( g \subseteq \kappa \) and range \( g \) cofinal in \( \alpha \) as follows: \( g\left( \beta \right) = \inf \{ f\... | Yes |
Theorem 5.52. (a) For every cardinal \( \kappa ,{\kappa }^{ + } \) is regular. | Proof. (a) By theorem \( {5.50},\operatorname{cf}\left( {\kappa }^{ + }\right) = {\kappa }^{ + } \) or \( \operatorname{cf}\left( {\kappa }^{ + }\right) \leq \kappa \) . If the former, we’re done, so suppose the latter. Let \( \lambda = \operatorname{cf}\left( {\kappa }^{ + }\right) \) . There is an increasing cofinal ... | Yes |
Theorem 5.54. Suppose \( I \neq \varnothing \) and \( {\kappa }_{i} < {\lambda }_{i} \) for all \( i \in I \) . Then \( {\sum }_{i \in I}{\kappa }_{i} < {\Pi }_{i \in I}{\lambda }_{i} \) . | Proof. Let \( \left\{ {{x}_{i} : i \in I}\right\} \) be pairwise disjoint, each \( \left| {x}_{i}\right| = {\kappa }_{i} \), and let \( {f}_{i} : {x}_{i} \rightarrow {\lambda }_{i} \smallsetminus \{ 0\} \) be 1-1. For \( y \in {x}_{i} \) we define \( {f}_{y} \in {\Pi }_{j \in I}{\lambda }_{i} \) by \( {f}_{y}\left( i\r... | Yes |
Theorem 5.55. For each infinite cardinal \( \kappa \)\n\n\[ \text{(a)}\kappa < {\kappa }^{\operatorname{cf}\left( \kappa \right) }\text{.}\]\n\n\[ \text{(b)}\kappa < {cf}\left( {2}^{\kappa }\right) \text{.} \] | Proof. For (a): Let \( f \) be a 1-1 increasing cofinal map from \( \operatorname{cf}\left( \kappa \right) \) to \( \kappa \) . Then \( \kappa = \left| {\mathop{\bigcup }\limits_{{\alpha < \operatorname{cf}\left( \kappa \right) }}{f}_{\alpha }}\right| \leq \) \( {\sum }_{\alpha < \operatorname{cf}\left( \kappa \right) ... | Yes |
Example 5.60. How big is \( \mathbb{R} \) ? | We already answered this in theorem 5.10, but here’s a short proof. By section 4.7, every real corresponds to a set of rationals. So \( \left| \mathbb{R}\right| \leq {\omega }^{\omega } = {2}^{\omega } \) . And, since the Cantor set is a subset of \( \mathbb{R},\left| \mathbb{R}\right| \geq {2}^{\omega } \) . So \( \le... | Yes |
If \( D \) is dense in a linear order \( X \), then \( \left| X\right| \leq {2}^{\left| D\right| } \) . (Here, \( D \) is dense in \( X \) iff for every \( x < y \in X \) there is some \( d \in D \) with \( x \leq d \leq y \) .) | This follows from the following: if \( x \neq y \) then either \( \{ d \in D : d < x\} \neq \{ d \in D : d < y\} \) or \( x \notin D, y \in D \) and \( y = S\left( x\right) \) . The reader is invited to fill in the details. | No |
Let \( \mathbb{P} \) be the set of irrationals. What’s \( \left| \mathbb{P}\right| ? | \( \mathbb{Q} \) is countable, and \( {2}^{\omega } = \left| \mathbb{R}\right| = \left| {\mathbb{Q} \cup \mathbb{P}}\right| = \left| \mathbb{Q}\right| + \left| \mathbb{P}\right| \), so \( \left| \mathbb{P}\right| = {2}^{\omega } \) | Yes |
Let \( K \) be an infinite field of size \( \kappa \) . How big is \( P\left\lbrack K\right\rbrack \), the ring of polynomials in one variable over \( K \) ? | A polynomial over \( K \) has the form \( {\sum }_{i \leq n}{k}_{i}{x}^{i} \) where \( n \in \omega \) and each \( {k}_{i} \in K \) . So \( \left| {P\left\lbrack K\right\rbrack }\right| = \mathop{\bigcup }\limits_{{n \in \omega }}{K}^{n} \) . Each \( \left| {K}^{n}\right| = {\kappa }^{n} = \kappa \), so \( \left| {P\le... | Yes |
How many open sets of reals are there? | Recall that an open set of reals is a union of open intervals with rational endpoints. Let \( \mathcal{I} \) be the set of open intervals with rational endpoints. \( \left| \mathcal{I}\right| = {\left| \mathbb{Q}\right| }^{2} = \omega \) . Given an open set \( u \), we define \( f\left( u\right) = \{ I \in \mathcal{I} ... | Yes |
Define \( C\left( \mathbb{R}\right) \) to be the set of continuous functions from \( \mathbb{R} \) to \( \mathbb{R} \) . How big is \( C\left( \mathbb{R}\right) \) ? | Every continuous function from \( \mathbb{R} \) to \( \mathbb{R} \) is determined by its values on \( \mathbb{Q} \), so there are at most \( {\left| \mathbb{R}\right| }^{\left| \mathbb{Q}\right| } = {\left( {2}^{\omega }\right) }^{\omega } = {2}^{\omega } \) many such functions. There are exactly that many, since for e... | Yes |
For any \( x \), let \( {\left\lbrack x\right\rbrack }^{\kappa } \) denote the subsets of \( x \) of size exactly \( \kappa \) . If \( \kappa \) is infinite, how big is \( {\left\lbrack \kappa \right\rbrack }^{\kappa } \)? | Given \( a \in {\left\lbrack \kappa \right\rbrack }^{\kappa } \), let \( {\chi }_{a} \) be the characteristic function of \( a \) . There are at most \( {2}^{\kappa } \) many \( {\chi }_{a} \)’s, so \( {\left\lbrack \kappa \right\rbrack }^{\kappa } \leq {2}^{\kappa } \) . Since \( \kappa \times \kappa = \kappa \), ther... | Yes |
Theorem 6.7. ZFC cannot prove that there is a strongly inaccessible cardinal. | Proof. If there is a strongly inaccessible cardinal, there’s a smallest one, \( \kappa .{V}_{\kappa } \vDash \) ZFC. If ZFC proved that there is a strongly inaccessible cardinal, then there would be \( x \in {V}_{\kappa } \) with \( {V}_{\kappa } \vDash x \) is strongly inaccessible, i.e.\n\n1. \( {V}_{\kappa } \vDash ... | No |
Proposition 6.10. (a) The set of even natural numbers is definable from \( \omega \) . | For (a): \( \exists {x}_{0}x = {x}_{0} + {x}_{0} \) | Yes |
Proposition 6.11. \( \left| a\right| \leq \left| {\operatorname{Def}\left( a\right) }\right| \leq \omega \cdot \left| a\right| \) . | Proof. A formula \( \Phi \) with parameters in \( a \) can be represented as a finite sequence whose elements are either elements in \( a \) (the parameters) or elements in the language of set theory (which is countable). Hence there are at most \( \omega \cdot \left| a\right| = \sup \{ \left| a\right| ,\omega \} \) ma... | No |
Corollary 6.12. If \( a \) is infinite, then \( \operatorname{Def}\left( a\right) \neq \mathcal{P}\left( a\right) \) . | In fact, if \( a \) is infinite, then \( \left| {\mathcal{P}\left( a\right) \smallsetminus \operatorname{Def}\left( a\right) }\right| = {2}^{\left| a\right| } > \left| {\operatorname{Def}\left( a\right) }\right| = \left| a\right| \) . | Yes |
Proposition 6.15. Each \( {L}_{\alpha } \) is transitive. | Proof. Suppose \( {L}_{\beta } \) is transitive for each \( \beta < \alpha \) . If \( \alpha \) is a limit, we are done, so suppose \( {L}_{\alpha } = \) \( \operatorname{Def}\left( {L}_{\beta }\right) \) for some \( \beta \) . If \( y \in x \in {L}_{\alpha }, x \subseteq {L}_{\beta } \), so \( y \in {L}_{\beta } \) . ... | No |
Proposition 6.16. Every ordinal is an element of \( L \) . | Proof. By proposition 6.10(b), it suffices to show that every ordinal is a subset of \( L \) . By induction, suppose every \( \beta < \alpha \) is a subset of \( L \) . If \( \alpha \) is a limit, we’re done. If \( \alpha = \beta + 1 \) for some \( \beta \), then \( \beta \in {L}_{\gamma } \) for some \( \gamma \), hen... | Yes |
Proposition 6.18. \( L \vDash \) extensionality, regularity, infinity, pairing, union, and power set. | Proof. Regularity, extensionality and infinity are an exercise. Pairing and union follow from proposition 6.10(e) and (f). For power set, suppose \( a \in L \) . By the axiom of replacement there is some ordinal \( \beta \) so \( \mathcal{P}\left( a\right) \cap L \subset {L}_{\beta } \) . By the formula \ | No |
Proposition 6.19. If \( L \vDash \kappa \) is weakly inaccessible, then \( L \vDash \kappa \) is strongly inaccessible. | A cardinal \( \kappa \) in the real universe \( V \) is always a cardinal in \( L : L \) can’t have a function from a smaller ordinal \( \alpha \) onto \( \kappa \) if there isn’t one in \( V \) . Similarly, a weakly inaccessible cardinal in \( V \) is weakly inaccessible (hence strongly inaccessible) in \( L \) . | No |
Theorem 6.21. If \( {0}^{\# } \) exists, then \( {\omega }_{1} \) is strongly inaccessible in \( L \) . | Note that the \( {\omega }_{1} \) of theorem 6.21 is the real \( {\omega }_{1} \) . If \( {0}^{\# } \) exists, then not only is \( {\omega }_{1}^{L} \) countable, but there are many ordinals that \( L \) thinks are cardinals between it and the real \( {\omega }_{1} \), in fact \( L \) thinks there are strongly inaccess... | No |
Corollary 6.23. If \( {0}^{\# } \) does not exist, then for every singular strong limit cardinal \( \kappa {2}^{\kappa } = {\kappa }^{ + } \) . | Proof. Let \( \kappa \) be a singular strong limit. By exercise 38 of chapter 5 it suffices to show that \( {\kappa }^{\mathrm{{cf}}\left( \kappa \right) } = {\kappa }^{ + } \) . Suppose \( x \subset \kappa ,\left| x\right| = \operatorname{cf}\left( \kappa \right) = \lambda < \kappa \) . By theorem \( {6.22}, x \subset... | No |
Partition \( {\left\lbrack \omega \right\rbrack }^{2} \) into two pieces: the pairs whose product is even, and the pairs whose product is odd. | Any set consisting solely of even numbers is homogeneous. Any set consisting solely of odd numbers is homogeneous. A set with exactly one odd element and the rest even elements is homogeneous. Any set with at least one even and at least two odd numbers is non-homogeneous. | No |
Theorem 7.6. Every infinite partial order has either an infinite antichain or an infinite set of pairwise compatible elements. | Proof. Let \( A \) be an infinite countable subset of the given partial order, and let \( {\left\lbrack A\right\rbrack }^{2} = {P}_{0} \cup {P}_{1} \) where \( \{ x, y\} \in {P}_{0} \) iff \( x, y \) are incompatible; otherwise \( \{ x, y\} \in {P}_{1} \) . Let \( H \) be an infinite homogeneous set for this partition.... | Yes |
Theorem 7.7. An infinite partial order with no infinite antichain has either an infinite chain, or an infinite pairwise incomparable pairwise compatible set. | Proof. Let \( A \) be an infinite countable subset of the given partial order, and let \( {\left\lbrack A\right\rbrack }^{2} = {P}_{0} \cup {P}_{1} \cup {P}_{2} \) where \( \{ x, y\} \in {P}_{0} \) iff \( x, y \) are incompatible; \( \{ x, y\} \in {P}_{1} \) iff \( x, y \) are comparable; \( {P}_{2} = {\left\lbrack X\r... | Yes |
Proposition 7.8. Let \( \kappa ,\lambda ,\rho ,\sigma ,\tau \) be cardinals and suppose \( \kappa \rightarrow {\left( \lambda \right) }_{\sigma }^{\rho } \). Then\n\n(a) If \( \tau > \kappa \) then \( \tau \rightarrow {\left( \lambda \right) }_{\sigma }^{\rho } \).\n\n(b) If \( \tau < \lambda \) then \( \kappa \rightar... | The proof is left to the exercises. | No |
Proposition 7.9. If \( \kappa \) is infinite and \( \kappa \rightarrow {\left( \kappa \right) }_{2}^{2} \) then \( \kappa \rightarrow {\left( \kappa \right) }_{m}^{2} \) for all \( m \in \omega \) . | Proof. The proof is by induction on \( m \) . If \( \kappa \rightarrow {\left( \kappa \right) }_{m - 1}^{2} \) then, given a partition \( {P}_{1},\ldots {P}_{m} \) of \( {\left\lbrack \kappa \right\rbrack }^{2} \) into \( m \) pieces, let \( {P}_{1}^{ * } = {P}_{1} \cup {P}_{2} \), and for \( i > 2 \) let \( {P}_{i}^{ ... | Yes |
Proposition 7.10. \( \omega \rightarrow {\left( \omega \right) }_{2}^{2} \) . | Proof. Suppose \( {\left\lbrack \omega \right\rbrack }^{2} = {P}_{1} \cup {P}_{2} \) where \( {P}_{1} \cap {P}_{2} = \varnothing \) . We recursively build a sequence of natural numbers \( \left\{ {{k}_{j} : j < \omega }\right\} \) and a sequence of infinite sets \( \left\{ {{A}_{j} : j < \omega }\right\} \) so that eac... | Yes |
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