Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 12. A family \( \mathfrak{F} \) is normal if and only if its closure \( {\mathfrak{F}}^{ - } \) with respect to the distance function (65) is compact. | It is also customary to say that \( \mathfrak{F} \) is relatively compact if \( {\mathfrak{F}}^{ - } \) is compact. Thus, normal and relatively compact families are the same. We shall now relate the notion of normal families to total boundedness. If \( \mathfrak{F} \) is normal, then \( {\mathfrak{F}}^{ - } \) is compa... | Yes |
Theorem 13. The family \( \mathfrak{F} \) is totally bounded if and only if to every compact set \( E \subset \Omega \) and every \( \varepsilon > 0 \) it is possible to find \( {f}_{1},\ldots ,{f}_{n}\epsilon \mathfrak{F} \) such that every \( {f\epsilon }\mathfrak{F} \) satisfies \( d\left( {f,{f}_{j}}\right) < \vare... | If \( \mathfrak{F} \) is totally bounded there exist \( {f}_{1},\ldots ,{f}_{n} \) such that, for any \( f \in \mathfrak{F} \) , \( \rho \left( {f,{f}_{j}}\right) < \varepsilon \) for some \( {f}_{j}.\; \) By (65) this implies \( {\delta }_{k}\left( {f,{f}_{j}}\right) < {2}^{k}\varepsilon \), or \( \delta \left( {f,{f}... | Yes |
Theorem 14. A family \( \mathfrak{F} \) of continuous functions with values in a metric space \( S \) is normal in the region \( \Omega \) of the complex plane if and only if\n\n(i) &is equicontinuous on every compact set \( E \subset \Omega \) ;\n\n(ii) for any \( z \in \Omega \) the values \( f\left( z\right), f \in ... | We give two proofs of the necessity of (i). Assume that \( \mathfrak{F} \) is normal and determine \( {f}_{1},\ldots ,{f}_{n} \) as in Theorem 13. Because each of these functions is uniformly continuous on \( E \) we can find a \( \delta > 0 \) such that \( d\left( {{f}_{j}\left( z\right) ,{f}_{j}\left( {z}_{0}\right) ... | Yes |
Theorem 15. A family \( \mathfrak{F} \) of analytic functions is normal with respect to \( \mathbf{C} \) if and only if the functions in \( \mathfrak{F} \) are uniformly bounded on every compact set. | To prove the sufficiency we prove equicontinuity. Let \( C \) be the boundary of a closed disk in \( \Omega \), of radius \( r \) . If \( z,{z}_{0} \) are inside \( C \) we obtain by Cauchy's integral theorem\n\n\[ f\left( z\right) - f\left( {z}_{0}\right) = \frac{1}{2\pi i}{\int }_{C}\left( {\frac{1}{\zeta - z} - \fra... | Yes |
Theorem 16. A locally bounded family of analytic functions has locally bounded derivatives. | This follows at once by the Cauchy representation of the derivative. If \( C \) is the boundary of a closed disk in \( \Omega \), of radius \( r \), then\n\n\[ \n{f}^{\prime }\left( z\right) = \frac{1}{2\pi i}{\int }_{C}\frac{f\left( \zeta \right) {d\zeta }}{{\left( \zeta - z\right) }^{2}}. \n\]\n\nHence \( \left| {{f}... | Yes |
Theorem 17. A family of analytic or meromorphic functions \( f \) is normal in the classical sense if and only if the expressions\n\n(58)\n\n\[ \rho \left( f\right) = \frac{2\left| {{f}^{\prime }\left( z\right) }\right| }{1 + {\left| f\left( z\right) \right| }^{2}} \] \n\nare locally bounded. | The geometric meaning of the quantity \( \rho \left( f\right) \) is rather evident. Indeed, by use of the formula in Chap. 1, Sec. 2.4\n\n\[ d\left( {f\left( {z}_{1}\right), f\left( {z}_{2}\right) }\right) = \frac{2\left| {f\left( {z}_{1}\right) - f\left( {z}_{2}\right) }\right| }{{\left\lbrack \left( 1 + {\left| f\lef... | Yes |
Theorem 1. Given any simply connected region \( \Omega \) which is not the whole plane, and a point \( {z}_{0} \in \Omega \), there exists a unique analytic function \( f\left( z\right) \) in \( \Omega \) , normalized by the conditions \( f\left( {z}_{0}\right) = 0,{f}^{\prime }\left( {z}_{0}\right) > 0 \), such that \... | The uniqueness is easily proved, for if \( {f}_{1} \) and \( {f}_{2} \) are two such functions, then \( {f}_{1}\left\lbrack {{f}_{2}^{-1}\left( w\right) }\right\rbrack \) defines a one-to-one mapping of \( \left| w\right| < 1 \) onto itself. We know that such a mapping is given by a linear transformation \( S \) (Chap.... | Yes |
Theorem 2. Let \( f \) be a topological mapping of a region \( \Omega \) onto a region \( {\Omega }^{\prime } \) . If \( \left\{ {z}_{n}\right\} \) or \( z\left( t\right) \) tends to the boundary of \( \Omega \), then \( \left\{ {f\left( {z}_{n}\right) }\right\} \) or \( f\left( {z\left( t\right) }\right) \) tends to t... | Indeed, let \( K \) be a compact set in \( {\Omega }^{\prime } \) . Then \( {f}^{-1}\left( K\right) \) is a compact set in \( \Omega \), and there exists \( {n}_{0} \) (or \( {t}_{0} \) ) such that \( {z}_{n} \) (or \( z\left( t\right) \) ) is not in \( {f}^{-1}\left( K\right) \) for \( n > {n}_{0} \) (or \( t > {t}_{0... | Yes |
Theorem 3. Suppose that the boundary of a simply connected region \( \Omega \) contains a line segment \( \gamma \) as a one-sided free boundary arc. Then the function \( f\left( z\right) \) which maps \( \Omega \) onto the unit disk can be extended to a function which is analytic and one to one on \( \Omega \cup \gamm... | For the proof we consider a disk around \( {x}_{0}{\epsilon \gamma } \) which is so small that the half disk in \( \Omega \) does not contain the point \( {z}_{0} \) with \( f\left( {z}_{0}\right) = 0 \) . Then \( \log f\left( z\right) \) has a single-valued branch in the half disk, and its real part tends to 0 as \( z... | Yes |
Theorem 4. If the boundary of \( \Omega \) contains a free one-sided analytic arc \( \gamma \), then the mapping function has an analytic extension to \( \Omega \cup \gamma \), and \( \gamma \) is mapped on an arc of the unit circle. | We trust the reader to make the last statement more precise and to complete the proof. | No |
Theorem 6. A continuous function \( u\left( z\right) \) which satisfies condition (8) is necessarily harmonic. | Again, the condition need be satisfied only for sufficiently small \( r \) . If \( u \) satisfies (8), so does the difference between \( u \) and any harmonic function. Suppose that the disk \( \left| {z - {z}_{0}}\right| \leqq \rho \) is contained in \( \Omega \), the region where \( u \) is defined. By use of Poisson... | Yes |
Theorem 7. Consider a sequence of functions \( {u}_{n}\left( z\right) \), each defined and harmonic in a certain region \( {\Omega }_{n} \) . Let \( \Omega \) be a region such that every point in \( \Omega \) has a neighborhood contained in all but a finite number of the \( {\Omega }_{n} \), and assume moreover that in... | For the proof, suppose first that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{u}_{n}\left( {z}_{0}\right) = \infty \) for at least one point \( {z}_{0} \in \Omega \) . By assumption there exist \( r \) and \( m \) such that the functions \( {u}_{n}\left( z\right) \) are harmonic and form a nondecreasing sequence... | Yes |
Theorem 8. A continuous function \( v\left( z\right) \) is subharmonic in \( \Omega \) if and only if it satisfies the inequality\n\n(12)\n\n\[ v\left( {z}_{0}\right) \leqq \frac{1}{2\pi }{\int }_{0}^{2\pi }v\left( {{z}_{0} + r{e}^{i\theta }}\right) {d\theta } \]\n\nfor every disk \( \left| {z - {z}_{0}}\right| \leqq r... | The sufficiency follows by the fact that (12), rather than the mean-value property, is what is actually needed in order to show that \( v \) cannot have a maximum without being constant. Since \( v - u \) satisfies the same inequality, it follows that \( v \) is subharmonic.\n\nIn order to prove the necessity we form t... | Yes |
Lemma 2. Suppose that there exists a harmonic function \( \omega \left( z\right) \) in \( \Omega \) whose continuous boundary values \( \omega \left( \zeta \right) \) are strictly positive except at one point \( {\zeta }_{0} \) where \( \omega \left( {\zeta }_{0}\right) = 0 \) . Then, if \( f\left( \zeta \right) \) is ... | The lemma will be proved if we show that \( \mathop{\lim }\limits_{{z \rightarrow {\zeta }_{0}}}u\left( z\right) \leqq f\left( {\zeta }_{0}\right) + \varepsilon \) and \( \mathop{\lim }\limits_{{z \rightarrow {\zeta }_{0}}}u\left( z\right) \geqq f\left( {\zeta }_{0}\right) - \varepsilon \) for all \( \varepsilon > 0 \)... | Yes |
Theorem 10. The function \( F\left( z\right) \) effects a one-to-one conformal mapping of \( \Omega \) onto the annulus \( 1 < \left| w\right| < {e}^{{\lambda }_{1}} \) minus \( n - 2 \) concentric arcs situated on the circles \( \left| w\right| = {e}^{{\lambda }_{i}}, i = 2,\ldots, n - 1 \) . | The proof is by use of the argument principle. We know that \( F\left( z\right) \) is analytic with a constant modulus on each contour. The number of roots of the equation \( F\left( z\right) = {w}_{0} \) is given by\n\n(15)\n\n\[ \frac{1}{2\pi i}{\int }_{{C}_{1}}\frac{{F}^{\prime }\left( z\right) {dz}}{F\left( z\right... | Yes |
Lemma 3. The period \( {P}_{k}\left( {z}_{0}\right) \) equals the harmonic measure \( {\omega }_{k}\left( {z}_{0}\right) \) multiplied \( {by}{2\pi } \) . | The proof is another application of Theorem 21, Chap. 4. We express the fact that the integral of \( {\omega }_{k} * {dg} - g * d{\omega }_{k} \) over \( C - c \) must vanish. The integral over \( C \) reduces to \( {P}_{k}\left( {z}_{0}\right) \), and by the same computation as above the integral over \( c \) equals \... | Yes |
Theorem 1. A discrete module consists either of zero alone, of the integral multiples \( {n\omega } \) of a single complex number \( \omega \neq 0 \), or of all linear combinations \( {n}_{1}\omega + {n}_{2}{\omega }_{2} \) with integral coefficients of two numbers \( {\omega }_{1},{\omega }_{2} \) with nonreal ratio \... | As soon as \( M \) contains a number \( \omega \neq 0 \) it also contains one, call it \( {\omega }_{1} \), whose absolute value is a minimum. Indeed, if \( r \) is large enough the disk \( \left| z\right| \leqq r \) contains a point from \( M \), other than 0 . Because the points are isolated there are only a finite n... | Yes |
There exists a basis \( \left( {{\omega }_{1},{\omega }_{2}}\right) \) such that the ratio \( \tau = {\omega }_{2}/{\omega }_{1} \) satisfies the following conditions: (i) Im \( \tau > 0 \) ,(ii) \( - \frac{1}{2} < \) Re \( \tau \leqq \frac{1}{2} \) , (iii) \( \left| \tau \right| \geqq 1 \) ,(iv) \( \operatorname{Re}\t... | If we select \( {\omega }_{1} \) and \( {\omega }_{2} \) as in the proof of Theorem 1, then \( \left| {\omega }_{1}\right| \leqq \) \( \left| {\omega }_{2}\right| ,\left| {\omega }_{2}\right| \leqq \left| {{\omega }_{1} + {\omega }_{2}}\right| \), and \( \left| {\omega }_{2}\right| \leqq \left| {{\omega }_{1} - {\omega... | Yes |
Theorem 3. An elliptic function without poles is a constant. | If \( f\left( z\right) \) has no poles, it is bounded on the closure of \( {P}_{a} \), and hence in the whole plane. By Liouville’s theorem (Chap. 4, Sec. 2.3) it must reduce to a constant. | Yes |
Theorem 4. The sum of the residues of an elliptic function is zero. | We may choose \( a \) so that none of the poles fall on the boundary of \( {P}_{a} \) . If the boundary \( \partial {P}_{a} \) is traced in the positive sense, the sum of the residues at the poles in \( {P}_{a} \) equals\n\n\[ \frac{1}{2\pi i}{\int }_{\partial {P}_{a}}f\left( z\right) {dz} \]\n\nBecause \( f \) has per... | Yes |
Theorem 5. A nonconstant elliptic function has equally many poles as it has zeros. | The poles and zeros of \( f \) are simple poles of \( {f}^{\prime }/f \), which is itself an elliptic function. The multiplicities are the residues of \( {f}^{\prime }/f \), counted positive for zeros and negative for poles. The theorem now follows from Theorem 4. | Yes |
Theorem 6. The zeros \( {a}_{1},\ldots ,{a}_{n} \) and poles \( {b}_{1},\ldots ,{b}_{n} \) of an elliptic function satisfy \( {a}_{1} + \cdots + {a}_{n} \equiv {b}_{1} + \cdots + {b}_{n}\left( {\;\operatorname{mod}\;M}\right) \) . | This is proved by considering the integral\n\n(8)\n\n\[ \frac{1}{2\pi i}{\int }_{\partial {P}_{a}}\frac{z{f}^{\prime }\left( z\right) }{f\left( z\right) }{dz} \]\n\nwhere we may again assume that there are no zeros or poles on the boundary. By the calculus of residues the integral equals \( {a}_{1} + \cdots \) . + \( {... | Yes |
Theorem 8. Every point \( \tau \) in the upper half plane is equivalent under the congruence subgroup mod 2 to exactly one point in \( \widetilde{\Omega } \cup {\Omega }^{\prime } \) . | We refer to Fig. 7-4. The reader is asked to verify that the region \( \Delta \) is mapped on the shaded regions in the figure by means of the linear transformations \( \tau , - 1/\tau ,\tau - 1,1/\left( {1 - \tau }\right) ,\left( {\tau - 1}\right) /\tau ,\tau /\left( {1 - \tau }\right) \) which we shall denote by \( {... | No |
Theorem 1. Two analytic continuations \( {\bar{\gamma }}_{1} \) and \( {\bar{\gamma }}_{2} \) of a global analytic function \( \mathbf{f} \) along the same arc \( \gamma \) are either identical, or \( {\bar{\gamma }}_{1}\left( t\right) \neq {\bar{\gamma }}_{2}\left( t\right) \) for all \( t \) . | The proof is a triviality. Because \( \pi \) is a local homeomorphism the image of \( {\bar{\gamma }}_{1} - {\bar{\gamma }}_{2} \) cannot contain a point of the zero section without being contained in it. | Yes |
Theorem 3. If \( P\left( {w, z}\right) \) and \( Q\left( {w, z}\right) \) are relatively prime polynomials, there are only a finite number of values \( {z}_{0} \) for which the equations \( P\left( {w,{z}_{0}}\right) = 0 \) and \( Q\left( {w,{z}_{0}}\right) = 0 \) have a common root. | We suppose that \( P \) and \( Q \) are ordered according to decreasing powers of \( w \) and set \( Q\left( {w, z}\right) = {b}_{0}\left( z\right) {w}^{m} + \cdots + {b}_{m}\left( z\right) \) where \( {b}_{0}\left( z\right) \) is not identically zero. If \( P \) is divided by \( Q \), the division algorithm yields a q... | Yes |
Lemma 1. There exists an open disk \( \Delta \), containing \( {z}_{0} \), and \( n \) function elements \( \left( {{f}_{1},\Delta }\right) ,\left( {{f}_{2},\Delta }\right) ,\ldots ,\left( {{f}_{n},\Delta }\right) \) with these properties:\n\n(a) \( P\left( {{f}_{i}\left( z\right), z}\right) = 0 \) in \( \Delta \) ;\n\... | The polynomial \( P\left( {w,{z}_{0}}\right) \) has simple zeros at \( w = {w}_{i} \) . We determine \( \varepsilon > 0 \) so that the disks \( \left| {w - {w}_{i}}\right| \leqq \varepsilon \) do not overlap and denote the circles \( \left| {w - {w}_{i}}\right| = \varepsilon \) by \( {C}_{i} \) . Then \( P\left( {w,{z}... | Yes |
An analytic function is an algebraic function if it has a finite number of branches and at most algebraic singularities. | Every algebraic function \( w = \mathbf{f}\left( z\right) \) satisfies an irreducible equation \( P\left( {w, z}\right) = 0 \), unique up to a constant factor, and every such equation determines a corresponding algebraic function uniquely. | Yes |
Theorem 5 (Picard). An entire function with more than one finite lacunary value reduces to a constant. | We recall that an entire function \( f\left( z\right) \) is one which is analytic in the whole plane. If \( a \) and \( b \) are distinct finite values and if \( f\left( z\right) \) is different from \( a \) and \( b \) for all \( z \), we are required to show that \( f\left( z\right) \) is constant. Consider \( {f}_{1... | Yes |
If \( {z}_{0} \) is a regular singular point for the equation (10), there exist linearly independent solutions of the form \( {\left( z - {z}_{0}\right) }^{{\alpha }_{1}}{g}_{1}\left( z\right) \) and \( {\left( z - {z}_{0}\right) }^{{\alpha }_{2}}{g}_{2}\left( z\right) \) with \( {g}_{1}\left( 0\right) ,{g}_{2}\left( 0... | If one solution is known it is not difficult to find another, linearly independent of the first. The methods which lead to a second solution belong more properly in a textbook on differential equations. It is also impossible to treat the case of irregular singularities in this book. | No |
Theorem 1 设 \( 1 < p \) ,\n\n\[ \n\left\{ \begin{array}{ll} p < \infty , & \text{ 如果 }n = 1, \\ p < {p}_{0}\left( {n,\gamma }\right) , & \text{ 如果 }n \geq 2, \end{array}\right. \n\] \n\n其中 \( {p}_{0}\left( {n,\gamma }\right) = \frac{n + 5 - {2\gamma } + \sqrt{{\left( n + 5 - 2\gamma \right) }^{2} + {8n} - 8}}{2\left( {... | ## 3 定理的证明\n\n定义泛函如下\n\n\[ \nU\left( t\right) = {\int }_{{R}^{n}}{u}_{t}\left( {t, x}\right) \mathrm{d}x \n\] \n\n(4) \n\n(3) 式中,取 \( \phi \) 满足 \( \phi \equiv 1,\left\{ {\left( {s, x}\right) \in \left\lbrack {0, t}\right\rbrack \times {R}^{n} : \left| x\right| \leq R + s}\right\} \) . 于是,可得\n\n\[ \n{\int }_{{R}^{n}}{u... | Yes |
(1) \( a = a \) .\n\n(2) \( a = b \rightarrow b = a \) .\n\n(3) \( a = b \land b = c \rightarrow a = c \) . | Proof.\n\n(1) \( \left( {\forall x}\right) \left\lbrack {x \in a \leftrightarrow x \in a}\right\rbrack \) .\n\n(2) \( \left( {\forall x}\right) \left\lbrack {x \in a \leftrightarrow x \in b}\right\rbrack \rightarrow \left( {\forall x}\right) \left\lbrack {x \in b \leftrightarrow x \in a}\right\rbrack \) .\n\n(3) \( \le... | Yes |
Proposition 3.3. \( a = b \rightarrow \left\lbrack {a \in c \leftrightarrow b \in c}\right\rbrack \) . | Proof. Axiom 1 and Proposition 3.2(2). | No |
Proposition 4.4. Each wff in the wider sense \( \varphi \), is reducible to one and only one wff \( {\varphi }^{ * } \) determined from \( \varphi \) by the rules (1)-(7) of Definition 4.3. | Proof (By induction on \( n \) the number of logical symbols plus class symbols, in \( \varphi \) ). If \( n = 0 \), i.e., if \( \varphi \) has no logical symbols or class symbols, then \( \varphi \) must be of the form \( a \in b \) . By (1) of Definition 4.3, \( {\varphi }^{ * } \) is \( a \in b \) .\n\nAs our induct... | Yes |
Proposition 4.7. If \( A, B \), and \( C \) are terms then\n\n(1) \( A = A \) ,\n\n(2) \( A = B \rightarrow B = A \) ,\n\n(3) \( A = B \land B = C \rightarrow A = C \) . | The proof is similar to that of Proposition 3.2 and is left to the reader. | No |
Proposition 4.8. If \( A \) and \( B \) are terms and \( \varphi \) is a wff in the wider sense, then\n\n\[ A = B \rightarrow \left\lbrack {\varphi \left( A\right) \leftrightarrow \varphi \left( B\right) }\right\rbrack .\n\] | The proof is by induction. It is similar to the proof of Theorem 3.4 and is\n\nleft to the reader. | No |
Proposition 4.9. \( a = \{ x \mid x \in a\} \) . | Proof. \( \left( {\forall x}\right) \left\lbrack {x \in a \leftrightarrow x \in a}\right\rbrack \) . | Yes |
Proposition 4.14. \( {\mathcal{P}}_{i}\left( \mathrm{{Ru}}\right) \) . | Proof. From Proposition 4.12\n\n\[ \mathcal{M}\left( \mathrm{{Ru}}\right) \rightarrow \left\lbrack {\mathrm{{Ru}} \in \mathrm{{Ru}} \leftrightarrow \mathrm{{Ru}} \notin \mathrm{{Ru}}}\right\rbrack .\n\]\n\nTherefore \( \mathrm{{Ru}} \) is a proper class. | No |
Corollary 5.8. \( \mathcal{M}\left( {a \cup b}\right) \) . | Proof. Proposition 5.7, the Axiom of Unions, and the Axiom of Pairing. | No |
Proposition 5.11 (Zermelo's Schema of Separation). \[ \mathcal{M}\left( {a \cap A}\right) \] | Proof. Applying Axiom 5 to the wff \( b \in A \land b = c \) where \( b \) and \( c \) do not occur in \( A \), we have that \[ \left( {\forall x, y, z}\right) \left\lbrack {x \in A \land x = y}\right\rbrack \land \left\lbrack {x \in A \land x = z}\right\rbrack \rightarrow y = z. \] Therefore \[ \mathcal{M}\left( {\{ y... | Yes |
Proposition 5.15. \( a - a = 0 \) . | Proof. \( a - a = \{ x \in a \mid x \notin a\} \)\n\n\[ = \{ x \mid x \neq x\} \]\n\n\[ = 0\text{.} \] | Yes |
Corollary 5.16. \( \mathcal{M}\left( 0\right) \) . | Proof. Propositions 5.15 and 5.13. | No |
Proposition 5.18. \( \neg \left\lbrack {{a}_{1} \in {a}_{2} \in \cdots \in {a}_{n} \in {a}_{1}}\right\rbrack \) . | Proof. Let \( a = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{n}}\right\} \) . Suppose that \( {a}_{1} \in {a}_{2} \in \cdots \in {a}_{n} \in {a}_{1} \) . Then \( \left( {\forall x}\right) \left\lbrack {x \in a \rightarrow x \cap a \neq 0}\right\rbrack \) . This contradicts Regularity. | Yes |
Corollary 5.19. \( a \notin a \) . | Proof. Proposition 5.18 with \( n = 1 \) . | No |
Proposition 5.22. \( \left( {\forall x}\right) \left\lbrack {x \subseteq A \rightarrow x \in A}\right\rbrack \rightarrow A = V \) . | Proof. Assume that \( \left( {\forall x}\right) \left\lbrack {x \subseteq A \rightarrow x \in A}\right\rbrack \) . If \( B = V - A \) and if \( B \neq 0 \) then by (strong) Regularity there exists a set \( a \) such that\n\n\[ a \in B \land a \cap B = 0 \]\n\nthat is\n\n\[ \left( {\forall y}\right) \left\lbrack {y \in ... | No |
Proposition 6.2. \( \mathcal{M}\left( {a \times b}\right) \) . | Proof.\n\n\[ c \in a \times b \rightarrow \left( {\exists x, y}\right) \left\lbrack {x \in a \land y \in b \land c = \langle x, y\rangle }\right\rbrack .\n\]\n\n\[ \rightarrow \left( {\exists x, y}\right) \left\lbrack {\{ x\} \subseteq a \cup b\land \{ x, y\} \subseteq a \cup b \land c = \langle x, y\rangle }\right\rbr... | Yes |
Proposition 6.7. \( {\mathcal{U}}_{n}\left( A\right) \rightarrow \mathcal{M}\left( {{A}^{a}a}\right) \) | Proof. From Definition 6.4(2)\n\n\[ \n{\mathcal{U}}_{n}\left( A\right) \leftrightarrow \left( {\forall x}\right) \left( {\forall y}\right) \left( {\forall z}\right) \left\lbrack {\langle x, y\rangle \in A\land \langle x, z\rangle \in A \rightarrow y = z}\right\rbrack . \n\] \n\nThen from the Axiom Schema of Replacement... | Yes |
(1) \( \mathcal{M}\left( {a}^{-1}\right) \) . | Proof. (1) If \( A = \{ \langle \langle x, y\rangle ,\langle y, x\rangle \rangle \mid x, y \in V\} \), then \( A \) is single valued and hence by Proposition 6.7, \( {A}^{\alpha }a \) is a set. But \( {A}^{\alpha }a = {a}^{-1} \) ; therefore \( {a}^{-1} \) is a set. | No |
(1) \( \mathcal{M}\left( {A \times B}\right) \leftrightarrow \mathcal{M}\left( {B \times A}\right) \) . | Proof. (1) Suppose \( C = \{ \langle \langle x, y\rangle ,\langle y, x\rangle \rangle \mid x, y \in V\} \) . Then \( C \) is single valued, \( {C}^{u}\left( {A \times B}\right) = \left( {B \times A}\right) \) and \( {C}^{u}\left( {B \times A}\right) = \left( {A \times B}\right) \) . | Yes |
(1) \( \\langle b, c\\rangle \\in A \\land \\left( {\\exists !y}\\right) \\left\\lbrack {\\langle b, y\\rangle \\in A}\\right\\rbrack \\rightarrow {A}^{\\prime }b = c \) . | (1) From Definition 6.11, \( \\langle b, c\\rangle \\in A \\land \\left( {\\exists !y}\\right) \\left\\lbrack {\\langle b, y\\rangle \\in A}\\right\\rbrack \) implies\n\n\[ \n a \\in {A}^{c}b \\leftrightarrow a \\in c \n\]\n\ni.e., \( {A}^{c}b = c \) . | Yes |
Corollary 6.13. \( \mathcal{M}\left( {{A}^{4}b}\right) \) . | Proof. Proposition 6.12 assures us that \( \left( {\exists y}\right) \left\lbrack {{A}^{i}b = y}\right\rbrack \) . | No |
Proposition 6.17. \( {\mathcal{U}}_{\mathcal{H}}\left( A\right) \rightarrow \mathcal{M}\left( {A \upharpoonright a}\right) \) | Proof. If \( A \) is single valued, then certainly \( A \upharpoonright a \) is single valued. Since, by Definition \( {6.6}\left( 1\right), A \upharpoonright a \) is a relation, it follows that \( A \upharpoonright a \) is a function on \( \mathcal{D}\left( {A \upharpoonright a}\right) \). Furthermore, since \( \mathc... | Yes |
Proposition 6.20. \( {\left( {R}^{-1}\right) }^{a}\{ a\} = \{ x \mid {xRa}\} \) . | Proof. \( {\left( {R}^{-1}\right) }^{a}\{ a\} = \left\{ {x\mid \langle a, x\rangle \in {R}^{-1}}\right\} \n\n\[ \n= \{ x \mid \langle x, a\rangle R\} \n\] \n\n\[ \n= \{ x \mid {xRa}\} \text{.} \n\] | Yes |
Proposition 6.23. \( R\operatorname{Fr}A \land {a}_{1} \in A \land \cdots \land {a}_{n} \in A \rightarrow \neg \left\lbrack {{a}_{1}R{a}_{2} \land {a}_{2}R{a}_{3} \land \cdots \land {a}_{n}R{a}_{1}}\right\rbrack . | The proof is left to the reader. | No |
\[ R\text{Wfwe}A \land B \subseteq A \land B \neq 0 \rightarrow \left( {\exists x \in B}\right) \left\lbrack {B \cap {\left( {R}^{-1}\right) }^{\alpha }\{ x\} = 0}\right\rbrack \text{.} \] | Proof. If \( B \) is not empty, then \( B \) contains an element \( b \) . If \( B \cap {\left( {R}^{-1}\right) }^{a}\{ b\} = 0 \) , then \( b \) is the set we seek. But suppose that \( B \cap {\left( {R}^{-1}\right) }^{\alpha }\{ b\} \neq 0 \) . If in addition \( B \subseteq A \) and \( R \) is a well-founded well ord... | Yes |
Proposition 6.27. \( R \) Wfwe \( A \land B \subseteq A \land \left( {\forall x \in A}\right) \left\lbrack {A \cap {\left( {R}^{-1}\right) }^{\alpha }\{ x\} \subseteq B \rightarrow x \in B}\right\rbrack \rightarrow A = B \). | Proof. If \( A - B \neq 0 \) then by Proposition 6.26\n\n\[ \left( {\exists x \in A - B}\right) \left\lbrack {\left( {A - B}\right) \cap {\left( {R}^{-1}\right) }^{a}\{ x\} = 0}\right\rbrack \]\n\nthen\n\n\[ A \cap {\left( {R}^{-1}\right) }^{a}\{ x\} \leqq B. \]\n\nSince \( x \in A \) it follows from hypothesis that \(... | Yes |
(1) \( \left( {I \upharpoonright A}\right) {\operatorname{Isom}}_{R, R}\left( {A, A}\right) \) . | The proof is left to the reader. | No |
Proposition 6.31. If \( H{\operatorname{Isom}}_{{R}_{1},{R}_{2}}\left( {{A}_{1},{A}_{2}}\right) \land B \subseteq {A}_{1} \land a \in {A}_{1} \), then\n\n(1) \( B \cap {\left( {R}_{1}^{-1}\right) }^{a}\{ a\} = 0 \leftrightarrow {H}^{a}B \cap {\left( {R}_{2}^{-1}\right) }^{a}\left\{ {{H}^{c}a}\right\} = 0 \) ,\n\n(2) \(... | Proof\n\n(1) \( b \in B \cap {\left( {R}_{1}^{-1}\right) }^{a}\{ a\} \rightarrow b \in B \land b{R}_{1}a \)\n\n\[ \rightarrow {H}^{i}b \in {H}^{u}B \land {H}^{i}b{R}_{2}{H}^{i}a \]\n\n\[ \rightarrow {H}^{ \cdot }b \in {H}^{\cdot a}B \cap {\left( {R}_{2}^{-1}\right) }^{a}\{ {H}^{ \cdot }a\} . \]\n\n\[ b \in {H}^{ \lessd... | Yes |
Proposition 6.33. If\n\n\[ \begin{matrix} H : {A}_{1}\xrightarrow[\mathrm{{onto}}]{1 - 1}{A}_{2} \land {R}_{2} = \{ \langle {H}^{\prime }x,{H}^{\prime }y\rangle |x \in {A}_{1} \land y \in {A}_{1} \land \langle x, y\rangle \in {R}_{1}\} \end{matrix} \]\n\nthen\n\n\[ H{\operatorname{Isom}}_{{R}_{1},{R}_{2}}\left( {{A}_{1... | The proof is left to the reader. | No |
Proposition 7.2. \( \operatorname{Tr}\left( A\right) \land B \in A \rightarrow B \subset A \) . | Proof. If \( B \in A \), then \( B \) is a set. If in addition \( A \) is transitive, then by Definition 7.1, \( B \subseteq A \) . But \( B = A \) implies that \( A \in A \) which contradicts Corollary 5.19. Consequently \( B \subset A \) . | Yes |
Proposition 7.4. \( \operatorname{Ord}\left( A\right) \rightarrow E \) We \( A \) . | Proof. Obvious from Definitions 7.3, 6.24(2), 6.21, and Axiom 6′. | No |
Proposition 7.6. \( \operatorname{Ord}\left( A\right) \land a \in A \rightarrow \operatorname{Ord}\left( a\right) \) | Proof. Since \( A \) is transitive \( a \in A \) implies \( a \subseteq A \) . Consequently if \( b, c \in a \), then \( b, c \in A \) and hence, since \( A \) is an ordinal,\n\n\[ b \in c \vee b = c \vee c \in b \]\n\nthat is,\n\n\[ \left( {\forall x, y \in a}\right) \left\lbrack {x \in y \vee x = y \vee y \in x}\righ... | Yes |
Proposition 7.7. \( \operatorname{Ord}\left( A\right) \land \operatorname{Tr}\left( B\right) \rightarrow \left\lbrack {B \subset A \leftrightarrow B \in A}\right\rbrack \) . | Proof. Since \( A \) is an ordinal, \( A \) is transitive and so by Proposition 7.2\n\n\[ B \in A \rightarrow B \subset A\text{.}\]\n\nConversely if \( B \subset A \), then \( A - B \neq 0 \) . From Proposition 7.5, \( A - B \) has an E-minimal element \( b \), that is,\n\n\[ \left( {b \in A - B}\right) \land \left( {A... | Yes |
Corollary 7.8. \( \operatorname{Ord}\left( A\right) \land \operatorname{Ord}\left( B\right) \rightarrow \left\lbrack {B \subset A \leftrightarrow B \in A}\right\rbrack \) . | Proof. \( \operatorname{Ord}\left( B\right) \rightarrow \operatorname{Tr}\left( B\right) \) . | No |
Proposition 7.9. \( \operatorname{Ord}\left( A\right) \land \operatorname{Ord}\left( B\right) \rightarrow \operatorname{Ord}\left( {A \cap B}\right) \) | Proof. We first note that \( A \cap B \subseteq A \) . Furthermore, since \( A \) and \( B \) are transitive\n\n\[ a \in A \cap B \rightarrow a \in A \land a \in B \]\n\n\[ \rightarrow a \subset A \land a \subset B \]\n\n\[ \rightarrow a \subset A \cap B\text{.} \]\n\nTherefore \( A \cap B \) is a transitive subclass o... | Yes |
Proposition 7.10. \( \operatorname{Ord}\left( A\right) \land \operatorname{Ord}\left( B\right) \rightarrow \left\lbrack {A \in B \vee A = B \vee B \in A}\right\rbrack \) . | Proof. We first observe that \( A \cap B \subseteq A \land A \cap B \subseteq B \) . If \( A \cap B \subset A \land \) \( A \cap B \subset B \), then \( A \cap B \in A \land A \cap B \in B \) (Propositions 7.9 and 7.7) hence \( A \cap B \in A \cap B \) . But this contradicts Proposition 5.18. Therefore \( A \cap B = A ... | Yes |
Proposition 7.12. Ord(On). | Proof. From Proposition 7.6,\n\n\[ a \in \mathrm{{On}} \rightarrow a \leqq \mathrm{{On}} \]\n\ni.e., On is transitive. From Propositions 7.6 and 7.10\n\n\[ \left( {\forall x, y \in \mathrm{{On}}}\right) \left\lbrack {x \in y \vee x = y \vee y \in x}\right\rbrack . \]\n\nTherefore, by Definition 7.3, On is an ordinal. | Yes |
Proposition 7.13. \( {\mathcal{P}}_{i}\left( \mathrm{{On}}\right) \) . | Proof. Were On a set it would follow that On \( \in \) On. But this contradicts Proposition 5.18. | Yes |
Corollary 7.14. \( \operatorname{Ord}\left( A\right) \rightarrow A \in \) On \( \vee A = \) On. | Proof. From Propositions 7.10 and 7.12, \( A \in \mathrm{{On}} \vee A = \mathrm{{On}} \vee \mathrm{{On}} \in A \) . But since On is a proper class we cannot have On \( \in A \) . | Yes |
Corollary 7.15. \( \operatorname{Ord}\left( A\right) \rightarrow A \subseteq \) On. | Proof. Corollary 7.14, Proposition 7.12 and Corollary 7.8. | No |
Theorem 7.17 (The Principle of Transfinite Induction). If (1) \( A \subseteq \) On and (2) \( \left( {\forall \alpha }\right) \left\lbrack {\alpha \subseteqq A \rightarrow \alpha \in A}\right\rbrack \), then \( A = \) On. | Proof. To prove that \( A = \) On, given that \( A \subseteq \) On, it is sufficient to prove that \( \mathrm{{On}} \subseteqq A \) . Suppose that \( \mathrm{{On}} \) is not a subclass of \( A \) . Then \( \mathrm{{On}} - A \neq 0 \) and hence by Propositions 7.12 and 7.5. \( \left( {\exists \alpha \in \mathrm{{On}} - ... | Yes |
Proposition 7.19. \( A \subseteq \mathrm{{On}} \rightarrow \operatorname{Ord}\left( {\cup \left( A\right) }\right) \). | Proof. If \( A \subseteq \) On, then since by Proposition 7.6, elements of ordinals are ordinals, it follows that\n\n\[ \cup \left( A\right) \subseteq \text{On.} \]\n\nFurthermore if \( a \in \cup \left( A\right) \) then, from the definition of union, there exists a set \( b \) such that\n\n\[ a \in b \land b \in A\tex... | Yes |
Proposition 7.20. \( A \subseteq \) On \( \land \alpha \in A \rightarrow \alpha \leqq \cup \left( A\right) \) . | Proof. By Proposition 7.19 \( \cup \left( A\right) \) is an ordinal. Furthermore\n\n\[ \alpha \in A \rightarrow \alpha \subseteq \cup \left( A\right) \]\n\n\[ \rightarrow \alpha \leqq \cup \left( A\right) \text{.} \] | Yes |
Proposition 7.21. \( A \subseteq \) On \( \land \left( {\forall \beta \in \mathrm{A}}\right) \left\lbrack {\beta \leqq \alpha }\right\rbrack \rightarrow \cup \left( A\right) \leqq \alpha \) . | Proof. If \( \beta \in \cup \left( A\right) \), then \( \left( {\exists \gamma }\right) \left\lbrack {\beta < \gamma \land \gamma \in A}\right\rbrack \) . Therefore \( \beta < \gamma \land \gamma \leqq \alpha \) . Hence \( \beta < \alpha \), that is, \( \cup \left( A\right) \subseteq \alpha \) . | Yes |
Proposition 7.24. \( \left( {\forall \alpha }\right) \left\lbrack {\alpha + 1 \in \text{On}}\right\rbrack \) . | Proof. Since \( \left( {\forall \alpha }\right) \left\lbrack {\alpha \in \mathrm{{On}} \land \alpha \subseteq \mathrm{{On}}}\right\rbrack \) it follows that \( \alpha + 1 \subseteq \mathrm{{On}} \) . Furthermore if \( a \in \alpha + 1 \), then \( a \in \alpha \) or \( a = \alpha \) and hence \( a \subseteqq \alpha \) .... | Yes |
Proposition 7.26. \( a \subset \) On \( \rightarrow \left( {\forall \alpha \in a}\right) \left\lbrack {\alpha < \cup \left( a\right) + 1}\right\rbrack \) . | Proof. From Propositions 7.19 and 7.20, \( \cup \left( a\right) \) is an ordinal and \( \alpha \leqq \cup \left( a\right) \) . From Proposition 7.23 we then conclude that \( \alpha < \cup \left( a\right) + 1 \) . | Yes |
Proposition 7.30 (Peano's Postulates).\n\n(1) \( 0 \in \omega \) .\n\n(2) \( \left( {\forall i}\right) \left\lbrack {i + 1 \in \omega }\right\rbrack \) .\n\n(3) \( \left( {\forall i}\right) \left\lbrack {i + 1 \neq 0}\right\rbrack \) .\n\n(4) \( \left( {\forall i}\right) \left( {\forall j}\right) \left\lbrack {i + 1 = ... | Proof. (1) From Definition 7.27 we have that \( 0 \in {K}_{\mathrm{I}} \) . Therefore \( 0 \cup \{ 0\} \subseteq {K}_{\mathrm{I}} \) hence \( 0 \in \omega \) .\n\n(2) Since \( i + 1 \in {K}_{\mathrm{I}} \) and since \( i \in \omega \rightarrow i + 1 \subseteq {K}_{\mathrm{I}} \) it follows that \( \left( {i + 1}\right)... | Yes |
Corollary 7.31 (The Principle of Finite Induction). If \( A \subseteq \omega \land 0 \in A \land \) \( \left( {\forall i}\right) \left\lbrack {i \in A \rightarrow i + 1 \in A}\right\rbrack \) then \( A = \omega \) . | Proof. Obvious from Proposition 7.30. | No |
Proposition 7.32. \( \operatorname{Ord}\left( \omega \right) \). | Proof. Since \( \omega \subseteq {K}_{\mathrm{I}} \) and \( {K}_{\mathrm{I}} \subseteq \) On it follows that \( \omega \subseteq \) On. Furthermore if \( a \in b \) and \( b \in \omega \), then \( b \) is an ordinal and \( b + 1 \subseteq {K}_{\mathrm{I}} \) . Then \( a \subseteq b, b \subseteq b + 1 \) and \( b + 1 \s... | Yes |
Proposition 7.34. \( F{\mathcal{F}}_{\mathcal{H}}\omega \rightarrow \left( {\exists n}\right) \left\lbrack {{F}^{c}\left( {n + 1}\right) \notin {F}^{c}n}\right\rbrack \) . | Proof. From the Axiom of Infinity, \( F \) is a function whose domain is a set. Therefore the range of \( F,\mathcal{W}\left( F\right) \), is a set. Then from the Axiom of Regularity there exists a set \( a \) in \( \mathcal{W}\left( F\right) \) such that\n\n\[ \mathcal{W}\left( F\right) \cap a = 0. \]\n\nBut since \( ... | Yes |
Proposition 7.39. \( \operatorname{Ord}\left( A\right) \land \operatorname{Ord}\left( B\right) \land F{\operatorname{Isom}}_{E, E}\left( {A, B}\right) \rightarrow A = B \) . | Proof. It is sufficient to prove that \( F \) is the identity function I restricted to \( A \) . This we prove by transfinite induction. If \( \beta \in A \land \left( {\forall \alpha < \beta }\right) \left\lbrack {{F}^{\epsilon }\alpha = \alpha }\right\rbrack \) then \( {F}^{\alpha }\beta = \beta \) . Furthermore sinc... | Yes |
Proposition 7.40. \[ \operatorname{Ord}\left( A\right) \land \operatorname{Ord}\left( B\right) \land {F}_{1}{\operatorname{Isom}}_{E, R}\left( {A, C}\right) \land {F}_{2}{\operatorname{Isom}}_{E, R}\left( {B, C}\right) \rightarrow A = B. \] | Proof. Proposition 7.39 and the fact that \( {F}_{2} \circ {F}_{1}^{-1}{\operatorname{Isom}}_{E, E}\left( {A, B}\right) \) . | No |
Corollary 7.42. \( \left( {\exists !f}\right) \left\lbrack {f{\mathcal{F}}_{n}\alpha \land \left( {\forall \beta < \alpha }\right) \left\lbrack {{f}^{c}\beta = {G}^{c}\left( {f \upharpoonright \beta }\right) }\right\rbrack }\right\rbrack \) . | The proof is left to the reader. | No |
Proposition 7.44. If \( G = \{ \langle x, y\rangle \mid \left\lbrack {x = 0 \land y = a}\right\rbrack \vee \lbrack x \neq 0 \land \sup \left( {\mathcal{D}\left( x\right) }\right) \neq \mathcal{D}\left( x\right) \land y = {H}^{\iota }{x}^{\iota }\sup \left( {\mathcal{D}\left( x\right) }\right) \rbrack \vee \lbrack x \ne... | Proof. (1) \( {F}^{\prime }0 = {G}^{\prime }\left( {F \upharpoonright 0}\right) = {G}^{\prime }0 = a \) .\n\n(2) \( {F}^{c}\left( {\beta + 1}\right) = {G}^{c}\left( {F \upharpoonright \left( {\beta + 1}\right) }\right) \) . Since \( \mathcal{D}\left( {F \upharpoonright \left( {\beta + 1}\right) }\right) = \beta + 1 \) ... | Yes |
Corollary 7.45 (Principle of Finite Recursion). \\\\[ \\left( {\\exists !f}\\right) \\left\\lbrack {f{\\mathcal{F}}_{n}\\omega \\land {f}^{i}0 = a \\land \\left( {\\forall k}\\right) \\left\\lbrack {{f}^{i}\\left( {k + 1}\\right) = {H}^{i}{f}^{i}k}\\right\\rbrack }\\right\\rbrack . \\\\] | Proof. If in Proposition 7.44 we restrict \\( F \\) to \\( \\omega \\) then \\( F \\upharpoonright \\omega \\) is a function on \\( \\omega \\) and hence is a set. Therefore \\( \\left( {\\exists f}\\right) \\left\\lbrack {f = F \\upharpoonright \\omega }\\right\\rbrack \\) . Then \\( f{\\mathcal{F}}_{\\mathcal{H}}\\om... | Yes |
Proposition 7.48. If \( F{\mathcal{F}}_{\mathcal{R}} \) On \( \land \left( {\forall \alpha }\right) \left\lbrack {{F}^{\iota }\alpha \in \left\lbrack {A - {F}^{\iota }\alpha }\right\rbrack }\right\rbrack \), then (1) \( \mathcal{W}\left( \mathrm{F}\right) \subseteq A \), (2) \( {\mathcal{{Un}}}_{2}\left( F\right) \), (... | Proof. (1) Since \( \left( {\forall \alpha }\right) \left\lbrack {{F}^{\alpha }\alpha \in \left( {A - {F}^{\alpha }\alpha }\right) }\right\rbrack \) it follows that \( \left( {\forall \alpha }\right) \left\lbrack {{F}^{\alpha }\alpha \in A}\right\rbrack \) and hence \( \mathcal{W}\left( F\right) \subseteq A \). (2) If ... | Yes |
Proposition 7.49. If \( F{\mathcal{F}}_{\mathcal{H}} \) On \( \land \left( {\forall \alpha }\right) \left\lbrack {A - {F}^{\alpha }\alpha \neq 0 \rightarrow {F}^{\alpha }\alpha \in A - {F}^{\alpha }\alpha }\right\rbrack \land \) \( \mathcal{M}\left( A\right) \) then\n\n\[ \left( {\exists \alpha }\right) \left\lbrack {\... | Proof. If \( \left( {\forall \alpha }\right) \left\lbrack {A - {F}^{\alpha }\alpha \neq 0}\right\rbrack \) then \( \left( {\forall \alpha }\right) \left\lbrack {{F}^{\epsilon }\alpha \in A - {F}^{\alpha }\alpha }\right\rbrack \) and by Proposition \( {7.48A} \) is a proper class. Since by hypothesis \( A \) is a set we... | Yes |
Proposition 7.50. If \( R \) is well founded on \( A \) and well orders \( A \), if\n\n(1) \( G = \{ \langle x, y\rangle \mid y \in \left\lbrack {A - \mathcal{W}\left( x\right) }\right\rbrack \land \left\lbrack {\left( {A - \mathcal{W}\left( x\right) }\right) \cap {\left( {R}^{-1}\right) }^{\alpha }\{ y\} = 0}\right\rb... | Proof. As the first step in our proof we will show that \( G \) is single valued. For this purpose suppose that \( \left\langle {x,{y}_{1}}\right\rangle ,\left\langle {x,{y}_{2}}\right\rangle \in G \) . Then\n\n\[ {y}_{1} \in \left\lbrack {A - \mathcal{W}\left( x\right) }\right\rbrack \land {y}_{2} \in \left\lbrack {A ... | Yes |
Corollary 7.52. If \( A \) is a proper class of ordinals, if\n\n(1) \( G = \left\{ {\langle x, y\rangle \mid y \in \left\lbrack {A - \mathcal{W}\left( x\right) }\right\rbrack \land \left\lbrack {\left( {A - \mathcal{W}\left( x\right) }\right) \cap {\left( {E}^{-1}\right) }^{\alpha }\{ y\} }\right\rbrack = 0}\right\} \)... | Proof. \( E \) We \( A \land E \) Wfr \( A \) . | No |
Proposition 7.53. \( R \) We \( A \land \mathcal{M}\left( A\right) \rightarrow \left( {\exists !\alpha }\right) \left( {\exists !f}\right) \left\lbrack {f{\operatorname{Isom}}_{E, R}\left( {\alpha, A}\right) }\right\rbrack \) . | Proof. If \( G = \{ \langle x, y\rangle \mid y \in \left\lbrack {A - \mathcal{W}\left( x\right) }\right\rbrack \land \left\lbrack {\left( {A - \mathcal{W}\left( x\right) }\right) \cap {\left( {R}^{-1}\right) }^{\alpha }\{ y\} }\right\rbrack = 0\} \) , \( F\mathcal{F}n \) On, and if \( \left( {\forall \alpha }\right) \l... | No |
Corollary 7.54. \( A \subseteq \) On \( \land \mathcal{M}\left( A\right) \rightarrow \left( {\exists !\alpha }\right) \left( {\exists !f}\right) \left\lbrack {f{\operatorname{Isom}}_{E, E}\left( {\alpha, A}\right) }\right\rbrack \) . | Proof. \( A \subseteq \) On \( \rightarrow E \) We \( A \) . | No |
(1) Le We \( {\mathrm{{On}}}^{2} \land \)\n\n\[ \left\lbrack {B \subseteq {\mathrm{{On}}}^{2} \land B \neq 0 \rightarrow \left( {\exists x \in B}\right) \left\lbrack {B \cap {\left( {\mathrm{{Le}}}^{-1}\right) }^{u}\{ x\} = 0}\right\rbrack }\right\rbrack . | Proof. (1) The proof is left to the reader. | No |
(1) \( {R}_{0}{\mathrm{{WeOn}}}^{2} \land \left\lbrack {B \subseteqq {\mathrm{{On}}}^{2} \land B \neq 0 \rightarrow \left( {\exists x \in B}\right) \left\lbrack {B \cap {\left( {R}_{0}^{-1}\right) }^{\alpha }\{ x\} = 0}\right\rbrack }\right\rbrack \) . | Proof. (1) The proof is left to the reader. | No |
Proposition 8.3. \( 0 + \alpha = \alpha + 0 = \alpha \) . | Proof. By definition \( \alpha + 0 = \alpha \) . If \( 0 + \alpha = \alpha \), then \( 0 + \left( {\alpha + 1}\right) = \left( {0 + \alpha }\right) \) \( + 1 = \alpha + 1 \) . If \( \alpha \in {K}_{\mathrm{{II}}} \) and \( \left( {\forall \beta }\right) \left\lbrack {\beta < \alpha \rightarrow 0 + \beta = \beta }\right... | Yes |
Proposition 8.4. \( \alpha < \beta \rightarrow \gamma + \alpha < \gamma + \beta \) . | Proof (By transfinite induction on \( \beta \) ). For \( \beta = \alpha + 1 \) we have \( \gamma + \alpha < \) \( \left( {\gamma + \alpha }\right) + 1 = \gamma + \left( {\alpha + 1}\right) \) . If \( \alpha < \beta \rightarrow \gamma + \alpha < \gamma + \beta \) and if \( \alpha < \beta + 1 \) then \( \alpha < \beta \v... | Yes |
Corollary 8.5. \( \gamma + \alpha = \gamma + \beta \leftrightarrow \alpha = \beta \) . | Proof. If \( \alpha = \beta \), then by Theorem 3.4\n\n\[ x \in \gamma + \alpha \leftrightarrow x \in \gamma + \beta . \]\n\nConsequently \( \gamma + \alpha = \gamma + \beta \) . That is\n\n(1) \( \alpha = \beta \rightarrow \gamma + \alpha = \gamma + \beta \) .\n\nFrom Proposition 8.4\n\n(2) \( \alpha < \beta \rightarr... | Yes |
Proposition 8.6. \( \left( {\forall \alpha \in A}\right) \left( {\exists \beta \in B}\right) \left\lbrack {\alpha \leqq \beta }\right\rbrack \rightarrow \sup \left( A\right) \leqq \sup \left( B\right) \) . | Proof. If \( \gamma \in \sup \left( A\right) \) then \( \left( {\exists \alpha }\right) \left\lbrack {\gamma \in \alpha \land \alpha \in A}\right\rbrack \) . But \( \alpha \in A \rightarrow \left( {\exists \beta }\right) \lbrack \beta \in B \land \) \( \alpha \leqq \beta \rbrack \) i.e., \( \left( {\exists \beta }\righ... | Yes |
Proposition 8.7. \( \alpha \leqq \beta \rightarrow \alpha + \gamma \leqq \beta + \gamma \) . | Proof (By transfinite induction on \( \gamma \) ). If \( \alpha \leqq \beta \), then \( \alpha + 0 \leqq \beta + 0 \) . If \( \alpha + \gamma \) \( \leqq \beta + \gamma \) then \( \alpha + \left( {\gamma + 1}\right) \leqq \beta + \left( {\gamma + 1}\right) \) . If \( \gamma \in {K}_{\text{II }} \) and \( \left( {\foral... | Yes |
Proposition 8.8. \( \alpha \leqq \beta \rightarrow \left( {\exists !\gamma }\right) \left\lbrack {\alpha + \gamma = \beta }\right\rbrack \) . | Proof. Since \( \alpha \geqq 0 \) it follows from Propositions 8.7 and 8.3 that \( \alpha + \beta \) \( \geqq 0 + \beta = \beta \) . Thus there exists a smallest ordinal \( \gamma \) such that \( \alpha + \gamma \geqq \beta \) . If \( \gamma \in {K}_{1} \) then \( \gamma = 0 \vee \left( {\exists \delta }\right) \left\l... | Yes |
Proposition 8.10. \( n < \omega \land \omega \leqq \alpha \rightarrow n + \alpha = \alpha \) . | Proof (By transfinite induction on \( \alpha \) ). If \( \alpha = \omega \) we have\n\n\[ n + \omega = \mathop{\bigcup }\limits_{{\gamma < \omega }}\left( {n + \gamma }\right) \]\n\nBy Proposition 8.9, \( \gamma < \omega \rightarrow n + \gamma < \omega \) . Hence\n\n\[ \mathop{\bigcup }\limits_{{\gamma < \omega }}\left... | Yes |
Proposition 8.11. \( \beta \in {K}_{\mathrm{{II}}} \rightarrow \alpha + \beta \in {K}_{\mathrm{{II}}} \) . | Proof. If \( \beta \in {K}_{\text{II }} \), then \( \beta \neq 0 \) . Therefore \( \alpha + \beta \neq 0 \) . Thus \( \alpha + \beta \in {K}_{\text{II }} \) or \( \left( {\exists \delta }\right) \left\lbrack {\alpha + \beta = \delta + 1}\right\rbrack \) . But if \( \beta \in {K}_{\text{II }} \) then\n\n\[ \alpha + \bet... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.