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Proposition 1. Let \( f = g/h \in {\mathbf{C}}_{p}\left( x\right) \) be a rational function and let \( {S}_{r} \) be the sphere \( \{ x : \left| x\right| = r\} \) of radius \( r > 0 \) . Then:\n\n(a) If \( f \) has no pole on \( {S}_{r} \), then \( \left| {f\left( x\right) }\right| \leq {M}_{r}f\;\left( {x \in {S}_{r}}... | Proof. (a) If a critical sphere \( {S}_{r}\left( {r > 0}\right) \) contains no pole of \( f \), its denominator does not vanish on this sphere and \( r \) is a regular value for the denominator: \( \left| {h\left( x\right) }\right| = {M}_{r}h \) is constant on \( {S}_{r} \),\n\n\[ \left| {f\left( x\right) }\right| = \f... | Yes |
Proposition 2. Let \( f = g/h \in {\mathbf{C}}_{p}\left( x\right) ,{S}_{r} \) as before and consider an open ball \( D = {B}_{ < r}\left( a\right) \) of maximal radius in the sphere \( {S}_{r} \) (hence \( \left| a\right| = r \) ). If \( f \) has no pole in \( D \), then\n\n\[ \n{M}_{r}f = \mathop{\sup }\limits_{{x \in... | Proof. For \( s > r = \left| a\right| \), the spheres \( {S}_{s} \) and \( {S}_{s}\left( a\right) = \{ x : \left| {x - a}\right| = s\} \) coincide. Hence \( {M}_{s}f = {M}_{s, a}f \) (growth modulus with respect to the center \( a \) ): This is\n\nobvious for regular values of \( s \) and by continuity also for all val... | Yes |
Proposition 3. Let \( f = g/h \in {\mathbf{C}}_{p}\left( x\right) \) be a rational function, \( {S}_{r} \) as before. Then:\n\n(a) If \( f \) has no pole on \( {S}_{r} \), then \( {M}_{r}f = \mathop{\sup }\limits_{{\left| x\right| = r}}\left| {f\left( x\right) }\right| \) .\n\n(b) If \( f \) has no zero on \( {S}_{r} \... | Proof. Observe that if \( f \) has no pole and no zero in \( {S}_{r} \), then \( r \) is regular and \( \left| {f\left( x\right) }\right| = {M}_{r}f \) is constant on \( {S}_{r} \) . Now (a) follows from Propositions 1 and 2. For (b), replace \( f \) by \( 1/f \) and apply the previous result.\n\n(c) Choose a pole \( \... | Yes |
Proposition 1. Let \( 0 \neq f = g/h \in {\mathbf{C}}_{p}\left( x\right) \) . Assume \( \deg g < \deg h \) and that \( f \) has all its poles in a ball \( B = {B}_{ < \sigma } \) for some \( \sigma > 0 \) . Then for any subset \( D \) disjoint from \( B,\parallel f{\parallel }_{D} \leq \parallel f{\parallel }_{{B}^{c}}... | Proof. Since \( f \) has all its poles \( {\alpha }_{i} \) in \( B \), we have \( {\sigma }_{p} = \mathop{\max }\limits_{i}\left| {\alpha }_{i}\right| < \sigma \) and\n\n\[ \left| {f\left( x\right) }\right| \leq {M}_{\left| x\right| }f\;\left( {\left| x\right| > {\sigma }_{p}}\right) . \]\n\nOn the other hand, since \(... | Yes |
Proposition 2. Let \( f \in {\mathbf{C}}_{p}\left( x\right) \) be a rational function and let \( {f}_{B} \) be the sum of the principal parts of \( f \) corresponding to its poles in \( B = {B}_{ < \sigma } \) . If \( D \) is a maximal open ball \( {B}_{ < \sigma }\left( a\right) \) in the sphere \( \left| x\right| = \... | Proof. We may assume \( {f}_{B} \neq 0 \) and let us introduce \( {f}_{0} \mathrel{\text{:=}} f - {f}_{B} \in {\mathbf{C}}_{p}\left( x\right) \) , which is regular in \( B \) (but may have poles in the sphere \( \left| x\right| = \sigma \) ). Hence\n\n\[r \mapsto {M}_{r}{f}_{0}\;\text{is increasing (may be constant) fo... | Yes |
Proposition 1. Any \( f \in R{\left( D\right) }^{ \times } \) can be uniquely factorized as\n\n\[ f = {f}_{0} \cdot \mathop{\prod }\limits_{{1 \leq i \leq \ell }}{f}_{i}\;\text{ (Motzkin factorization),} \]\n\nwhere \( {f}_{0} \in R{\left( {B}_{ \leq r}\right) }^{ \times } \) and for \( 1 \leq i \leq \ell \)\n\n\[ {f}_... | Proof. The only possibility consists in collecting the zeros and poles of \( f \) in \( {B}_{i} \) and defining \( {f}_{i} \) as the product of the corresponding factors \( {\left( x - a\right) }^{{\mu }_{a}}(a \in {B}_{i} \) , \( {\mu }_{a} \in \mathbf{Z} \) positive for zeros and negative for poles of \( f \) ). With... | Yes |
Proposition 2. Assume \( \parallel f - 1{\parallel }_{D} < 1 \) . Then \( f \) has as many zeros as poles in each ball \( {B}_{i} \subset {D}^{c} \) . | Proof. The assumption implies \( \left| {f\left( x\right) - 1}\right| < 1 \) for all \( x \in D \), hence \( \left| {f\left( x\right) }\right| = 1 \) is constant in \( D \) . Consider a ball \( {B}_{i} \) and consider the growth modulus centered at \( {b}_{i} \in {B}_{i} \) . Without loss of generality, we may assume \... | Yes |
Proposition 3. If \( \parallel f - 1{\parallel }_{D} < 1 \), then the principal part \( {P}_{i}f \) of \( f \) relative to the ball \( {B}_{i} \) and the Motzkin factor \( {f}_{i} = {h}_{i} \) defined in Proposition 1 are related by | Proof. Let \( S \) denote the set of zeros and poles of \( f \), and let \( f = {f}_{0} \cdot \mathop{\prod }\limits_{{1 \leq i \leq \ell }}{f}_{i} \) be the Motzkin factorization of \( f \) . By Proposition 2, we have\n\n\[ {f}_{i} = \mathop{\prod }\limits_{{a \in S \cap {B}_{i}}}{\left( \frac{x - a}{x - {b}_{i}}\righ... | Yes |
Proposition 1. When \( D \subset {\mathbf{C}}_{p} \) is a closed and bounded subset, each \( f \in R\left( D\right) \) is bounded on \( D \), and \( H\left( D\right) \) is the closure of \( R\left( D\right) \) in the Banach algebra \( {C}_{b}\left( D\right) \) for the sup norm. | Proof. Recall (3.2): The functions\n\n\[ \n{x}^{n},\;\frac{1}{{\left( x - a\right) }^{m}}\;\left( {n \geq 0, a \notin D, m \geq 1}\right) \n\] \n\nconstitute a basis of the vector space \( R\left( D\right) \) . When \( D \) is bounded, the functions \( {x}^{n} \) \( \left( {n \geq 0}\right) \) are bounded on \( D \) . ... | Yes |
Proposition 2. Let \( D \subset {\mathbf{C}}_{p} \) be a closed, bounded, and infraconnected set. Assume \( 0 \in {B}_{D} \) and let \( 0 \leq d\left( {0, D}\right) \leq r \leq \delta \left( D\right) \). Then\n\n\[ \n{M}_{r}f \leq \parallel f{\parallel }_{D}\;\left( {f \in R\left( D\right) }\right) .\n\]\n\nIf the sphe... | Proof. Let \( \sigma = d\left( {0, D}\right) ,\delta = \delta \left( D\right) \), so that \( \{ \left| x\right| : x \in D\} \) is dense in the interval \( \left\lbrack {\sigma ,\delta }\right\rbrack \) . If \( f \in R\left( D\right) \), let us show that there exists a sequence \( {x}_{n} \in D \) with\n\n\[ \n\left| {f... | Yes |
Proposition 1. For an odd prime \( p \), we have \( {S}_{p}^{2} = \pm p \) . | Proof. The square of the sum \( {S}_{p} \) is\n\n\[ \n{S}_{p}^{2} = \mathop{\sum }\limits_{{0 < v,\mu < p}}\left( \frac{v}{p}\right) \left( \frac{\mu }{p}\right) {\zeta }^{v + \mu } = \mathop{\sum }\limits_{{0 < v,\mu < p}}\left( \frac{v\mu }{p}\right) {\zeta }^{v + \mu }. \n\]\n\nFor fixed \( \mu \neq 0,{v\mu } \) goe... | Yes |
Corollary 1. For a prime \( p \geq 3 \), the complex absolute value of \( {S}_{p} \) is | \[ {\left| {S}_{p}\right| }_{\mathrm{C}} = \sqrt{p} \] | Yes |
Corollary 2. For a prime \( p \geq 3 \), the quadratic extension \( \mathbf{Q}\left( \sqrt{p}\right) \) is contained in the cyclotomic field \( \mathbf{Q}\left( {\zeta ,\sqrt{-1}}\right) \) . | Observe that if \( p = 2 \), we have \( {\left( 1 + \sqrt{-1}\right) }^{2} = 2\sqrt{-1} \), so that \( \sqrt{2} \in \mathbf{Q}\left( \sqrt[4]{-1}\right) \) and the quadratic extension \( \mathbf{Q}\left( \sqrt{2}\right) \) is also contained in a cyclotomic one. | No |
Proposition 2. Let \( G \) be a group and \( K \) a field. Any set of distinct homomorphisms \( G \rightarrow {K}^{ \times } \) is linearly independent in the \( K \) -vector space of functions \( G \rightarrow K \) . | Proof. Since linear independence of any family is a property of its finite subsets, it is enough to prove that all finite sets of distinct homomorphisms are linearly independent. We argue by induction on the number of homomorphisms \( {\psi }_{i} \) . Since homomorphisms are nonzero maps, the independence assertion is ... | Yes |
Proposition 1. Let \( f = \mathop{\sum }\limits_{{m \geq 1}}{f}_{m}{x}^{m} \in B\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \), so that \( f\left( 0\right) = 0 \) . Then\n\n\[ \n{H}_{p}f \in A\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \Rightarrow m{f}_{m} \in A\;\left( {m \geq 1}\right) .\n\] | Proof. The coefficients of \( {H}_{p}f = \sum {h}_{m}{x}^{m} \in A\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) are given by\n\n\[ \n{h}_{m} = {f}_{m} - \frac{1}{p}\mathop{\sum }\limits_{I}{\sigma }^{i}{f}_{m/{p}^{i}} \in A \n\]\n\nwith the convention \( {f}_{m/{p}^{i}} = 0 \) if \( i > {\operatorname{ord}}... | Yes |
Proposition 2. Let \( g = \mathop{\sum }\limits_{{m \geq 1}}{g}_{m}{x}^{m} \) and \( h = \mathop{\sum }\limits_{{m \geq 1}}{h}_{m}{x}^{m} \) be two formal power series with zero constant term. Then\n\n\[ \n{H}_{p}\left( {g \circ h}\right) = {H}_{p}\left( g\right) \left( h\right) + \frac{1}{p}\mathop{\sum }\limits_{I}\m... | Proof. By definition,\n\n\[ \n{H}_{p}\left( {g \circ h}\right) = g \circ h - \frac{1}{p}\mathop{\sum }\limits_{I}{\sigma }_{ * }^{i}\left( {g \circ h}\right) \left( {x}^{{p}^{i}}\right) ,\n\]\n\nwhile\n\n\[ \n{H}_{p}\left( g\right) \left( h\right) = g\left( h\right) - \frac{1}{p}\mathop{\sum }\limits_{I}{\sigma }_{ * }... | Yes |
Proposition 4. Let \( A \) be a ring, \( I \) an ideal of \( A \) containing a prime \( p \), and \( x \) and \( y \) two elements of \( A \) satisfying \( x \equiv y\;\left( {\;\operatorname{mod}\;{I}^{r}}\right) \) for some integer \( r \geq 1 \) . Then \[ {p}^{v} \mid m\; \Rightarrow \;{x}^{m} \equiv {y}^{m}\;\left(... | Proof. (1) Let us write \( x = y + z \) with \( z \in {I}^{r} \) . Hence \[ {x}^{p} = {\left( y + z\right) }^{p} = {y}^{p} + {zp}\left( \cdots \right) + {z}^{p} \] with \[ {zp}\left( \cdots \right) \in {zI} \subset {I}^{r} \cdot I = {I}^{r + 1} \] and \[ {z}^{p} \in {I}^{pr} \subset {I}^{2r} \subset {I}^{r + 1}. \] Thi... | Yes |
Corollary 1. The Lucas sequence\n\n\\[ \n{\\ell }_{0} = 2,\\;{\\ell }_{1} = 1,\\;{\\ell }_{n + 1} = {\\ell }_{n} + {\\ell }_{n - 1}\\;\\left( {n \\geq 1}\\right) ,\n\\]\n\nis a p-Honda sequence for any prime \\( p \\) . | Proof. Let \\( M = \\left( \\begin{array}{ll} 1 & 1 \\\\ 1 & 0 \\end{array}\\right) \\in {M}_{2}\\left( \\mathbf{Z}\\right) \\) . The characteristic polynomial of \\( M \\) is \\( {x}^{2} - x - 1 \\), hence \\( {M}^{2} - M - I = 0 \\) (Hamilton-Cayley). We deduce\n\n\\[ \n{M}^{n + 2} = {M}^{n + 1} + {M}^{n}\\;\\left( {... | Yes |
Corollary 2. The Perrin sequence\n\n\\[ \n{a}_{0} = 3,\;{a}_{1} = 0,\;{a}_{2} = 2,\;{a}_{n + 2} = {a}_{n} + {a}_{n - 1}\\;\\left( {n \geq 1}\\right) ,\n\\]\n\nis a p-Honda sequence for any prime \\( p \\) . | Proof. Let \\( M = \\left( \\begin{array}{lll} 0 & 1 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0 \\end{array}\\right) \\in {M}_{3}\\left( \\mathbf{Z}\\right) \\) . The characteristic polynomial of \\( M \\) is \\( - {x}^{3} + x + 1 \\), hence \\( {M}^{3} - M - I = 0 \\) (Hamilton-Cayley). We deduce\n\n\\[ \n{M}^{n + 3} = {M}^{n ... | Yes |
Theorem 1.1 (Cantor). For any sequence \( \left\{ {a}_{n}\right\} \) of real numbers and for any interval I there exists a point \( p \) in I such that \( p \neq {a}_{n} \) for every \( n \) . | One proof runs as follows. Let \( {I}_{1} \) be a closed subinterval of \( I \) such that \( {a}_{1} \notin {I}_{1} \) . Let \( {I}_{2} \) be a closed subinterval of \( {I}_{1} \) such that \( {a}_{2} \notin {I}_{2} \) . Proceeding inductively, let \( {I}_{n} \) be a closed subinterval of \( {I}_{n - 1} \) such that \(... | Yes |
Theorem 1.2. Any subset of a nowhere dense set is nowhere dense. The union of two (or any finite number) of nowhere dense sets is nowhere dense. The closure of a nowhere dense set is nowhere dense. | Proof. The first statement is obvious. To prove the second, note that if \( {A}_{1} \) and \( {A}_{2} \) are nowhere dense, then for each interval \( I \) there is an interval \( {I}_{1} \subset I - {A}_{1} \) and an interval \( {I}_{2} \subset {I}_{1} - {A}_{2} \) . Hence \( {I}_{2} \subset I - \left( {{A}_{1} \cup {A... | No |
Theorem 1.3 (Baire). The complement of any set of first category on the line is dense. No interval in \( R \) is of first category. The intersection of any sequence of dense open sets is dense. | Proof. The three statements are essentially equivalent. To prove the first, let \( A = \bigcup {A}_{n} \) be a representation of \( A \) as a countable union of nowhere dense sets. For any interval \( I \), let \( {I}_{1} \) be a closed subinterval of \( I - {A}_{1} \) . Let \( {I}_{2} \) be a closed subinterval of \( ... | Yes |
Theorem 1.4. Any subset of a set of first category is of first category. The union of any countable family of first category sets is of first category. | It is obvious that the class of first category sets has these closure properties. However, the closure of a set of first category is not in general of first category. In fact, the closure of a linear set \( A \) is of first category if and only if \( A \) is nowhere dense. | No |
Theorem 1.5 (Borel). If a finite or infinite sequence of intervals \( {I}_{n} \) covers an interval \( I \), then \( \sum \left| {I}_{n}\right| \geqq \left| I\right| \) . | Proof. Assume first that \( I = \left\lbrack {a, b}\right\rbrack \) is closed and that all of the intervals \( {I}_{n} \) are open. Let \( \left( {{a}_{1},{b}_{1}}\right) \) be the first interval that contains \( a \) . If \( {b}_{1} \leqq b \) , let \( \left( {{a}_{2},{b}_{2}}\right) \) be the first interval of the se... | Yes |
For any real algebraic number \( z \) of degree \( n > 1 \) there exists a positive integer \( M \) such that | \[ \left| {z - \frac{p}{q}}\right| > \frac{1}{M{q}^{n}} \] for all integers \( p \) and \( q, q > 0 \) . Proof. Let \( f\left( x\right) \) be a polynomial of degree \( n \) with integer coefficients for which \( f\left( z\right) = 0 \) . Let \( M \) be a positive integer such that \( \left| {{f}^{\prime }\left( x\right... | Yes |
Theorem 2.3. Every Liouville number is transcendental. | Proof. Suppose some Liouville number \( z \) is algebraic, of degree \( n \) . Then \( n > 1 \), since \( z \) is irrational. By Lemma 2.2 there exists a positive integer \( M \) such that\n\n(3)\n\n\[ \left| {z - p/q}\right| > 1/M{q}^{n} \]\n\nfor all integers \( p \) and \( q \) with \( q > 0 \) . Choose a positive i... | Yes |
Theorem 2.4. The set \( E \) of Liouville numbers has s-dimensional Hausdorff measure zero, for every \( s > 0 \) . | Proof. It suffices to find, for each \( \varepsilon > 0 \) and for each positive integer \( m \) , a sequence of intervals \( {I}_{n} \) such that\n\n\[ E \cap \left( {-m, m}\right) \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{I}_{n},\;\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left| {I}_{n}\right| }^{s} < \var... | Yes |
Theorem 3.1. If \( A \subset B \) then \( {m}^{ * }\left( A\right) \leqq {m}^{ * }\left( B\right) \) . | This is obvious, since any sequence \( \left\{ {I}_{i}\right\} \) that covers \( B \) also covers \( A \) . | No |
Theorem 3.2. If \( A = \bigcup {A}_{i} \) then \( {m}^{ * }\left( A\right) \leqq \sum {m}^{ * }\left( {A}_{i}\right) \) . | This property of outer measure is called countable subadditivity. For any \( \varepsilon > 0 \) there is a sequence of intervals \( {I}_{ij}\left( {j = 1,2,\ldots }\right) \) that covers \( {A}_{i} \) such that \( \mathop{\sum }\limits_{j}\left| {I}_{ij}\right| \leqq {m}^{ * }\left( {A}_{i}\right) + \varepsilon /{2}^{i... | Yes |
Theorem 3.3. For any interval \( I,{m}^{ * }\left( I\right) = \left| I\right| \) . | Proof. The inequality \( {m}^{ * }\left( I\right) \leqq \left| I\right| \) is clear, since \( I \) covers itself. To prove the inverse inequality, let \( \varepsilon \) be an arbitrary positive number and let \( \left\{ {I}_{i}\right\} \) be an open covering of \( I \) such that \( \sum \left| {I}_{i}\right| < {m}^{ * ... | Yes |
Lemma 3.4. If \( {F}_{1} \) and \( {F}_{2} \) are disjoint bounded closed sets, then \( {m}^{ * }\left( {{F}_{1} \cup {F}_{2}}\right) = {m}^{ * }\left( {F}_{1}\right) + {m}^{ * }\left( {F}_{2}\right) \) . | Proof. There is a positive number \( \delta \) such that no interval of diameter less than \( \delta \) meets both \( {F}_{1} \) and \( {F}_{2} \) . For any \( \varepsilon > 0 \) there is a sequence of intervals \( {I}_{i} \) of diameter less than \( \delta \) such that \( {F}_{1} \cup {F}_{2} \subset \bigcup {I}_{i} \... | Yes |
Lemma 3.5. If \( {F}_{1},\ldots ,{F}_{n} \) are disjoint bounded closed sets, then \( {m}^{ * }\left( {\mathop{\bigcup }\limits_{1}^{n}{F}_{i}}\right) = \mathop{\sum }\limits_{1}^{n}{m}^{ * }\left( {F}_{i}\right) \) | This follows from Lemma 3.4 by induction on \( n \) . | No |
Lemma 3.6. For any bounded open set \( G \) and \( \varepsilon > 0 \) there exists a closed set \( F \) such that \( F \subset G \) and \( {m}^{ * }\left( F\right) > {m}^{ * }\left( G\right) - \varepsilon \) . | Proof. \( G \) can be represented as the union of a sequence of nonoverlapping intervals \( {I}_{i} \) . By definition, \( {m}^{ * }\left( G\right) \leqq \sum \left| {I}_{i}\right| \) . Determine \( n \) so that \( \mathop{\sum }\limits_{1}^{n}\left| {I}_{i}\right| > {m}^{ * }\left( G\right) - \varepsilon /2 \), and le... | Yes |
Lemma 3.7. If \( F \) is a closed subset of a bounded open set \( G \), then \( {m}^{ * }\left( {G - F}\right) = {m}^{ * }\left( G\right) - {m}^{ * }\left( F\right) \) . | Proof. By Lemma 3.6, for any \( \varepsilon > 0 \) there is a closed subset \( {F}_{1} \) of the open set \( G - F \) such that \( {m}^{ * }\left( {F}_{1}\right) > {m}^{ * }\left( {G - F}\right) - \varepsilon \) . By Lemma 3.4 and Theorem 3.1,\n\n\[ \n{m}^{ * }\left( F\right) + {m}^{ * }\left( {G - F}\right) < {m}^{ * ... | Yes |
Lemma 3.9. If \( A \) is measurable, then \( {A}^{\prime } \) is measurable. | For if \( F \subset A \subset G \), then \( {F}^{\prime } \supset {A}^{\prime } \supset {G}^{\prime } \) and \( {F}^{\prime } - {G}^{\prime } = G - F \) . | No |
Lemma 3.10. If \( A \) and \( B \) are measurable, then \( A \cap B \) is measurable. | Proof. Let \( {F}_{1} \) and \( {F}_{2} \) be closed sets, and let \( {G}_{1} \) and \( {G}_{2} \) be open sets, such that \( {F}_{1} \subset A \subset {G}_{1},{F}_{2} \subset B \subset {G}_{2},{m}^{ * }\left( {{G}_{1} - {F}_{1}}\right) < \varepsilon /2,{m}^{ * }\left( {{G}_{2} - {F}_{2}}\right) < \varepsilon /2 \) . T... | Yes |
Lemma 3.11. A bounded set \( A \) is measurable if for each \( \varepsilon > 0 \) there exists a closed set \( F \subset A \) such that \( {m}^{ * }\left( F\right) > {m}^{ * }\left( A\right) - \varepsilon \) . | Proof. For any \( \varepsilon > 0 \) let \( F \) be a closed subset of \( A \) such that \( {m}^{ * }\left( F\right) \) \( > {m}^{ * }\left( A\right) - \varepsilon /2 \) . Since \( {m}^{ * }\left( A\right) < \infty \) there exists a covering sequence of open intervals \( {I}_{i} \) of diameter less than 1 such that \( ... | Yes |
Lemma 3.12. Any interval and any nullset is measurable. | Proof. The first statement follows at once from Lemma 3.11 and Theorem 3.3. If \( {m}^{ * }\left( A\right) = 0 \), then for each \( \varepsilon > 0 \) there is a covering sequence of open intervals \( {I}_{i} \) such that \( \sum \left| {I}_{i}\right| < \varepsilon \) . Take \( G = \bigcup {I}_{i} \) and \( F = \varnot... | Yes |
Lemma 3.13. Let \( \\left\\{ {A}_{i}\\right\\} \) be a disjoint sequence of measurable sets all contained in some interval \( I \) . If \( A = \\bigcup {A}_{i} \), then \( A \) is measurable and \( {m}^{ * }\\left( A\\right) = \\sum {m}^{ * }\\left( {A}_{i}\\right) \) | Proof. For any \( \\varepsilon > 0 \) there exist closed sets \( {F}_{i} \\subset {A}_{i} \) such that \( {m}^{ * }\\left( {F}_{i}\\right) \) \( > {m}^{ * }\\left( {A}_{i}\\right) - \\varepsilon /{2}^{i + 1} \) . By countable subadditivity, \( {m}^{ * }\\left( A\\right) \\leqq \\mathop{\\sum }\\limits_{1}^{\\infty }{m}... | Yes |
Lemma 3.14. For any disjoint sequence of measurable sets \( {A}_{i} \), the set \( A = \bigcup {A}_{i} \) is measurable and \( {m}^{ * }\left( A\right) = \sum {m}^{ * }\left( {A}_{i}\right) \) . | Proof. Let \( {I}_{j}\left( {j = 1,2,\ldots }\right) \) be a sequence of disjoint intervals whose union is the whole of \( r \) -space such that any bounded set is covered by finitely many. By Lemmas 3.10 and 3.12, the sets \( {A}_{ij} = {A}_{i} \cap {I}_{j} \) are measurable. They are also disjoint. Put \( {B}_{j} = \... | Yes |
Theorem 3.17. If \( {A}_{i} \) is measurable, and \( {A}_{i} \subset {A}_{i + 1} \) for each \( i \), then the set \( A = \bigcup {A}_{i} \) is measurable and \( m\left( A\right) = \lim m\left( {A}_{i}\right) \) . If \( {A}_{i} \) is measurable and \( {A}_{i} \supset {A}_{i + 1} \) for each \( i \), then the set \( A =... | Proof. In the first case, put \( {B}_{1} = {A}_{1} \) and \( {B}_{i} = {A}_{i} - {A}_{i - 1} \) for \( i > 1 \) . Then \( \left\{ {B}_{i}\right\} \) is a disjoint sequence of measurable sets, with \( A = \bigcup {B}_{i} \) . Hence\n\n\[ m\left( A\right) = \sum m\left( {B}_{i}\right) = \lim \mathop{\sum }\limits_{1}^{n}... | Yes |
Theorem 3.17. If \( {A}_{i} \) is measurable, and \( {A}_{i} \subset {A}_{i + 1} \) for each \( i \), then the set \( A = \bigcup {A}_{i} \) is measurable and \( m\left( A\right) = \lim m\left( {A}_{i}\right) \) . If \( {A}_{i} \) is measurable and \( {A}_{i} \supset {A}_{i + 1} \) for each \( i \), then the set \( A =... | Proof. In the first case, put \( {B}_{1} = {A}_{1} \) and \( {B}_{i} = {A}_{i} - {A}_{i - 1} \) for \( i > 1 \) . Then \( \left\{ {B}_{i}\right\} \) is a disjoint sequence of measurable sets, with \( A = \bigcup {B}_{i} \) . Hence\n\n\[ m\left( A\right) = \sum m\left( {B}_{i}\right) = \lim \mathop{\sum }\limits_{1}^{n}... | Yes |
Theorem 3.18. The outer measure of any set \( A \) is expressed by the formula\n\[ \n{m}^{ * }\left( A\right) = \inf \{ m\left( G\right) : A \subset G, G\text{ open }\} .\n\]\nIf \( A \) is measurable, then\n\n\[ \n{m}^{ * }\left( A\right) = \sup \{ m\left( F\right) : A \supset F, F\text{ bounded and closed }\} .\n\]\n... | Proof. The first statement is clear, since the union of any covering sequence of open intervals is an open superset of \( A \) . To prove the second, let \( \alpha \) be any real number less than \( m\left( A\right) \), and let \( {A}_{i} = A \cap {\left( -i, i\right) }^{r} \) . By Theorem 3.17, \( m\left( A\right) = \... | Yes |
Theorem 3.19. If \( A \) is congruent by translation to a measurable set \( B \) , then \( A \) is measurable and \( m\left( A\right) = m\left( B\right) \) . | This is clear from the definitions and from the fact that congruent intervals have equal volume. Measurability and measure are also preserved by rotations and reflections of \( r \) -space, but we shall not prove this. | No |
Theorem 3.21. For any measurable set \( A \), let \( \phi \left( A\right) \) denote the set of points of \( R \) where \( A \) has density 1 . Then \( \phi \) has the following properties, where \( A \sim B \) means that \( A\bigtriangleup B \) is a nullset:\n\n1) \( \phi \left( A\right) \sim A \) ,\n\n2) \( A \sim B \... | Proof. The first assertion is just the Lebesgue density theorem. The second and third are immediate consequences of the definition of \( \phi \) . To prove 4), note that for any interval \( I \) we have \( I - \left( {A \cap B}\right) = \left( {I - A}\right) \) \( \cup \left( {I - B}\right) \) . Hence \( m\left( I\righ... | Yes |
Theorem 4.1. A set \( A \) has the property of Baire if and only if it can be represented in the form \( A = F\bigtriangleup Q \), where \( F \) is closed and \( Q \) is of first category. | Proof. If \( A = G\bigtriangleup P, G \) open and \( P \) of first category, then \( N = \bar{G} - G \) is a nowhere dense closed set, and \( Q = N\bigtriangleup P \) is of first category. Let \( F = \bar{G} \) . Then \( A = G\bigtriangleup P = \left( {\bar{G}\bigtriangleup N}\right) \bigtriangleup P = \bar{G}\bigtrian... | Yes |
Theorem 4.2. If \( A \) has the property of Baire, then so does its complement. | Proof. For any two sets \( A \) and \( B \) we have \( {\left( A\bigtriangleup B\right) }^{\prime } = {A}^{\prime }\bigtriangleup B \) . Hence if \( A = G\bigtriangleup P \), then \( {A}^{\prime } = {G}^{\prime }\bigtriangleup P \), and the conclusion follows from Theorem 4.1. | Yes |
Theorem 4.3. The class of sets having the property of Baire is a \( \sigma \) -algebra. It is the \( \sigma \) -algebra generated by the open sets together with the sets of first category. | Proof. Let \( {A}_{i} = {G}_{i}\bigtriangleup {P}_{i}\left( {i = 1,2,\ldots }\right) \) be any sequence of sets having the property of Baire. Put \( G = \bigcup {G}_{i}, P = \bigcup {P}_{i} \), and \( A = \bigcup {A}_{i} \) . Then \( G \) is open, \( P \) is of first category, and \( G - P \subset A \subset G \cup P \)... | Yes |
Theorem 4.4. A set has the property of Baire if and only if it can be represented as a \( {G}_{\delta } \) set plus a set of first category (or as an \( {F}_{\sigma } \) set minus a set of first category). | Proof. Since the closure of any nowhere dense set is nowhere dense, any set of first category is contained in an \( {F}_{\sigma } \) set of first category. If \( G \) is open and \( P \) is of first category, let \( Q \) be an \( {F}_{\sigma } \) set of first category that contains \( P \) . Then the set \( E = G - Q \... | Yes |
Theorem 4.5. Any open set \( H \) is of the form \( H = G - \bar{N} \), where \( G \) is regular open and \( N \) is nowhere dense. | Proof. Let \( G = {H}^{-\prime - \prime } \) and \( N = G - H \) . Then \( G \) is regular open, \( N \) is nowhere dense, and \( H = G - N \) . We have \( \bar{N} \subset \bar{G} - H \) . Therefore \( G - \bar{N} \) \( \supset G - \left( {\bar{G} - H}\right) = G \cap H = H \) . Also, \( H = G - N \supset G - \bar{N} \... | Yes |
Theorem 4.6. Any set having the property of Baire can be represented in the form \( A = G\bigtriangleup P \), where \( G \) is a regular open set and \( P \) is of first category. This representation is unique in any space in which every non-empty open set is of second category (that is, not of first category). | Proof. The existence of such a representation follows from Theorem 4.5; in any representation we can always replace the open set by the interior of its closure. To prove uniqueness, suppose \( G\bigtriangleup P = H\bigtriangleup Q \) , where \( G \) is a regular open set, \( H \) is open, and \( P \) and \( Q \) are of... | Yes |
Theorem 4.7. The intersection of any two regular open sets is a regular open set. | Proof. Let \( G = {G}^{-\prime - \prime } \) and \( H = {H}^{-\prime - \prime } \) . Since \( G \cap H \) is open, it follows\n\nthat\n\[ G \cap H \subset {\left( G \cap H\right) }^{-\prime - \prime } \subset {G}^{-\prime - \prime } = G.\]\n\nSimilarly,\n\[ G \cap H \subset {\left( G \cap H\right) }^{-\prime - \prime }... | Yes |
Theorem 4.8. For any linear set \( A \) of second category having the property of Baire, and for any measurable set \( A \) with \( m\left( A\right) > 0 \), there exists a positive number \( \delta \) such that \( \left( {x + A}\right) \cap A \neq \varnothing \) whenever \( \left| x\right| < \delta \) . | Proof. In the first case, let \( A = G\bigtriangleup P \) . Since \( G \) is non-empty, it contains an interval \( I \) . For any \( x \), we have\n\n\[ \left( {x + A}\right) \cap A \supset \left\lbrack {\left( {x + I}\right) \cap I}\right\rbrack - \left\lbrack {P \cup \left( {x + P}\right) }\right\rbrack . \]\n\nIf \(... | Yes |
Lemma 5.1. Any uncountable \( {G}_{\delta } \) subset of \( R \) contains a nowhere dense closed set \( C \) of measure zero that can be mapped continuously onto \( \left\lbrack {0,1}\right\rbrack \) . | Proof. Let \( E = \bigcap {G}_{n},{G}_{n} \) open, be an uncountable \( {G}_{\delta } \) set. Let \( F \) denote the set of all condensation points of \( E \) that belong to \( E \), that is, all points \( x \) in \( E \) such that every neighborhood of \( x \) contains uncountably many points of \( E.F \) is non-empty... | Yes |
Lemma 5.2. The class of uncountable closed subsets of \( R \) has power \( c \) . | Proof. The class of open intervals with rational endpoints is countable, and every open set is the union of some subclass. Hence there are at most \( c \) open sets, and therefore (by complementation) at most \( c \) closed sets. On the other hand, there are at least \( c \) uncountable closed sets, since there are tha... | Yes |
Theorem 5.3 (F. Bernstein). There exists a set \( B \) of real numbers such that both \( B \) and \( {B}^{\prime } \) meet every uncountable closed subset of the line. | By the well ordering principle and Lemma 5.2, the class \( \mathcal{F} \) of uncountable closed subsets of the line can be indexed by the ordinal numbers less than \( {\omega }_{c} \), where \( {\omega }_{c} \) is the first ordinal having \( c \) predecessors, say \( \mathcal{F} = \left\{ {{F}_{\alpha } : \alpha < {\om... | Yes |
Theorem 5.4. Any Bernstein set \( B \) is non-measurable and lacks the property of Baire. Indeed, every measurable subset of either \( B \) or \( {B}^{\prime } \) is a nullset, and any subset of \( B \) or \( {B}^{\prime } \) that has the property of Baire is of first category. | Proof. Let \( A \) be any measurable subset of \( B \) . Any closed set \( F \) contained in \( A \) must be countable (since every uncountable closed set meets \( {B}^{\prime } \) ), hence \( m\left( F\right) = 0 \) . Therefore \( m\left( A\right) = 0 \), by Theorem 3.18. Similarly, if \( A \) is a subset of \( B \) h... | Yes |
Theorem 5.5. Any set with positive outer measure has a non-measurable subset. Any set of second category has a subset that lacks the property of Baire. | Proof. If \( A \) has positive outer measure and \( B \) is a Bernstein set, Theorem 5.4 shows that the subsets \( A \cap B \) and \( A \cap {B}^{\prime } \) cannot both be measurable. If \( A \) is of second category, these two subsets cannot both have the property of Baire. | Yes |
Theorem 5.6 (Ulam). A finite measure \( \mu \) defined for all subsets of a set \( X \) of power \( {\aleph }_{1} \) vanishes identically if it is equal to zero for every one-element subset. | Proof. By hypothesis, there exists a well ordering of \( X \) such that for each \( y \) in \( X \) the set \( \{ x : x < y\} \) is countable. Let \( f\left( {x, y}\right) \) be a one-to-one mapping of this set onto a subset of the positive integers. Then \( f \) is an integer-valued function defined for all pairs \( \... | Yes |
Theorem 6.2. There exists a strategy by which \( \left( A\right) \) can be sure to win if and only if \( {I}_{1} \cap B \) is of first category for some interval \( {I}_{1} \subset {I}_{0} \) . | Proof. If such an interval exists, \( \left( A\right) \) can start by choosing it for \( {I}_{1} \) . Then, by an obvious strategy, he can insure that \( \bigcap {I}_{n} \) is disjoint to \( B \) . Since the intersection is non-empty, this is a winning strategy for \( \left( A\right) \) . On the other hand, if \( \left... | Yes |
Theorem 6.3. If the set \( A \) has the property of Baire, then (B) or (A) possesses a winning strategy according as \( A \) is of first or second category. | Proof. Let \( A = G\bigtriangleup P \), where \( G \) is open and \( P \) is of first category. If \( G \) is empty, then \( \left( B\right) \) has a winning strategy, by Theorem 6.1. If \( G \) is not empty, \( \left( A\right) \) has only to choose \( {I}_{1} \subset G \) to insure that he will be able to win. \( ▱ \) | No |
Theorem 7.2. For any \( {F}_{\sigma } \) set \( E \) there exists a bounded function \( f \) having \( E \) for its set of points of discontinuity. | Proof. Let \( E = \bigcup {F}_{n} \), where \( {F}_{n} \) is closed. We may assume that \( {F}_{n} \subset {F}_{n + 1} \) for all \( n \) . Let \( {A}_{n} \) denote the set of rational points interior to \( {F}_{n} \) . For any set \( A \), the function \( {\chi }_{A} \) defined by\n\n\[ \n{\chi }_{A}\left( x\right) = ... | Yes |
Theorem 7.3. If \( f \) can be represented as the limit of an everywhere convergent sequence of continuous functions, then \( f \) is continuous except at a set of points of first category. | Proof. It suffices to show that, for each \( \varepsilon > 0 \), the set \( F = \{ x : \omega \left( x\right) \geqq {5\varepsilon }\} \) is nowhere dense. Let \( f\left( x\right) = \lim {f}_{n}\left( x\right) ,{f}_{n} \) continuous, and define\n\n\[ \n{E}_{n} = \mathop{\bigcap }\limits_{{i, j \geqq n}}\left\{ {x : \lef... | Yes |
Theorem 7.4. Let \( f \) be a real-valued function on \( R \) . The set of points of discontinuity of \( f \) is of first category if and only if \( f \) is continuous at a dense set of points. | This is an immediate consequence of Theorem 7.1 and the fact that an \( {F}_{\sigma } \) set is of first category if and only if its complement is dense. | Yes |
Lemma 7.6. If \( \omega \left( x\right) < \varepsilon \) for each \( x \) in \( I \), then \( F\left( I\right) < \varepsilon \left| I\right| \) . | Proof. Suppose the contrary. Then \( F\left( I\right) \geqq \varepsilon \left| I\right| \), and so \( F\left( {I}_{1}\right) \geqq \varepsilon \left| I\right| /2 \) for at least one of the intervals \( {I}_{1} \) obtained by bisecting \( I \) . Similarly, \( F\left( {I}_{2}\right) \geqq \varepsilon \left| {I}_{1}\right... | Yes |
Any continuous function on a closed interval is integrable. | It may be noted that the above proof of this fact did not involve the notion of uniform continuity. | No |
The set of points of discontinuity of any monotone function \( f \) is countable. Any countable set is the set of points of discontinuity of some monotone function. | If \( f \) is monotone, there can be at most \( \left| {f\left( b\right) - f\left( a\right) }\right| /\varepsilon \) points in \( \left( {a, b}\right) \) where \( \omega \left( x\right) \geqq \varepsilon \) . Hence the set of points of discontinuity of \( f \) is countable. On the other hand, let \( \left\{ {x}_{i}\rig... | Yes |
Theorem 8.1. A real-valued function \( f \) on \( R \) has the property of Baire if and only if there exists a set \( P \) of first category such that the restriction of \( f \) to \( R - P \) is continuous. | Proof. Let \( {U}_{1},{U}_{2},\ldots \) be a countable base for the topology of \( R \) , for example, the open intervals with rational endpoints. If \( f \) has the property of Baire, then \( {f}^{-1}\left( {U}_{i}\right) = {G}_{i}\bigtriangleup {P}_{i} \), where \( {G}_{i} \) is open and \( {P}_{i} \) is of first cat... | Yes |
Theorem 8.2 (Lusin). A real-valued function \( f \) on \( R \) is measurable if and only if for each \( \varepsilon > 0 \) there exists a set \( E \) with \( m\left( E\right) < \varepsilon \) such that the restriction of \( f \) to \( R - E \) is continuous. | Proof. Let \( {U}_{1},{U}_{2},\ldots \) be a countable base for the topology of \( R \) . If \( f \) is measurable, then for each \( i \) there exists a closed set \( {F}_{i} \) and an open set \( {G}_{i} \) such that\n\n\[ \n{F}_{i} \subset {f}^{-1}\left( {U}_{i}\right) \subset {G}_{i}\;\text{ and }\;m\left( {{G}_{i} ... | Yes |
Theorem 8.3. If a sequence of measurable functions \( {f}_{n} \) converges to \( f \) at each point of a set \( E \) of finite measure, then for each \( \varepsilon > 0 \) there is a set \( F \subset E \) with \( m\left( F\right) < \varepsilon \) such that \( {f}_{n} \) converges to \( f \) uniformly on \( E - F \) . | Proof. For any two positive integers \( n \) and \( k \) let\n\n\[ \n{E}_{n, k} = \mathop{\bigcup }\limits_{{i = n}}^{\infty }\left\{ {x \in E : \left| {{f}_{i}\left( x\right) - f\left( x\right) }\right| \geqq 1/k}\right\} .\n\]\nThen \( {E}_{n, k} \supset {E}_{n + 1, k} \) and \( {\bigcap }_{n = 1}^{\infty }{E}_{n, k}... | Yes |
Theorem 9.1. If \( X \) is a topologically complete metric space, and if \( A \) is of first category in \( X \), then \( X - A \) is dense in \( X \) . | Proof. Let \( A = \bigcup {A}_{n} \), where \( {A}_{n} \) is nowhere dense, let \( \varrho \) be a metric with respect to which \( X \) is complete, and let \( {S}_{0} \) be a non-empty open set. Choose a nested sequence of balls \( {S}_{n} \) of radius \( {r}_{n} < 1/n \) such that \( \overline{{S}_{n}} \subset {S}_{n... | Yes |
Theorem 9.2. In a Baire space \( X \), a set \( E \) is residual if and only if \( E \) contains a dense \( {G}_{\delta } \) subset of \( X \) . | Proof. Suppose \( B = \bigcap {G}_{n},{G}_{n} \) open, is a \( {G}_{\delta } \) subset of \( E \) that is dense in \( X \) . Then each \( {G}_{n} \) is dense, and \( X - E \subset X - B = \bigcup \left( {X - {G}_{n}}\right) \) is of first category. Conversely, if \( X - E = \bigcup {A}_{n} \), where \( {A}_{n} \) is no... | Yes |
Theorem 12.1 (Alexandroff). Any non-empty \( {G}_{\delta } \) subset of a complete metric space is topologically complete, that is, the subset can be remetrized so as to be complete. | Proof of Theorem 12.1. Let \( X \) be a non-empty \( {G}_{\delta } \) subset of a complete metric space \( \left( {Y,\varrho }\right) \), say \( X = \bigcap {G}_{i},{G}_{i} \) open in \( Y \) . Put \( {F}_{i} = Y - {G}_{i} \) and let\n\n\[ d\left( {x,{F}_{i}}\right) = \inf \left\{ {\varrho \left( {x, y}\right) : y \in ... | No |
Lemma 12.2. [18, p. 316]. Let \( \left( {X,\varrho }\right) \) be a metric space, and suppose that there exists a sequence \( \left\{ {f}_{i}\right\} \) of real-valued continuous functions on \( X \) with the property that a Cauchy sequence \( \left\{ {x}_{n}\right\} \) is convergent whenever each of the sequences \( \... | Proof. Define a new distance function in \( X \) by\n\n\[ \sigma \left( {x, y}\right) = \varrho \left( {x, y}\right) + \mathop{\sum }\limits_{{i = 1}}^{\infty }\left( {1/{2}^{i}}\right) \min \left( {1,\left| {{f}_{i}\left( x\right) - {f}_{i}\left( y\right) }\right| }\right) .\n\]\n\nTo verify the triangle axiom it suff... | Yes |
Theorem 12.3. If a subset \( X \) of a metric space \( \left( {Z,\varrho }\right) \) is homeomorphic to a complete metric space \( \left( {Y,\sigma }\right) \), then \( X \) is a \( {G}_{\delta } \) subset of \( Z \) . | Proof. Let \( f \) be a homeomorphism of \( X \) onto \( Y \) . For each \( x \in X \) , and each \( n \), there is a positive number \( \delta \left( {x, n}\right) \) such that \( \sigma \left( {f\left( x\right), f\left( {x}^{\prime }\right) }\right) < 1/n \) whenever \( \varrho \left( {x,{x}^{\prime }}\right) < \delt... | Yes |
Theorem 13.2. For any uncountable closed set \( A \) contained in \( I = \left\lbrack {0,1}\right\rbrack \) , there exists an \( h \in H \) such that \( h\left( A\right) \) has positive measure. | Proof. By Lemma 5.1, there exists a closed set \( F \subset A \) and a continuous map \( f \) of \( F \) onto \( \left\lbrack {0,1}\right\rbrack \) . For each \( x \in I \), define\n\n\[ h\left( x\right) = x/2 + m\left( {f\left( {\left\lbrack {0, x}\right\rbrack \cap F}\right) }\right) /2. \]\n\nThen \( h \) is a stric... | Yes |
Theorem 15.1 (Kuratowski-Ulam). If \( E \) is a plane set of first category, then \( {E}_{x} \) is a linear set of first category for all \( x \) except a set of first category. If \( E \) is a nowhere dense subset of the plane \( X \times Y \), then \( {E}_{x} \) is a nowhere dense subset of \( Y \) for all \( x \) ex... | Proof. The two statement are essentially equivalent. For if \( E = \bigcup {E}_{i} \) , then \( {E}_{x} = \mathop{\bigcup }\limits_{i}{\left( {E}_{i}\right) }_{x} \) . Hence the first statement follows from the second. If \( E \) is nowhere dense, so is \( \bar{E} \), and \( {E}_{x} \) is nowhere dense whenever \( {\le... | Yes |
Theorem 15.2. If \( E \) is a subset of \( X \times Y \) with the property of Baire, then \( {E}_{x} \) has the property of Baire for all \( x \) except a set of first category in \( X \) . | Proof. Let \( E = G\bigtriangleup P \), where \( G \) is open and \( P \) is of first category. Then \( {E}_{x} = {G}_{x} \vartriangle {P}_{x} \), for all \( x \) . Every section of an open set is open, hence \( {E}_{x} \) has the property of Baire whenever \( {P}_{x} \) is of first category. By Theorem 15.1, this is t... | Yes |
Theorem 15.3. A product set \( A \times B \) is of first category in \( X \times Y \) if and only if at least one of the sets \( A \) or \( B \) is of first category. | Proof. If \( G \) is a dense open subset of \( X \), then \( G \times Y \) is a dense open subset of \( X \times Y \) . Hence \( A \times B \) is nowhere dense in \( X \times Y \) whenever \( A \) is nowhere dense in \( X \) . Since \( \left( {\bigcup {A}_{i}}\right) \times B = \bigcup \left( {{A}_{i} \times B}\right) ... | Yes |
Theorem 15.4. If \( E \) is a subset of \( X \times Y \) that has the property of Baire, and if \( {E}_{x} \) is of first category for all \( x \) except a set of first category, then \( E \) is of first category. | Proof. Suppose the contrary. Then \( E = G\bigtriangleup P \), where \( P \) is of first category and \( G \) is an open set of second category. There exist open sets \( U \) and \( V \) such that \( U \times V \subset G \) and \( U \times V \) is of second category. (This is clear in the case of the plane. In general ... | Yes |
Theorem 15.5. There exists a plane set \( E \) of second category such that no three points of \( E \) are collinear. | Proof. The class of plane \( {G}_{\delta } \) sets of second category has power \( c \) . Let \( \left\{ {{E}_{\alpha } : \alpha < {\omega }_{c}}\right\} \) be a well ordering of this class, where \( {\omega }_{c} \) is the first ordinal preceded by \( c \) ordinals. Suppose points \( {p}_{\beta } \), with no three col... | Yes |
Theorem 16.1 (Banach Category Theorem). In a topological space \( X \) , the union of any family of open sets of first category is of first category. | Proof. Let \( G \) be the union of a family \( \mathcal{G} \) of non-empty open sets of first category. Let \( \mathcal{F} = \left\{ {{U}_{\alpha } : \alpha \in A}\right\} \) be a maximal family of disjoint nonempty open sets with the property that each is contained in some member of \( \mathcal{G} \) . Then the closed... | Yes |
Lemma 16.2 (Montgomery). Let \( \\left\\{ {{G}_{\\alpha } : \\alpha \\in A}\\right\\} \) be a well-ordered family of open subsets of a metric space \( X \), and for each \( \\alpha \\in A \) let \( {F}_{\\alpha } \) be a closed subset of\n\n\[ \n{H}_{\\alpha } = {G}_{\\alpha } - \\mathop{\\bigcup }\\limits_{{\\beta < \... | Proof. For each \( \\alpha \\in A \) and each positive integer \( n \), let\n\n\[ \n{F}_{\\alpha, n} = \\left\\{ {x \\in {F}_{\\alpha } : d\\left( {x, X - {G}_{\\alpha }}\\right) \\geqq 1/n}\\right\\} .\n\]\n\nThen \( {F}_{\\alpha, n} \) is a closed set, and \( {F}_{\\alpha } = \\left( {\\mathop{\\bigcup }\\limits_{{n ... | Yes |
Theorem 16.3. Let \( \mu \) be a finite Borel measure in a metric space \( X \) . If \( G \) is the union of a family \( \mathcal{G} \) of open sets of measure zero, and if card \( \mathcal{G} \) has measure zero, then \( \mu \left( G\right) = 0 \) . | Proof. Let \( \left\{ {{G}_{\alpha } : \alpha \in A}\right\} \) be a well ordering of \( \mathcal{G} \), and put \( {H}_{\alpha } = {G}_{\alpha } \n- \( \mathop{\bigcup }\limits_{{\beta < \alpha }}{G}_{\beta } \) for each \( \alpha \in A \) . Each of the sets \( {H}_{\alpha } \) is the difference of two open sets, ther... | Yes |
Theorem 16.4. If \( X \) is a metric space with a base whose cardinal has measure zero, and if \( \mu \) is a finite Borel measure in \( X \), then the union of any family of open sets of measure zero has measure zero. | Proof. Let \( \mathcal{B} \) be a base whose cardinal has measure zero. For any family \( \mathcal{G} \) of open sets of measure zero, let \( {\mathcal{B}}_{0} \) be the set of all members of \( \mathcal{B} \) that are contained in some member of \( \mathcal{G} \) . Then \( \mu \left( {\bigcup \mathcal{G}}\right) = \mu... | Yes |
Theorem 16.5. Let \( X \) be a metric space with a base whose cardinal has measure zero. Let \( \mu \) be a nonatomic Borel measure in \( X \) such that\n\n(i) every set of infinite measure has a subset with positive finite measure, and\n\n(ii) every set of measure zero is contained in a \( {G}_{\delta } \) set of meas... | Proof. By selecting a point from each member of the given base, we obtain a dense set \( S \) of at most the same cardinality. For each positive integer \( n \), let \( {F}_{n} \) be a maximal subset of \( S \) with the property that \( \varrho \left( {x, y}\right) \) \( \geqq 1/n \) for any two distinct points of \( {... | Yes |
Theorem 17.2. An S-measurable mapping \( T \) of \( X \) into \( X \) has the recurrence property if and only if \( T \) is nondissipative. | Proof. Suppose \( T \) is nondissipative. Consider any \( E \in S \), and let \( F = E - \mathop{\bigcup }\limits_{1}^{\infty }{T}^{-k}E \) . Since \( T \) is \( S \) -measurable and \( S \) is a \( \sigma \) -ring, \( F \) belongs to \( S \) . For any integers \( 0 \leqq i < j \) we have\n\n\[ \n{T}^{-j}F \cap {T}^{-i... | Yes |
Theorem 19.5. Let \( X \) be a set of power \( {\aleph }_{1} \), and let \( K \) be a class of subsets of \( X \) with the following properties:\n\n(a) \( K \) is a \( \sigma \) -ideal,\n\n(b) the union of \( K \) is \( X \),\n\n(c) \( K \) has a subclass \( G \) of power \( \leqq {\aleph }_{1} \) with the property tha... | Proof. Let \( A = \{ \alpha : 0 \leqq \alpha < \Omega \} \) be the set of ordinals of first or second class, that is, all ordinals less than the first ordinal, \( \Omega \), that has uncountably many predecessors. Then \( A \) has power \( {\aleph }_{1} \), and there exists a mapping \( \alpha \rightarrow {G}_{\alpha }... | Yes |
Theorem 19.6. Let \( X \) be a set of power \( {\aleph }_{1} \) . Let \( K \) and \( L \) be two classes of subsets of \( X \) each of which has properties (a) to (d) of Theorem 19.5. Suppose further that \( X \) is the union of two complementary sets \( M \) and \( N \) , with \( M \in K \) and \( N \in L \) . Then th... | Proof. Let \( {X}_{\alpha }\left( {0 \leqq \alpha < \Omega }\right) \) be a decomposition of \( X \) corresponding to \( K \), as constructed in the proof of Theorem 19.5. We may assume that \( M \) belongs to the generating class \( G \), and that \( {G}_{0} \) is taken equal to \( M \) . Then \( {X}_{0} = M \), becau... | Yes |
Proposition 20.1. Any linear set \( E \) of second category has a subset \( N \) of power \( c \) such that every uncountable subset of \( N \) is of second category. | Proof. Let \( \left\{ {{X}_{\alpha } : \alpha < \Omega }\right\} \) be the decomposition of \( X \) corresponding to the class \( K \) of first category sets in the proof of Theorem 19.5. Let \( N \) be a set obtained by selecting just one point from each non-empty set of the form \( E \cap {X}_{\alpha } \) . Since \( ... | Yes |
Proposition 20.2. There exists a one-to-one mapping \( f \) of the line onto a subset of itself such that \( f\left( E\right) \) is of second category whenever \( E \) is uncountable. | Proof. Let \( f \) be any one-to-one mapping of the line onto a Lusin set. \( ▱ \) | No |
Proposition 20.3. Any linear set \( E \) of second category contains \( c \) disjoint subsets of second category. | Proof. Let \( f \) be a one-to-one mapping of \( R \) onto a Lusin set contained in \( E \) . Then the conclusion follows from the fact that the line and plane have the same cardinality. | No |
Theorem 21.3. If \( E \) is a measurable tail set in \( X \), then either \( \mu \left( E\right) = 0 \) or \( \mu \left( E\right) = 1 \) . | We merely sketch the proof, specializing that of Halmos [12, p. 201]. Let \( {A}_{n} \) be a subset of \( {X}^{n} \), and put \( F = {A}_{n} \times {Y}^{n} \). Let \( E = {X}^{n} \times {B}_{n} \), where \( {B}_{n} \subset {Y}^{n} \), for each \( n \). Then \( E \cap F = {A}_{n} \times {B}_{n} \). In our case, \( {X}^{... | No |
Theorem 21.4. If \( E \) is a tail set in \( X \) having the property of Baire, then \( E \) is either of first category or residual. | This theorem is true ! For suppose that \( E \) is not residual. Then \( X - E \) is of the form \( G\bigtriangleup P, G \) open and non-empty, \( P \) of first category. \( G \) is a countable union of basic open sets of the form \( U = {A}_{n} \times {Y}^{n} \) (corresponding to the closed intervals used to define th... | Yes |
Theorem 22.1. Let \( \mu \) be a category measure in a regular Baire space \( X \) . For any open set \( G \) and \( \varepsilon > 0 \) there is a closed set \( F \) such that \( F \subset G \) and \( \mu \left( F\right) > \mu \left( G\right) - \varepsilon \), and for every closed set \( F \) there is an open set \( G ... | Proof. Let \( \mathcal{F} \) be a maximal disjoint family of non-empty open sets \( U \) such that \( \bar{U} \subset G \) . Each member of \( \mathcal{F} \) has positive measure, hence \( \mathcal{F} \) is countable, say \( \mathcal{F} = \left\{ {U}_{i}\right\} \) . Put \( U = \bigcup {U}_{i} \) . Then \( U \subset G ... | Yes |
Theorem 22.2. If \( X \) is a regular Baire space and \( \mu \) is a category measure in \( X \), then every set of first category in \( X \) is nowhere dense. | Proof. Let \( P = \bigcup {N}_{i},{N}_{i} \) nowhere dense, be any set of first category. Since \( \mu \left( {\bar{N}}_{i}\right) = 0 \), Theorem 22.1 implies that for any two positive integers \( i \) and \( j \) there is an open set \( {G}_{ij} \) such that \( {\bar{N}}_{i} \subset {G}_{ij} \) and \( \mu \left( {G}_... | Yes |
Theorem 22.3. If \( \mu \) is a category measure in a regular Baire space \( X \) , then for any set \( E \) having the property of Baire,\n\n\[ \n\mu \left( E\right) = \mu \left( \bar{E}\right) = \mu \left( {E}^{\prime - \prime }\right)\n\]\n\nand\n\n\[ \n\mu \left( E\right) = \left\{ \begin{array}{l} \inf \{ \mu \lef... | Proof. Let \( E = G\bigtriangleup P, G \) open and \( P \) of first category. Then \( P \) is nowhere dense, and so is \( \bar{P} \) . Since\n\n\[ G - \bar{P} \subset E \subset G \cup P,\]\n\nwe have\n\n\[ G - \bar{P} \subset {E}^{\prime - \prime } \subset E \subset \bar{E} \subset \bar{G} \cup \bar{P} \]\n\nThe first ... | Yes |
Theorem 22.6. A set \( N \subset X \) is nowhere dense relative to \( \mathcal{T} \) if and only if \( N \in \mathcal{N} \). Every nowhere dense set is closed. | Proof. If \( N \in \mathcal{N} \), then \( X - N = \phi \left( X\right) - N \in \mathcal{T} \), hence each member of \( \mathcal{N} \) is closed. If \( N \in \mathcal{N} \) and \( \phi \left( {A}_{1}\right) - {N}_{1} \subset N \) for some \( {A}_{1} \in S \) and \( \left. {{N}_{1} \in \mathcal{N}\text{, then}\phi \left... | Yes |
Theorem 22.7. A set \( A \subset X \) has the property of Baire if and only if \( A \in S \) . | Proof. If \( A \in S \), then \( A = \phi \left( A\right) \vartriangle \left( {\phi \left( A\right) \vartriangle A}\right) \) . Since \( \phi \left( A\right) \in \mathcal{T} \), and \( \phi \left( A\right) \bigtriangleup A \in \mathcal{N} \), it follows from Theorem 22.6 that \( A \) has the property of Baire. Converse... | Yes |
Theorem 22.8. A set \( G \subset X \) is regular open if and only if \( G = \phi \left( A\right) \) for some \( A \in S \) . | Proof. If \( A \in S \), then \( \phi \left( A\right) \) is open, and the closure of \( \phi \left( A\right) \) is of the form \( \phi \left( A\right) \cup N \) for some \( N \in \mathcal{N} \), by Theorem 22.6. Let \( \phi \left( {A}_{1}\right) - {N}_{1} \) be any open subset of \( \phi \left( A\right) \cup N \) . The... | Yes |
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