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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 \( \left| {f\left( x\right) }\right| \leq {M}_{r}f\;\left( {x \in {S}_{r}}\right) \) .\n\n(b) If \( f \) has no zero on \( {S}_{r} \), then \( \left| {f\left( x\right) }\right| \geq {M}_{r}f\;\left( {x \in {S}_{r}}\right) \) .
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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\[ \left| {f\left( x\right) }\right| = \frac{\left| g\left( x\right) \right| }{{M}_{r}h} \leq \frac{{M}_{r}g}{{M}_{r}h} = {M}_{r}f\;\left( {\left| x\right| = r}\right) .\n\]\n\n(b) Replace \( f \) by \( 1/f \) .
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Yes
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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\[ \n{M}_{r}f = \mathop{\sup }\limits_{{x \in D}}\left| {f\left( x\right) }\right| \mathrel{\text{:=}} \parallel f{\parallel }_{D}.\n\]
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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 values \( s \geq r \) . This proves\n\n\[ \n{M}_{r}f = {M}_{r, a}f\n\]\n\nSince \( f \) is regular in the ball \( D \), its growth modulus \( {M}_{t, a}f \) (with respect to the center \( a \) ) is an increasing function of \( t < r \) . By the maximum principle (2.5) for balls,\n\n\[ \n{M}_{t, a}f = \mathop{\sup }\limits_{{\left| {x - a}\right| \leq t}}\left| {f\left( x\right) }\right| \;\left( {t < r}\right) ,\n\]\n\nand by continuity of the function \( t \mapsto {M}_{t, a}f \),\n\n\[ \n{M}_{r, a}f = \mathop{\sup }\limits_{{t < r}}{M}_{t, a}f = \mathop{\sup }\limits_{{t < r}}\mathop{\sup }\limits_{{\left| {x - a}\right| \leq t}}\left| {f\left( x\right) }\right|\n\]\n\n\[ \n= \mathop{\sup }\limits_{{\left| {x - a}\right| < r}}\left| {f\left( x\right) }\right| = \parallel f{\parallel }_{D}.\n\]
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Yes
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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 \( {S}_{r} \), then \( {M}_{r}f = \mathop{\inf }\limits_{{\left| x\right| = r}}\left| {f\left( x\right) }\right| \) .\n\n(c) If \( f \) has both zeros and poles on \( {S}_{r} \), then\n\n\( \left| {f\left( x\right) }\right| \) assumes all values of \( \left| {\mathbf{C}}_{p}\right| \) on \( x \in {S}_{r} \) .
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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 \( \alpha \in {S}_{r} \) and a zero \( \beta \in {S}_{r} \) with\n\n\[ \left| {\alpha - \beta }\right| = \min \left| {{\alpha }_{i} - {\beta }_{j}}\right| \mathrel{\text{:=}} \delta \]\n\n(minimum taken over the zeros \( {\beta }_{j} \) and poles \( {\alpha }_{i} \) in \( {S}_{r} \) ). Then\n\n\[ {M}_{s,\alpha }f = {M}_{s,\beta }f\;\left( {s > \delta }\right) ,\]\n\nsince the spheres of radius \( s > \delta \) and centers \( \alpha \) (resp. \( \beta \) ) coincide. By continuity,\n\n\[ {M}_{\delta ,\alpha }f = {M}_{\delta ,\beta }f \mathrel{\text{:=}} M. \]\n\nNow, for each \( y \in {\mathbf{C}}_{p},\left| y\right| < M, f - y \) has a critical radius \( r < \delta \) and \( f\left( x\right) = y \) has a solution \( x \in {B}_{ < \delta }\left( \beta \right) \subset {S}_{r} \) . Similarly, for each \( y \in {\mathbf{C}}_{p},\left| y\right| > M, f\left( x\right) = y \) has a solution \( x \in {B}_{ < \delta }\left( \alpha \right) \subset {S}_{r} \) (consider \( 1/f \) ). Finally, as in (2.6), \( f \) also assumes some values \( y = f\left( x\right) \) where \( \left| y\right| = M \) .
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Yes
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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 \( B,\parallel f{\parallel }_{D} \leq \parallel f{\parallel }_{{B}^{c}} = {M}_{\sigma }f \) . If \( D = {B}_{ < \sigma }\left( a\right) \) is a maximal open ball in the sphere \( \left| x\right| = \sigma \), then\n\n\[ \parallel f{\parallel }_{D} = \parallel f{\parallel }_{{B}^{c}} = {M}_{\sigma }f. \]
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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 \( \deg g < \deg h \), the growth modulus \( {M}_{r}f \) decreases for \( r \geq {\sigma }_{p} \)\n\n\[ \left| {f\left( x\right) }\right| \leq {M}_{\sigma }f\;\left( {\left| x\right| \geq \sigma }\right) ,\]\n\nand for \( D \subset {B}^{c} \) we have\n\n\[ \parallel f{\parallel }_{D} \leq \parallel f{\parallel }_{{B}^{c}} \leq {M}_{\sigma }f. \]\n\nTaking a sequence \( {x}_{n} \in {B}^{c} \) with regular \( {r}_{n} = \left| {x}_{n}\right| \searrow \sigma \), so that \( \left| {f\left( {x}_{n}\right) }\right| = {M}_{{r}_{n}}f \nearrow \) \( {M}_{\sigma }f \), we see that \( \parallel f{\parallel }_{{B}^{c}} \geq \mathop{\sup }\limits_{n}\left| {f\left( {x}_{n}\right) }\right| = {M}_{\sigma }f \) . Finally, if \( D = {B}_{ < \sigma }\left( a\right) \) is a maximal open ball in the sphere \( \left| x\right| = \sigma \), then Proposition 2 of (3.3) shows that \( \parallel f{\parallel }_{D} = {M}_{\sigma }f \), since \( f \) has no pole on \( \left| x\right| = \sigma \) .
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Yes
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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 \( \left| x\right| = \sigma \) and \( D \) contains no pole of \( f \) , then\n\n\[{\begin{Vmatrix}{f}_{B}\end{Vmatrix}}_{D} = {M}_{\sigma }{f}_{B} \leq {M}_{\sigma }f = \parallel f{\parallel }_{D}.\]
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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\[r \mapsto {M}_{r}{f}_{0}\;\text{is increasing (may be constant) for}r \leq \sigma \text{.}\]\n\nOn the other hand, \( {M}_{r}{f}_{B} \) decreases (strictly) beyond\n\n\[{\sigma }_{p} \mathrel{\text{:=}} \max \{ \left| \alpha \right| : \alpha \text{ is a pole of }f\text{ in }B\} < \sigma .\]\n\nThere is at most one crossing point of \( {M}_{r}{f}_{0} \) and \( {M}_{r}{f}_{B} \) in the interval \( \left( {{\sigma }_{p},\sigma }\right) \) . Hence \( {M}_{r}{f}_{0} \neq {M}_{r}{f}_{B} \) with at most one exception \( r \in \left( {{\sigma }_{p},\sigma }\right) \) . For all regular values (all except finitely many), these \( {M}_{r} \) represent absolute values of the corresponding functions where in a sum, the strongest wins:\n\n\[{M}_{r}{f}_{B} \leq \max \left( {{M}_{r}{f}_{B},{M}_{r}{f}_{0}}\right) \overset{!}{ = }{M}_{r}\left( {{f}_{B} + {f}_{0}}\right) = {M}_{r}f\;\left( {{\sigma }_{p}\left( {f}_{B}\right) < r \nearrow \sigma }\right) .\]\n\nTaking an increasing sequence of regular values \( {r}_{n} \nearrow \sigma \), we conclude that\n\n\[{M}_{\sigma }{f}_{B} \leq {M}_{\sigma }f\]\n\nFinally, by Proposition 2 of (3.3) we have \( {M}_{\sigma }{f}_{B} = {\begin{Vmatrix}{f}_{B}\end{Vmatrix}}_{D},{M}_{\sigma }f = \parallel f{\parallel }_{D} \) .
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Yes
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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\[ {f}_{i} = {\left( x - {b}_{i}\right) }^{{m}_{i}}{h}_{i} \in R{\left( {B}_{i}^{c}\right) }^{ \times },\;{\begin{Vmatrix}{h}_{i} - 1\end{Vmatrix}}_{{B}_{i}^{c}} < 1,\;{h}_{i}\left( x\right) \rightarrow 1\left( {\left| x\right| \rightarrow \infty }\right) . \]
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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 \( {f}_{0} = f/\mathop{\prod }\limits_{{1 \leq i \leq \ell }}{f}_{i} \) all requirements are satisfied.
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Yes
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Proposition 2. Assume \( \parallel f - 1{\parallel }_{D} < 1 \) . Then \( f \) has as many zeros as poles in each ball \( {B}_{i} \subset {D}^{c} \) .
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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 \( i = 1,{b}_{1} = 0 \), since \( {b}_{1} \) is also a center of the ball \( {B}_{ \leq r} \) . Since \( D \) contains a maximal open ball \( {D}_{1} \) of the sphere \( \left| x\right| = {\sigma }_{1} \mathrel{\text{:=}} \sigma \) having no pole of \( f - 1 \) (in fact infinitely many such balls), Proposition 2 of (3.3) shows that\n\n\[ \n{M}_{\sigma }\left( {f - 1}\right) = \parallel f - 1{\parallel }_{{D}_{1}} \leq \parallel f - 1{\parallel }_{D} < 1.\n\]\n\nBy continuity of the growth modulus, we have \( {M}_{t}\left( {f - 1}\right) < 1 \) for all \( t \) close to \( \sigma \) . For regular values of \( t < \sigma \) close to \( \sigma \) we have\n\n\[ \n\left| {f\left( x\right) - 1}\right| = {M}_{t}\left( {f - 1}\right) < 1 \Rightarrow \left| {f\left( x\right) }\right| = 1\;\left( {\left| x\right| = t}\right) .\n\]\n\nHence \( {M}_{t}f = 1 \) for \( t \nearrow \sigma \) . But our study of Laurent series has shown (3.3) that for \( t < \sigma \) close to \( \sigma ,{M}_{t}f = {t}^{m} \), where \( m \) is the difference between the number of zeros and poles of \( f \) in \( {B}_{1} \) . Hence \( m = 0 \), as asserted.
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Yes
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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
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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\[ {f}_{i} = \mathop{\prod }\limits_{{a \in S \cap {B}_{i}}}{\left( \frac{x - a}{x - {b}_{i}}\right) }^{{\mu }_{a}} = \mathop{\prod }\limits_{{a \in S \cap {B}_{i}}}{\left( x - a\right) }^{{\mu }_{a}} \mathrel{\text{:=}} 1 + {\omega }_{i}. \]\n\nHence \( {\omega }_{i} = {f}_{i} - 1 \) is a rational function that is regular outside \( {B}_{i} \) and tends to 0 when \( \left| x\right| \rightarrow \infty : {\omega }_{i} \) is a sum of principal parts of poles in \( {B}_{i} \) . Moreover,\n\n\[ {\begin{Vmatrix}{\omega }_{i}\end{Vmatrix}}_{D} \leq {\begin{Vmatrix}{\omega }_{i}\end{Vmatrix}}_{{B}_{i}^{c}} < 1 \]\n\nSimilarly, \( {f}_{0} = c\left( {1 + {\omega }_{0}}\right) \), and replacing \( f \) by \( f/c \) (hence \( {f}_{0} \) by \( {f}_{0}/c \) ) we may assume \( c = 1 \) . Let us compare the additive and multiplicative decompositions\n\n\[ f = {P}_{0}f + \mathop{\sum }\limits_{{1 \leq i \leq \ell }}{P}_{i}f = \mathop{\prod }\limits_{{0 \leq j \leq \ell }}\left( {1 + {\omega }_{j}}\right) = \left( {1 + {\omega }_{i}}\right) \cdot \underset{1 + {\psi }_{i}\text{ regular in }{B}_{i}}{\underbrace{\mathop{\prod }\limits_{{j \neq i}}\left( {1 + {\omega }_{j}}\right) }} \]\n\nwith \( {\begin{Vmatrix}{\psi }_{i}\end{Vmatrix}}_{D} < 1 \) . Hence\n\n\[ f = \left( {1 + {\psi }_{i}}\right) + {\omega }_{i}\left( {1 + {\psi }_{i}}\right) \]\n\nand the principal part \( {P}_{i}f \) of \( f \) relative to the ball \( {B}_{i} \) is also the principal part of \( {\omega }_{i}\left( {1 + {\psi }_{i}}\right) \):\n\n\[ {P}_{i}f = {P}_{i}\left( {{\omega }_{i} + {\omega }_{i}{\psi }_{i}}\right) = {\omega }_{i} + {P}_{i}\left( {{\omega }_{i}{\psi }_{i}}\right) . \]\n\n(*) \n\nBy the rational Mittag-Leffler theorem (3.4),\n\n\[ {\begin{Vmatrix}{P}_{i}\left( {\omega }_{i}{\psi }_{i}\right) \end{Vmatrix}}_{D} \leq {\begin{Vmatrix}{\omega }_{i}{\psi }_{i}\end{Vmatrix}}_{D} < {\begin{Vmatrix}{\omega }_{i}\end{Vmatrix}}_{D} \]\n\nand in \( \left( *\right) \) the first term is dominant:\n\n\[ {\begin{Vmatrix}{P}_{i}f\end{Vmatrix}}_{D} = {\begin{Vmatrix}{\omega }_{i}\end{Vmatrix}}_{D} = {\begin{Vmatrix}{f}_{i} - 1\end{Vmatrix}}_{D} \]
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Yes
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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.
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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 \( D \) . Moreover, when \( D \) is closed and \( a \notin D \), the distance \( \mathop{\inf }\limits_{{x \in D}}\left| {x - a}\right| \) is positive, so that the functions \( 1/{\left( x - a\right) }^{m}\left( {m \geq 1}\right) \) are also bounded on \( D \) . This proves that all rational functions having no pole on \( D \) define bounded continuous functions \( D \rightarrow {\mathbf{C}}_{p} \), and the same is true for the analytic elements (IV.2.1). Since the closure of a subalgebra of \( {C}_{b}\left( D\right) \) is also a subalgebra, the statement follows.
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Yes
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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 sphere \( \left| x\right| = r \) meets \( D \), we have, more precisely,\n\n\[ \n{M}_{r}f \leq \parallel f{\parallel }_{{S}_{r} \cap D} \leq \parallel f{\parallel }_{D}\;\left( {f \in R\left( D\right) }\right) .\n\]
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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\[ \n\left| {f\left( {x}_{n}\right) }\right| = {M}_{\left| {x}_{n}\right| }f \rightarrow {M}_{r}f\;\left( {n \rightarrow \infty }\right) ,\n\]\n\nso that\n\n\[ \n\parallel f{\parallel }_{D} \geq \sup \left| {f\left( {x}_{n}\right) }\right| \geq \lim \left| {f\left( {x}_{n}\right) }\right| = {M}_{r}f.\n\]\n\nFirst case: \( D \) does not meet the sphere \( {S}_{r} = \{ \left| x\right| = r\} \) . Since \( D \) is infraconnected and \( r \in \overline{\{ \left| x\right| : x \in D\} } \), we can find points \( {x}_{n} \in D \) with \( \left| {x}_{n}\right| \rightarrow r \) monotonically. All except finitely many (that we may discard) are regular, and we have finished in this case.\n\nSecond case: There is a point \( a \in D \cap {S}_{r} \) (observe that in this case \( D \subset {S}_{r} \) may well happen for \( r = \delta \) !). We have \( {M}_{s, a}f = {M}_{s}f \) for \( s > r \), since the spheres of radius \( s \) and respective centers \( a \) or 0 coincide. By continuity we also have \( {M}_{r, a}f = {M}_{r}f \) . By density of the values \( \left| {x - a}\right| \left( {x \in D\text{infraconnected}}\right) \) in the interval \( \left\lbrack {0,\delta }\right\rbrack \) we can find a sequence \( {x}_{n} \in D \) such that \( \left| {{x}_{n} - a}\right| = {r}_{n} \) is regular for \( f \) (with respect to the center \( a \) ), \( {r}_{n} \nearrow r : \left| {x}_{n}\right| = \left| a\right| = r \) . Hence \( {x}_{n} \in {S}_{r} \cap D \),\n\n\[ \n\left| {f\left( {x}_{n}\right) }\right| \rightarrow {M}_{r, a}f = {M}_{r}f\n\]\n\n\[ \n\parallel f{\parallel }_{{S}_{r} \cap D} \geq \mathop{\sup }\limits_{n}\left| {f\left( {x}_{n}\right) }\right| \geq {M}_{r}f.\n\]\n\nThe proof is complete.
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Yes
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Proposition 1. For an odd prime \( p \), we have \( {S}_{p}^{2} = \pm p \) .
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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 \( \mu \neq 0,{v\mu } \) goes through all nonzero classes \( {\;\operatorname{mod}\;p} \), and we can replace \( v \) by \( {v\mu } \) in the double sum:\n\n\[ \n{S}_{p}^{2} = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v{\mu }^{2}}{p}\right) {\zeta }^{\left( {v + 1}\right) \mu } = \mathop{\sum }\limits_{{v,\mu }}\left( \frac{v}{p}\right) {\zeta }^{\left( {v + 1}\right) \mu }. \n\]\n\nWe consider separately the terms with \( v = p - 1 \) :\n\n\[ \n\mathop{\sum }\limits_{\mu }\left( \frac{-1}{p}\right) {\zeta }^{0} = \left( {p - 1}\right) \left( \frac{-1}{p}\right) \n\]\n\nand for \( v \neq p - 1 \) \n\n\[ \n\mathop{\sum }\limits_{{v \neq p - 1}}\left( \frac{v}{p}\right) \mathop{\sum }\limits_{{\mu \neq 0}}{\zeta }^{\mu \left( {v + 1}\right) } \n\]\n\nRecall that \n\n\[ \n\mathop{\sum }\limits_{{\mu \neq 0}}{\zeta }^{\mu \left( {v + 1}\right) } = \underset{ = 0\text{ because }v + 1 \neq 0}{\underbrace{\mathop{\sum }\limits_{{0 \leq \mu < p}}{\zeta }^{\mu \left( {v + 1}\right) }}} - 1 = - 1. \n\]\n\nHence \n\n\[ \n{S}_{p}^{2} = \left( {p - 1}\right) \left( \frac{-1}{p}\right) - \mathop{\sum }\limits_{{v \neq - 1}}\left( \frac{v}{p}\right) \n\]\n\n\[ \n= p\left( {-1/p}\right) - \underbrace{\left( \frac{-1}{p}\right) - \mathop{\sum }\limits_{{v ≢ - 1}}\left( \frac{v}{p}\right) } \n\]\n\n\[ \n= p\left( {-1/p}\right) - \underset{ = 0}{\underbrace{\mathop{\sum }\limits_{{0 < v < p}}\left( \frac{v}{p}\right) }}. \n\]\n\nThe announced formula is proved.
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Yes
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Corollary 1. For a prime \( p \geq 3 \), the complex absolute value of \( {S}_{p} \) is
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\[ {\left| {S}_{p}\right| }_{\mathrm{C}} = \sqrt{p} \]
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Yes
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Corollary 2. For a prime \( p \geq 3 \), the quadratic extension \( \mathbf{Q}\left( \sqrt{p}\right) \) is contained in the cyclotomic field \( \mathbf{Q}\left( {\zeta ,\sqrt{-1}}\right) \) .
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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.
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No
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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 \( G \rightarrow K \) .
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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 true for one homomorphism. Assume that \( n - 1 \) distinct homomorphisms are always independent and consider \( n \) distinct homomorphisms \( {\psi }_{i}\left( {1 \leq i \leq n}\right) \) . Starting from a linear dependence relation\n\n\[{\alpha }_{1}{\psi }_{1}\left( x\right) + \cdots + {\alpha }_{n}{\psi }_{n}\left( x\right) = 0\;\left( {x \in G,{\alpha }_{i} \in K}\right) ,\]\n\nwe multiply it by the value \( {\psi }_{1}\left( a\right) \) (for some \( a \in G \) ):\n\n\[{\alpha }_{1}{\psi }_{1}\left( a\right) {\psi }_{1}\left( x\right) + \cdots + {\alpha }_{n}{\psi }_{1}\left( a\right) {\psi }_{n}\left( x\right) = 0\;\left( {x \in G}\right) .\n\nOn the other hand, we may replace \( x \) by \( {ax} \) in the first equality, and since \( {\psi }_{i}\left( {ax}\right) = \) \( {\psi }_{i}\left( a\right) {\psi }_{i}\left( x\right) \), we obtain\n\n\[{\alpha }_{1}{\psi }_{1}\left( a\right) {\psi }_{1}\left( x\right) + \cdots + {\alpha }_{n}{\psi }_{n}\left( a\right) {\psi }_{n}\left( x\right) = 0\;\left( {x \in G}\right) .\n\nIf we subtract the two relations obtained, the first term disappears, and we get a shorter relation:\n\n\[{\alpha }_{2}\left( {{\psi }_{1}\left( a\right) - {\psi }_{2}\left( a\right) }\right) {\psi }_{2} + \cdots + {\alpha }_{n}\left( {{\psi }_{1}\left( a\right) - {\psi }_{n}\left( a\right) }\right) {\psi }_{n} = 0.\n\nBy the induction assumption, all the coefficients of this new relation vanish. If we choose \( a \in G \) such that \( {\psi }_{1}\left( a\right) - {\psi }_{n}\left( a\right) \neq 0 \) (this is possible since \( {\psi }_{1} \neq {\psi }_{n} \) ), we see that \( {\alpha }_{n} = 0 \) . Using the induction assumption again, we get \( {\alpha }_{i} = 0 \) \( \left( {1 \leq i < n}\right) \) .
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Yes
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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\[ \n{H}_{p}f \in A\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \Rightarrow m{f}_{m} \in A\;\left( {m \geq 1}\right) .\n\]
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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 \( i > {\operatorname{ord}}_{p}\left( m\right) \), namely if \( m/{p}^{i} \) is not an integer. This series of identities starts with \( {h}_{m} = {f}_{m} \in A \) when \( \left( {m, p}\right) = 1 \) . We proceed by induction on the order \( v \) of \( m \), the case \( v = 0 \) having just been treated. When \( p \mid m \), we have\n\n\[ \nm{f}_{m} - \frac{m}{p}\mathop{\sum }\limits_{I}{\sigma }^{i}{f}_{m/{p}^{i}} = m{h}_{m} \in {mA}, \n\]\n\nso that\n\n\[ \nm{f}_{m} \equiv \frac{m}{p}\mathop{\sum }\limits_{I}{\sigma }^{i}{f}_{m/{p}^{i}} = \mathop{\sum }\limits_{I}{p}^{i - 1}{\sigma }^{i}\left( \underset{ \in A\text{ by induction }}{\underbrace{\frac{m}{{p}^{i}}{f}_{m/{p}^{i}}}}\right) \;\left( {\;\operatorname{mod}\;{mA}}\right) \n\]\n\nand hence \( m{f}_{m} \in A \) as expected.
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Yes
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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\[ \n{H}_{p}\left( {g \circ h}\right) = {H}_{p}\left( g\right) \left( h\right) + \frac{1}{p}\mathop{\sum }\limits_{I}\mathop{\sum }\limits_{{m \geq 1}}\left( {{\sigma }^{i}{g}_{m}}\right) \cdot \left( {h{\left( x\right) }^{{p}^{i}m} - {\left( {\sigma }_{ * }^{i}h\left( {x}^{{p}^{i}}\right) \right) }^{m}}\right) .\n\]
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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\[ \n{H}_{p}\left( g\right) \left( h\right) = g\left( h\right) - \frac{1}{p}\mathop{\sum }\limits_{I}{\sigma }_{ * }^{i}g\left( {h}^{{p}^{i}}\right) .\n\]\n\nThe first terms are the same and cancel by subtraction. Using the obvious relation\n\n\[ \n{\sigma }_{ * }^{i}\left( {g \circ h}\right) = {\sigma }_{ * }^{i}\left( g\right) \circ {\sigma }_{ * }^{i}\left( h\right)\n\]\n\nand the expansion \( g = \mathop{\sum }\limits_{{m \geq 1}}{g}_{m}{x}^{m} \) we get the announced result.
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Yes
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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 \[ {p}^{v} \mid m\; \Rightarrow \;{x}^{m} \equiv {y}^{m}\;\left( {\;\operatorname{mod}\;{I}^{r + v}}\right) \;\left( {v \in \mathbf{N}}\right) . \]
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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 \[ {z}^{p} \in {I}^{pr} \subset {I}^{2r} \subset {I}^{r + 1}. \] This establishes the case \( m = p\left( {v = 1}\right) \) of the lemma. (2) The case \( m = {p}^{v} \) is treated by induction on \( v \), the basic step \( v \mapsto v + 1 \) being analogous to the first case already treated. Hence \[ {x}^{{p}^{v}} \equiv {y}^{{p}^{v}}\;{\left( {\;\operatorname{mod}\;I}\right) }^{r + v}\;\left( {t \geq 0}\right) . \] (3) Finally, if we raise a congruence to the power \( \ell = m/{p}^{v} \), it is preserved: If \( {x}^{\prime } \equiv {y}^{\prime }\;\left( {\;\operatorname{mod}\;I}\right) {}^{s} \), say \( {x}^{\prime } = {y}^{\prime } + {z}^{\prime } \) with \( {z}^{\prime } \in {I}^{s} \), then \[ {\left( {x}^{\prime }\right) }^{\ell } = {\left( {y}^{\prime }\right) }^{\ell } + {z}^{\prime }\left( \cdots \right) \in {\left( {y}^{\prime }\right) }^{\ell } + {I}^{s}. \]
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Yes
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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 \\) .
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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( {n \\geq 0}\\right) .\n\\]\n\nSince \\( \\operatorname{Tr}{I}_{2} = 2,\\operatorname{Tr}M = 1 \\), this proves that \\( {\\ell }_{n} = \\operatorname{Tr}{M}^{n} \\) is the Lucas sequence. ∎
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Yes
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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 \\) .
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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 + 1} + {M}^{n}\\;\\left( {n \geq 0}\\right) .\n\\]\n\nSince \\( \\operatorname{Tr}{I}_{3} = 3,\\operatorname{Tr}M = 0 \\), and \\( \\operatorname{Tr}{M}^{2} = 2 \\), this proves that \\( {a}_{n} = \\operatorname{Tr}{M}^{n} \\) is the Perrin sequence.
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Yes
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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 \( n \) .
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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 \( {a}_{n} \notin {I}_{n} \) . The nested sequence of closed intervals \( {I}_{n} \) has a non-empty intersection. If \( p \in \cap {I}_{n} \), then \( p \in I \) and \( p \neq {a}_{n} \) for every \( n \) .
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Yes
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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.
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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 \( {I}_{2} \subset I - \left( {{A}_{1} \cup {A}_{2}}\right) \) . This shows that \( {A}_{1} \cup {A}_{2} \) is nowhere dense. Finally, any open interval contained in \( {A}^{\prime } \) is also contained in \( {A}^{-\prime } \) .
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No
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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.
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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 \( {I}_{1} - {A}_{2} \), and so on. Then \( \bigcap {I}_{n} \) is a non-empty subset of \( I - A \), hence \( {A}^{\prime } \) is dense. To specify all the choices in advance, it suffices to arrange the (denumerable) class of closed intervals with rational endpoints into a sequence, take \( {I}_{0} = I \), and for \( n > 0 \) take \( {I}_{n} \) to be the first term of the sequence that is contained in \( {I}_{n - 1} - {A}_{n} \)
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Yes
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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.
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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.
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No
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Theorem 1.5 (Borel). If a finite or infinite sequence of intervals \( {I}_{n} \) covers an interval \( I \), then \( \sum \left| {I}_{n}\right| \geqq \left| I\right| \) .
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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 sequence that contains \( {b}_{1} \) . If \( {b}_{n - 1} \leqq b \) , let \( \left( {{a}_{n},{b}_{n}}\right) \) be the first interval that contains \( {b}_{n - 1} \) . This procedure must terminate with some \( {b}_{N} > b \) . Otherwise the increasing sequence \( \left\{ {b}_{n}\right\} \) would converge to a limit \( x \leqq b \), and \( x \) would belong to \( {I}_{k} \) for some \( k \) . All but a finite number of the intervals \( \left( {{a}_{n},{b}_{n}}\right) \) would have to precede \( {I}_{k} \) in the given sequence, namely, all those for which \( {b}_{n - 1} \in {I}_{k} \) . This is impossible, since no two of these intervals are equal. (Incidentally, this reasoning reproduces Borel's own proof of the \
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Yes
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For any real algebraic number \( z \) of degree \( n > 1 \) there exists a positive integer \( M \) such that
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\[ \left| {z - \frac{p}{q}}\right| > \frac{1}{M{q}^{n}} \] 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 \( \left| {{f}^{\prime }\left( x\right) }\right| \leqq M \) whenever \( \left| {z - x}\right| \leqq 1 \) . Then, by the mean value theorem, (1) \[ \left| {f\left( x\right) }\right| = \left| {f\left( z\right) - f\left( x\right) }\right| \leqq M\left| {z - x}\right| \;\text{ whenever }\;\left| {z - x}\right| \leqq 1. \] Now consider any two integers \( p \) and \( q \), with \( q > 0 \) . We wish to show that \( \left| {z - p/q}\right| > 1/M{q}^{n} \) . This is evidently true in case \( \left| {z - p/q}\right| > 1 \), so we may assume that \( \left| {z - p/q}\right| \leqq 1 \) . Then, by \( \left( 1\right) ,\left| {f\left( {p/q}\right) }\right| \leqq M\left| {z - p/q}\right| \), and therefore (2) \[ \left| {{q}^{n}f\left( {p/q}\right) }\right| \leqq M{q}^{n}\left| {z - p/q}\right| . \] The equation \( f\left( x\right) = 0 \) has no rational root (otherwise \( z \) would satisfy an equation of degree less then \( n \) ). Moreover, \( {q}^{n}f\left( {p/q}\right) \) is an integer. Hence the left member of (2) is at least 1 and we infer that \( \left| {z - p/q}\right| \geqq 1/M{q}^{n} \) . Equality cannot hold, because \( z \) is irrational. []
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Yes
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Theorem 2.3. Every Liouville number is transcendental.
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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 integer \( k \) such that \( {2}^{k} \geqq {2}^{n}M \) . Because \( z \) is a Liouville number there exist integers \( p \) and \( q \) , with \( q > 1 \), such that\n\n(4)\n\n\[ \left| {z - p/q}\right| < 1/{q}^{k} \]\n\nFrom (3) and (4) it follows that \( 1/{q}^{k} > 1/M{q}^{n} \) . Hence \( M > {q}^{k - n} \geqq {2}^{k - n} \geqq M \) , a contradiction.
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Yes
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Theorem 2.4. The set \( E \) of Liouville numbers has s-dimensional Hausdorff measure zero, for every \( s > 0 \) .
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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\[ E \cap \left( {-m, m}\right) \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{I}_{n},\;\mathop{\sum }\limits_{{n = 1}}^{\infty }{\left| {I}_{n}\right| }^{s} < \varepsilon ,\;\text{ and }\;\left| {I}_{n}\right| < \varepsilon . \]\n\nFor each positive integer \( n \), we have\n\n\[ E \cap \left( {-m, m}\right) \subset \mathop{\bigcup }\limits_{{q = 2}}^{\infty }\mathop{\bigcup }\limits_{{p = - {mq}}}^{{mq}}\left( {p/q - 1/{q}^{n}, p/q + 1/{q}^{n}}\right) . \]\n\nChoose \( n \) so as to satisfy simultaneously the following conditions:\n\n\[ 1/{2}^{n - 1} < \varepsilon ,\;{ns} > 2,\;\frac{\left( {{2m} + 1}\right) {2}^{s}}{{ns} - 2} < \varepsilon . \]\n\nThen each of the intervals \( \left( {p/q - 1/{q}^{n}, p/q + 1/{q}^{n}}\right) \) has length \( 2/{q}^{n} \leqq 2/{2}^{n} < \varepsilon \) , and we have\n\n\[ \mathop{\sum }\limits_{{q = 2}}^{\infty }\mathop{\sum }\limits_{{p = - {mq}}}^{{mq}}{\left( 2/{q}^{n}\right) }^{s} = \mathop{\sum }\limits_{{q = 2}}^{\infty }\frac{\left( {{2mq} + 1}\right) {2}^{s}}{{q}^{ns}} \]\n\n\[ \leqq \left( {{2m} + 1}\right) {2}^{s}\mathop{\sum }\limits_{{q = 2}}^{\infty }\frac{1}{{q}^{{ns} - 1}} \leqq \left( {{2m} + 1}\right) {2}^{s}{\int }_{1}^{\infty }\frac{dx}{{x}^{{ns} - 1}} \]\n\n\[ = \frac{\left( {{2m} + 1}\right) {2}^{s}}{{ns} - 2} < \varepsilon \]
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Yes
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Theorem 3.1. If \( A \subset B \) then \( {m}^{ * }\left( A\right) \leqq {m}^{ * }\left( B\right) \) .
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This is obvious, since any sequence \( \left\{ {I}_{i}\right\} \) that covers \( B \) also covers \( A \) .
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No
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Theorem 3.2. If \( A = \bigcup {A}_{i} \) then \( {m}^{ * }\left( A\right) \leqq \sum {m}^{ * }\left( {A}_{i}\right) \) .
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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 \( \mathop{\sum }\limits_{j}\left| {I}_{ij}\right| \leqq {m}^{ * }\left( {A}_{i}\right) + \varepsilon /{2}^{i} \) . Then \( A \subset \mathop{\bigcup }\limits_{{i, j}}{I}_{ij} \) and \( \mathop{\sum }\limits_{{i, j}}\left| {I}_{ij}\right| \leqq \mathop{\sum }\limits_{i}{m}^{ * }\left( {A}_{i}\right) + \varepsilon \) . Therefore \( {m}^{ * }\left( A\right) \leqq \sum {m}^{ * }\left( {A}_{i}\right) + \varepsilon \) . Letting \( \varepsilon \rightarrow 0 \), the required inequality follows.
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Yes
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Theorem 3.3. For any interval \( I,{m}^{ * }\left( I\right) = \left| I\right| \) .
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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 \( \sum \left| {I}_{i}\right| < {m}^{ * }\left( I\right) + \varepsilon \) . Let \( J \) be a closed subinterval of \( I \) such that \( \left| J\right| > \left| I\right| - \varepsilon \) . By the Heine-Borel theorem, \( J \subset \mathop{\bigcup }\limits_{1}^{k}{I}_{i} \) for some \( k \) . Let \( {K}_{1},\ldots ,{K}_{n} \) be an enumeration of the closed intervals into which \( {\bar{I}}_{1},\ldots ,{\bar{I}}_{k} \) are divided by all the \( \left( {r - 1}\right) \) -dimensional hyperplanes that contain an \( \left( {r - 1}\right) \) -dimensional face of one of the intervals \( {I}_{1},\ldots ,{I}_{k} \), or \( J \), and let \( {J}_{1},\ldots ,{J}_{m} \) be the closed intervals into which \( J \) is divided by these same hyperplanes. Then each interval \( {J}_{i} \) is equal to at least one of the intervals \( {K}_{j} \) . Consequently,\n\n\[ \n\left| J\right| = \mathop{\sum }\limits_{{i = 1}}^{m}\left| {J}_{i}\right| \leqq \mathop{\sum }\limits_{{j = 1}}^{n}\left| {K}_{j}\right| = \mathop{\sum }\limits_{{i = 1}}^{k}\left| {I}_{i}\right| < {m}^{ * }\left( I\right) + \varepsilon .\n\]\n\nTherefore \( \left| I\right| \leqq {m}^{ * }\left( I\right) + {2\varepsilon } \) . The desired inequality follows by letting \( \varepsilon \rightarrow 0 \) .
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Yes
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Lemma 3.4. If \( {F}_{1} \) and \( {F}_{2} \) are disjoint bounded closed sets, then \( {m}^{ * }\left( {{F}_{1} \cup {F}_{2}}\right) = {m}^{ * }\left( {F}_{1}\right) + {m}^{ * }\left( {F}_{2}\right) \) .
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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 \( {F}_{1} \cup {F}_{2} \subset \bigcup {I}_{i} \) and \( \sum \left| {I}_{i}\right| \) \( \leqq {m}^{ * }\left( {{F}_{1} \cup {F}_{2}}\right) + \varepsilon \) . Let \( \sum \left| {I}_{i}\right| \) denote the sum over those intervals that meet \( {F}_{1} \), and let \( \mathop{\sum }\limits^{{\prime \prime }}\left| {I}_{i}\right| \) denote the sum over the remaining intervals (which cover \( {F}_{2} \) ). Then\n\n\[ \n{m}^{ * }\left( {F}_{1}\right) + {m}^{ * }\left( {F}_{2}\right) \leqq \sum \left| {I}_{i}\right| + \sum {}^{\prime \prime }\left| {I}_{i}\right| = \sum \left| {I}_{i}\right| \leqq {m}^{ * }\left( {{F}_{1} \cup {F}_{2}}\right) + \varepsilon .\n\]\n\nLetting \( \varepsilon \rightarrow 0 \), we conclude that\n\n\[ \n{m}^{ * }\left( {F}_{1}\right) + {m}^{ * }\left( {F}_{2}\right) \leqq {m}^{ * }\left( {{F}_{1} \cup {F}_{2}}\right) .\n\]\n\nThe reverse inequality follows from Theorem 3.2.
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Yes
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Lemma 3.5. If \( {F}_{1},\ldots ,{F}_{n} \) are disjoint bounded closed sets, then \( {m}^{ * }\left( {\mathop{\bigcup }\limits_{1}^{n}{F}_{i}}\right) = \mathop{\sum }\limits_{1}^{n}{m}^{ * }\left( {F}_{i}\right) \)
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This follows from Lemma 3.4 by induction on \( n \) .
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No
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Lemma 3.6. For any bounded open set \( G \) and \( \varepsilon > 0 \) there exists a closed set \( F \) such that \( F \subset G \) and \( {m}^{ * }\left( F\right) > {m}^{ * }\left( G\right) - \varepsilon \) .
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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 let \( {J}_{i} \) be a closed interval contained in the interior of \( {I}_{i} \) such that \( \left| {J}_{i}\right| > \left| {I}_{i}\right| - \varepsilon /{2n}\left( {i = 1,2,\ldots, n}\right) \) . Then \( F = \mathop{\bigcup }\limits_{1}^{n}{J}_{i} \) is a closed subset of \( G \), and by Theorem 3.3 and Lemma 3.5, \( {m}^{ * }\left( F\right) \) \( = \mathop{\sum }\limits_{1}^{n}\left| {J}_{i}\right| > \mathop{\sum }\limits_{1}^{n}\left| {I}_{i}\right| - \varepsilon /2 > {m}^{ * }\left( G\right) - \varepsilon \) .
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Yes
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Lemma 3.7. If \( F \) is a closed subset of a bounded open set \( G \), then \( {m}^{ * }\left( {G - F}\right) = {m}^{ * }\left( G\right) - {m}^{ * }\left( F\right) \) .
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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\[ \n{m}^{ * }\left( F\right) + {m}^{ * }\left( {G - F}\right) < {m}^{ * }\left( F\right) + {m}^{ * }\left( {F}_{1}\right) + \varepsilon = {m}^{ * }\left( {F \cup {F}_{1}}\right) + \varepsilon \leqq {m}^{ * }\left( G\right) + \varepsilon .\n\]\n\nLetting \( \varepsilon \rightarrow 0 \), we conclude that\n\n\[ \n{m}^{ * }\left( F\right) + {m}^{ * }\left( {G - F}\right) \leqq {m}^{ * }\left( G\right) .\n\]\n\nThe reverse inequality follows from Theorem 3.2.
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Yes
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Lemma 3.9. If \( A \) is measurable, then \( {A}^{\prime } \) is measurable.
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For if \( F \subset A \subset G \), then \( {F}^{\prime } \supset {A}^{\prime } \supset {G}^{\prime } \) and \( {F}^{\prime } - {G}^{\prime } = G - F \) .
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No
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Lemma 3.10. If \( A \) and \( B \) are measurable, then \( A \cap B \) is measurable.
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Proof. Let \( {F}_{1} \) and \( {F}_{2} \) be closed sets, and let \( {G}_{1} \) and \( {G}_{2} \) be open sets, such that \( {F}_{1} \subset A \subset {G}_{1},{F}_{2} \subset B \subset {G}_{2},{m}^{ * }\left( {{G}_{1} - {F}_{1}}\right) < \varepsilon /2,{m}^{ * }\left( {{G}_{2} - {F}_{2}}\right) < \varepsilon /2 \) . Then \( F = {F}_{1} \cap {F}_{2} \subset A \cap B \subset {G}_{1} \cap {G}_{2} = G \), say, and\n\n\[ G - F \subset \left( {{G}_{1} - {F}_{1}}\right) \cup \left( {{G}_{2} - {F}_{2}}\right) . \]\n\nHence \( {m}^{ * }\left( {G - F}\right) \leqq {m}^{ * }\left( {{G}_{1} - {F}_{1}}\right) + {m}^{ * }\left( {{G}_{2} - {F}_{2}}\right) < \varepsilon \) .
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Yes
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Lemma 3.11. A bounded set \( A \) is measurable if for each \( \varepsilon > 0 \) there exists a closed set \( F \subset A \) such that \( {m}^{ * }\left( F\right) > {m}^{ * }\left( A\right) - \varepsilon \) .
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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 \( \sum \left| {I}_{i}\right| < {m}^{ * }\left( A\right) + \varepsilon /2 \) . Let \( G \) be the union of those intervals \( {I}_{i} \) that meet \( A \) . Then \( F \subset A \subset G, G \) is bounded, and by Lemma 3.7, \( {m}^{ * }\left( {G - F}\right) = {m}^{ * }\left( G\right) - {m}^{ * }\left( F\right) \leqq \sum \left| {I}_{i}\right| - {m}^{ * }\left( F\right) < {m}^{ * }\left( A\right) \) \( + \varepsilon /2 - {m}^{ * }\left( F\right) < \varepsilon \) . Hence \( A \) is measurable.
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Yes
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Lemma 3.12. Any interval and any nullset is measurable.
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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 \( F = \varnothing \) . Then \( F \) is closed, \( G \) is open, \( F \subset A \subset G \), and \( {m}^{ * }\left( {G - F}\right) \leqq \sum \left| {I}_{i}\right| < \varepsilon \) . Hence \( A \) is measurable.
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Yes
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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 \( {m}^{ * }\\left( A\\right) = \\sum {m}^{ * }\\left( {A}_{i}\\right) \)
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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, \( {m}^{ * }\\left( A\\right) \\leqq \\mathop{\\sum }\\limits_{1}^{\\infty }{m}^{ * }\\left( {A}_{i}\\right) \) . Determine \( k \) so that\n\n\[ \n\\mathop{\\sum }\\limits_{1}^{k}{m}^{ * }\\left( {A}_{i}\\right) > {m}^{ * }\\left( A\\right) - \\varepsilon /2 \n\]\n\nand put \( F = \\mathop{\\bigcup }\\limits_{1}^{k}{F}_{i} \) . Then, by Lemma 3.5,\n\n\[ \n{m}^{ * }\\left( F\\right) = \\mathop{\\sum }\\limits_{1}^{k}{m}^{ * }\\left( {F}_{i}\\right) > \\mathop{\\sum }\\limits_{1}^{k}{m}^{ * }\\left( {A}_{i}\\right) - \\varepsilon /2 > {m}^{ * }\\left( A\\right) - \\varepsilon . \n\]\n\nHence \( A \) is measurable, by Lemma 3.11. For any \( n \) we have\n\n\[ \n\\mathop{\\sum }\\limits_{1}^{n}{m}^{ * }\\left( {A}_{i}\\right) < \\mathop{\\sum }\\limits_{1}^{n}{m}^{ * }\\left( {F}_{i}\\right) + \\varepsilon /2 = {m}^{ * }\\left( {\\mathop{\\bigcup }\\limits_{1}^{n}{F}_{i}}\\right) + \\varepsilon /2 \\leqq {m}^{ * }\\left( A\\right) + \\varepsilon /2. \n\]\n\nLetting \( n \\rightarrow \\infty \) and then letting \( \\varepsilon \\rightarrow 0 \) we conclude that \( \\mathop{\\sum }\\limits_{1}^{\\infty }{m}^{ * }\\left( {A}_{i}\\right) \\leqq {m}^{ * }\\left( A\\right) \) . The reverse inequality follows from countable subadditivity.
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Yes
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Lemma 3.14. For any disjoint sequence of measurable sets \( {A}_{i} \), the set \( A = \bigcup {A}_{i} \) is measurable and \( {m}^{ * }\left( A\right) = \sum {m}^{ * }\left( {A}_{i}\right) \) .
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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 \( {B}_{j} = \left( {{\bigcup }_{i}{A}_{ij}}\right. \) . By Lemma 3.13, \( {B}_{j} \) is a measurable subset of \( {I}_{j} \) . The sets \( {B}_{j} \) are disjoint, and \( A = \bigcup {B}_{j} \) . For any \( \varepsilon > 0 \) there exist closed sets \( {F}_{j} \) and bounded open sets \( {G}_{j} \) such that \( {F}_{j} \subset {B}_{j} \subset {G}_{j} \) and \( {m}^{ * }\left( {{G}_{j} - {F}_{j}}\right) < \varepsilon /{2}^{j} \) . Let \( F = \bigcup {F}_{j} \) and \( G = \bigcup {G}_{j} \) . Then \( F \) is closed, since any convergent sequence contained in \( \bar{F} \) is bounded and therefore contained in the union of a finite number of the sets \( {F}_{i} \), which is a closed subset of \( F \) . Also, \( G \) is open. We have \( F \subset A \subset G \) and \( G - F = \bigcup \left( {{G}_{j} - F}\right) \) \( \subset \bigcup \left( {{G}_{j} - {F}_{j}}\right) \) . Hence \( {m}^{ * }\left( {G - F}\right) \leqq \sum {m}^{ * }\left( {{G}_{j} - {F}_{j}}\right) < \varepsilon \) . This shows that \( A \) is measurable. Since \( {A}_{i} = \mathop{\bigcup }\limits_{j}{A}_{ij} \), we have \( {m}^{ * }\left( {A}_{i}\right) \leqq \mathop{\sum }\limits_{j}{m}^{ * }\left( {A}_{ij}\right) \), and therefore\n\n\[ \mathop{\sum }\limits_{i}{m}^{ * }\left( {A}_{i}\right) \leqq \mathop{\sum }\limits_{{i, j}}{m}^{ * }\left( {A}_{ij}\right) = \mathop{\sum }\limits_{j}\mathop{\sum }\limits_{i}{m}^{ * }\left( {A}_{ij}\right) = \mathop{\sum }\limits_{j}{m}^{ * }\left( {B}_{j}\right) \]\n\nby Lemma 3.13. Also, for any \( n \) ,\n\n\[ \mathop{\sum }\limits_{1}^{n}{m}^{ * }\left( {B}_{j}\right) \leqq \mathop{\sum }\limits_{1}^{n}{m}^{ * }\left( {F}_{j}\right) + \mathop{\sum }\limits_{1}^{n}{m}^{ * }\left( {{G}_{j} - {F}_{j}}\right) \]\n\n\[ \leqq {m}^{ * }\left( {\mathop{\bigcup }\limits_{1}^{n}{F}_{j}}\right) + \varepsilon \leqq {m}^{ * }\left( A\right) + \varepsilon . \]\n\nLetting \( n \rightarrow \infty \) and then letting \( \varepsilon \rightarrow 0 \) we conclude that \( \mathop{\sum }\limits_{j}{m}^{ * }\left( {B}_{j}\right) \leqq {m}^{ * }\left( A\right) \) . Therefore \( \sum {m}^{ * }\left( {A}_{i}\right) \leqq {m}^{ * }\left( A\right) \) . The reverse inequality again follows from countable subadditivity.
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Yes
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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 \( A = \bigcap {A}_{i} \) is measurable, and \( m\left( A\right) \) \( = \lim m\left( {A}_{i}\right) \) provided \( m\left( {A}_{i}\right) < \infty \) for some \( i \) .
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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\[ m\left( A\right) = \sum m\left( {B}_{i}\right) = \lim \mathop{\sum }\limits_{1}^{n}m\left( {B}_{i}\right) = \lim m\left( {A}_{n}\right) ,\]\n\nwhere the limit may be equal to \( \infty \) .\n\nIn the second case, we may assume \( m\left( {A}_{1}\right) < \infty \) . Put \( {B}_{i} = {A}_{1} - {A}_{i} \) and \( B = {A}_{1} - A \) . Then \( {B}_{i} \subset {B}_{i + 1} \) and \( \bigcup {B}_{i} = B \) . Hence \( m\left( {A}_{1}\right) - m\left( A\right) = m\left( B\right) \) \( = \lim m\left( {B}_{i}\right) = \lim \left( {m\left( {A}_{1}\right) - m\left( {A}_{i}\right) }\right) = m\left( {A}_{1}\right) - \lim m\left( {A}_{i}\right) \), and so \( m\left( A\right) \) \( = \lim m\left( {A}_{i}\right) \), both members being finite.
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Yes
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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 \( A = \bigcap {A}_{i} \) is measurable, and \( m\left( A\right) \) \( = \lim m\left( {A}_{i}\right) \) provided \( m\left( {A}_{i}\right) < \infty \) for some \( i \) .
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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\[ m\left( A\right) = \sum m\left( {B}_{i}\right) = \lim \mathop{\sum }\limits_{1}^{n}m\left( {B}_{i}\right) = \lim m\left( {A}_{n}\right) ,\]\n\nwhere the limit may be equal to \( \infty \) .\n\nIn the second case, we may assume \( m\left( {A}_{1}\right) < \infty \) . Put \( {B}_{i} = {A}_{1} - {A}_{i} \) and \( B = {A}_{1} - A \) . Then \( {B}_{i} \subset {B}_{i + 1} \) and \( \bigcup {B}_{i} = B \) . Hence \( m\left( {A}_{1}\right) - m\left( A\right) = m\left( B\right) \) \( = \lim m\left( {B}_{i}\right) = \lim \left( {m\left( {A}_{1}\right) - m\left( {A}_{i}\right) }\right) = m\left( {A}_{1}\right) - \lim m\left( {A}_{i}\right) \), and so \( m\left( A\right) \) \( = \lim m\left( {A}_{i}\right) \), both members being finite.
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Yes
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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\nConversely, if this equation holds and \( {m}^{ * }\left( A\right) < \infty \), then \( A \) is measurable.
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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, \( m\left( A\right) = \lim m\left( {A}_{i}\right) \), hence we can choose \( i \) so that \( m\left( {A}_{i}\right) > \alpha \) . By measurability, \( {A}_{i} \) (which is bounded) contains a closed set \( F \) with \( m\left( F\right) > \alpha \), and \( F \) is also a subset of \( A \) . Conversely, if \( {m}^{ * }\left( A\right) < \infty \) and \( F \) is a closed subset of \( A \) with \( m\left( F\right) > {m}^{ * }\left( A\right) - \varepsilon /2 \), let \( G \) be an open superset of \( A \) such that \( m\left( G\right) < {m}^{ * }\left( A\right) + \varepsilon /2 \) . Then \( F \subset A \subset G \) and \( m\left( {G - F}\right) < \varepsilon \), hence \( A \) is measurable.
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Yes
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Theorem 3.19. If \( A \) is congruent by translation to a measurable set \( B \) , then \( A \) is measurable and \( m\left( A\right) = m\left( B\right) \) .
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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.
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No
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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) \( A \sim B \) implies \( \phi \left( A\right) = \phi \left( B\right) \) ,\n\n3) \( \phi \left( \varnothing \right) = \varnothing \) and \( \phi \left( R\right) = R \) ,\n\n4) \( \phi \left( {A \cap B}\right) = \phi \left( A\right) \cap \phi \left( B\right) \) ,\n\n5) \( A \subset B \) implies \( \phi \left( A\right) \subset \phi \left( B\right) \) .
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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 \( m\left( I\right) - m\left( {I \cap A \cap B}\right) \leqq m\left( I\right) - m\left( {I \cap A}\right) + m\left( I\right) - m\left( {I \cap B}\right) \) . Therefore\n\n\[ \frac{m\left( {I \cap A}\right) }{\left| I\right| } + \frac{m\left( {I \cap B}\right) }{\left| I\right| } - 1 \leqq \frac{m\left( {I \cap A \cap B}\right) }{\left| I\right| }.\]\n\nTaking \( I = \left\lbrack {x - h, x + h}\right\rbrack \) and letting \( h \rightarrow 0 \) it follows that \( \phi \left( A\right) \cap \phi \left( B\right) \) \( \subset \phi \left( {A \cap B}\right) \) . The opposite inclusion is obvious. Property 5) is a consequence of 4).
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Yes
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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.
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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 \( A = G\bigtriangleup P = \left( {\bar{G}\bigtriangleup N}\right) \bigtriangleup P = \bar{G}\bigtriangleup \left( {N\bigtriangleup P}\right) = F\bigtriangleup Q \) . Conversely, if \( A = F\bigtriangleup Q \), where \( F \) is closed and \( Q \) is of first category, let \( G \) be the interior of \( F \) . Then \( N = F - G \) is nowhere dense, \( P = N\bigtriangleup Q \) is of first category, and \( A = F \vartriangle Q = \left( {G \vartriangle N}\right) \vartriangle Q = G \vartriangle \left( {N \vartriangle Q}\right) = G \vartriangle P \) .
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Yes
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Theorem 4.2. If \( A \) has the property of Baire, then so does its complement.
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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.
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Yes
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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.
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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 \( G - P \subset A \subset G \cup P \) . Hence \( G\bigtriangleup A \subset P \) is of first category, and \( A = G\bigtriangleup \left( {G\bigtriangleup A}\right) \) has the property of Baire. This result, together with Theorem 4.2, shows that the class in question is a \( \sigma \) -algebra. It is evidently the smallest \( \sigma \) -algebra that includes all open sets and all sets of first category.
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Yes
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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).
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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 \( E = G - Q \) is a \( {G}_{\delta } \), and we have\n\n\[ G \vartriangle P = \left\lbrack {\left( {G - Q}\right) \vartriangle \left( {G \cap Q}\right) }\right\rbrack \vartriangle \left( {P \cap Q}\right) \]\n\n\[ = E \vartriangle \left\lbrack {\left( {G \vartriangle P}\right) \cap Q}\right\rbrack \text{.} \]\n\nThe set \( \left( {G\bigtriangleup P}\right) \cap Q \) is of first category and disjoint to \( E \) . Hence any set having the property of Baire can be represented as the disjoint union of a \( {G}_{\delta } \) set and a set of first category. Conversely, any set that can be so represented belongs to the \( \sigma \) -algebra generated by the open sets and the sets of first category; it therefore has the property of Baire. The parenthetical statement follows by complementation, with the aid of Theorem 4.2. \( ▱ \)
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Yes
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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.
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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, \( H = G - N \supset G - \bar{N} \) . Hence \( H \) \( = G - \bar{N} \) .
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Yes
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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).
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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 first category. Then \( H - \bar{G} \subset H \vartriangle G = P \vartriangle Q \) . Hence \( H - \bar{G} \) is an open set of first category, therefore empty. We have \( H \subset \bar{G} \), and therefore \( H \subset {G}^{-\prime - \prime } = G \) . Thus in the regular open representation the open set \( G \) is maximal. If both \( G \) and \( H \) are regular open, then each contains the other. Hence \( G = H \) and \( P = Q \) .
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Yes
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Theorem 4.7. The intersection of any two regular open sets is a regular open set.
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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\[ G \cap H \subset {\left( G \cap H\right) }^{-\prime - \prime } \subset {H}^{-\prime - \prime } = H.\]\n\nTherefore \( G \cap H = {\left( G \cap H\right) }^{-\prime - \prime } \).
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Yes
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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 \( \left| x\right| < \delta \) .
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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 \( \left| x\right| < \left| I\right| \), the right member represents an interval minus a set of first category; it is therefore non-empty. Hence we may take \( \delta = \left| I\right| \) .\n\nIn the second case, let \( F \) be a bounded closed subset of \( A \) with \( m\left( F\right) > 0 \) (Theorem 3.18). Enclose \( F \) in a bounded open set \( G \) with \( m\left( G\right) < \left( {4/3}\right) m\left( F\right) \) . \( G \) is the union of a sequence of mutually disjoint intervals. For at least one of these, say \( I \), we must have \( m\left( {F \cap I}\right) > \left( {3/4}\right) m\left( I\right) \) . Take \( \delta = m\left( I\right) /2 \) . If \( \left| x\right| < \delta \), then \( \left( {x + I}\right) \cup I \) is an interval of length less than \( \left( {3/2}\right) m\left( I\right) \) that contains both \( F \cap I \) and \( x + \left( {F \cap I}\right) \) . These sets cannot be disjoint, because \( m\left( {x + \left( {F \cap I}\right) }\right) = m\left( {F \cap I}\right) > \left( {3/4}\right) m\left( I\right) \) . Since \( \left( {x + A}\right) \cap A \supset \left\lbrack {x + \left( {F \cap I}\right) }\right\rbrack \) \( \cap \left\lbrack {F \cap I}\right\rbrack \), it follows that the left member is non-empty.
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Yes
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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 \( \left\lbrack {0,1}\right\rbrack \) .
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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; otherwise, the family of intervals that have rational endpoints and contain only countably many points of \( E \) would cover \( E \), and \( E \) would be countable. Similar reasoning shows that \( F \) has no isolated points. Let \( I\left( 0\right) \) and \( I\left( 1\right) \) be two disjoint closed intervals of length at most \( 1/3 \) whose interiors meet \( F \) and whose union is contained in \( {G}_{1} \) . Proceeding inductively, if \( {2}^{n} \) disjoint closed intervals \( I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \) \( \left( {{i}_{k} = 0\text{or 1}}\right) \) whose interiors all meet \( F \) and whose union is contained in \( {G}_{n} \) have been defined, let \( I\left( {{i}_{1},\ldots ,{i}_{n + 1}}\right) \left( {{i}_{n + 1} = 0}\right. \) or 1 \( ) \) be disjoint closed intervals of length at most \( 1/{3}^{n + 1} \) contained in \( {G}_{n + 1} \cap I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \) whose interiors meet \( F \) . From the fact that \( F \) has no isolated points and that \( E \subset {G}_{n + 1} \) it is clear that such intervals exist. Thus a family of intervals \( I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \) having the stated properties can be defined. Let\n\n\[ C = \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{{i}_{1}\ldots {i}_{n}}}I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \]\n\nThen \( C \) is a closed nowhere dense subset of \( E.C \) has measure zero for the same reason as the Cantor set. (In fact, \( C \) is homeomorphic to the Cantor set.) For each \( x \) in \( C \) there is a unique sequence \( \left\{ {i}_{n}\right\} ,{i}_{n} = 0 \) or 1, such that \( x \in I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \) for every \( n \), and every such sequence corresponds to some point of \( C \) . Let \( f\left( x\right) \) be the real number having the binary development. \( {i}_{1}{i}_{2}{i}_{3}\ldots \) . Then \( f \) maps \( C \) onto \( \left\lbrack {0,1}\right\rbrack \) . Hence \( C \) has power \( c \) . \( f \) is continuous because \( \left| {f\left( x\right) - f\left( {x}^{\prime }\right) }\right| \leqq 1/{2}^{n} \) when \( x \) and \( {x}^{\prime } \) both belong to \( C \cap I\left( {{i}_{1},\ldots ,{i}_{n}}\right) \) .
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Yes
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Lemma 5.2. The class of uncountable closed subsets of \( R \) has power \( c \) .
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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 that many closed intervals. Hence there are exactly \( c \) uncountable closed subsets of the line.
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Yes
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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.
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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 \( \mathcal{F} = \left\{ {{F}_{\alpha } : \alpha < {\omega }_{c}}\right\} \) . We may assume that \( R \), and therefore each member of \( \mathcal{F} \), has been well ordered. Note that each member of \( \mathcal{F} \) has power \( c \) , by Lemma 5.1, since any closed set is a \( {G}_{\delta } \) . Let \( {p}_{1} \) and \( {q}_{1} \) be the first two. members of \( {F}_{1} \) . Let \( {p}_{2} \) and \( {q}_{2} \) be the first two members of \( {F}_{2} \) different from both \( {p}_{1} \) and \( {q}_{1} \) . If \( 1 < \alpha < {\omega }_{c} \) and if \( {p}_{\beta } \) and \( {q}_{\beta } \) have been defined for all \( \beta < \alpha \), let \( {p}_{\alpha } \) and \( {q}_{\alpha } \) be the first two elements of \( {F}_{\alpha } - \mathop{\bigcup }\limits_{{\beta < \alpha }}\left\{ {{p}_{\beta },{q}_{\beta }}\right\} \) . This set is non-empty (it has power \( c \) ) for each \( \alpha \), and so \( {p}_{\alpha } \) and \( {q}_{\alpha } \) are defined for all \( \alpha < {\omega }_{c} \) . Put \( B = \left\{ {{p}_{\alpha } : \alpha < {\omega }_{c}}\right\} \) . Since \( {p}_{\alpha } \in B \cap {F}_{\alpha } \) and \( {q}_{\alpha } \in {B}^{\prime } \cap {F}_{\alpha } \) for each \( \alpha < {\omega }_{c} \), the set \( B \) has the property that both it and its complement meet every uncountable closed set. Let us call any set with this property a Bernstein set.
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Yes
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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.
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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 \( B \) having the property of Baire, then \( A = E \cup P \), where \( E \) is \( {G}_{\delta } \) and \( P \) is of first category. The set \( E \) must be countable, since every uncountable \( {G}_{\delta } \) set contains an uncountable closed set, by Lemma 5.1, and therefore meets \( {B}^{\prime } \) . Hence \( A \) is of first category. The same reasoning applies to \( {B}^{\prime } \).
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Yes
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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.
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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.
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Yes
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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.
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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 \( \left( {x, y}\right) \) of elements of \( X \) for which \( x < y \) . It has the property\n\n(1)\n\n\[ x < {x}^{\prime } < y\;\text{ implies }\;f\left( {x, y}\right) \neq f\left( {{x}^{\prime }, y}\right) . \]\n\nFor each \( x \) in \( X \) and each positive integer \( n \), define\n\n\[ {F}_{x}^{n} = \{ y : x < y, f\left( {x, y}\right) = n\} . \]\n\nWe may picture these sets as arranged in an array\n\n\[ {F}_{{x}_{1}}^{1}{F}_{{x}_{2}}^{1}\ldots {F}_{x}^{1}\ldots \]\n\n\[ {F}_{{x}_{1}}^{2}{F}_{{x}_{2}}^{2}\ldots {F}_{x}^{2}\ldots \]\n\n................\n\n\[ {F}_{{x}_{1}}^{n}{F}_{{x}_{2}}^{n}\ldots {F}_{x}^{n}\ldots \]\n\n................\n\nwith \( {\aleph }_{0} \) rows and \( {\aleph }_{1} \) columns. This array has the following properties: (2) The sets in any row are mutually disjoint.\n\n(3) The union of the sets in any column is equal to \( X \) minus a countable set.\n\nTo verify (2), suppose \( y \in {F}_{x}^{n} \cap {F}_{{x}^{\prime }}^{n} \), for some \( n \) and some \( y, x \), and \( {x}^{\prime } \) , with \( x \leqq {x}^{\prime } \) . Then \( x < y,{x}^{\prime } < y \), and \( f\left( {x, y}\right) = f\left( {{x}^{\prime }, y}\right) = n \) . Hence \( x = {x}^{\prime } \) , by (1). Therefore, for any fixed \( n \), the sets \( {F}_{x}^{n}\left( {x \in X}\right) \) are disjoint.\n\nTo verify (3), observe that if \( x < y \), then \( y \) belongs to one of the sets \( {F}_{x}^{n} \), namely, that one for which \( n = f\left( {x, y}\right) \) . Hence the union of the sets \( {F}_{x}^{n}\left( {n = 1,2,\ldots }\right) \) differs from \( X \) by the countable set \( \{ y : y \leqq x\} \) .\n\nBy (2), in any row there can be at most countably many sets for which \( \mu \left( {F}_{x}^{n}\right) > 0 \) (since \( \mu \left( X\right) \) is finite). Therefore there can be at most countably many such sets in the whole array. Since there are uncountably many columns, it follows that there exists an element \( x \) in \( X \) such that \( \mu \left( {F}_{x}^{n}\right) = 0 \) for every \( n \) . The union of the sets in this column has measure zero, and the complementary countable set also has measure zero. Therefore \( \mu \left( X\right) = 0 \), and so \( \mu \) is identically zero.
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Yes
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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 \( {I}_{1} \subset {I}_{0} \) .
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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 \( \left( A\right) \) has a winning strategy he can always modify it so as to insure that the intersection of the intervals \( {I}_{n} \) will consist of just one point of \( A \) . (For instance, this will be insured if he always chooses \( {I}_{{2n} + 1} \) as if \( {I}_{2n} \) had been a subinterval half as long.) This defines a winning strategy for the second player in the game \( \left\langle {{I}_{1} \cap B,{I}_{1} \cap A}\right\rangle \) . By Theorem 6.1, such a strategy can exist only if \( {I}_{1} \cap B \) is of first category.
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Yes
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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.
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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. \( ▱ \)
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No
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Theorem 7.2. For any \( {F}_{\sigma } \) set \( E \) there exists a bounded function \( f \) having \( E \) for its set of points of discontinuity.
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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\[ \n{\chi }_{A}\left( x\right) = \left\{ \begin{array}{lll} 1 & \text{ when } & x \in A \\ 0 & \text{ when } & x \notin A \end{array}\right.\n\]\n\nis called the indicator function (or characteristic function) of \( A \) . The function \( {f}_{n} = {\chi }_{{F}_{n}} - {\chi }_{{A}_{n}} = {\chi }_{{F}_{n} - {A}_{n}} \) has oscillation equal to 1 at each point of \( {F}_{n} \), and equal to 0 elsewhere. Let \( \left\{ {a}_{n}\right\} \) be a sequence of positive numbers such that \( {a}_{n} > \mathop{\sum }\limits_{{i > n}}{a}_{i} \) for every \( n \) . (For instance, let \( {a}_{n} = 1/n \) !.) Then the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}{f}_{n}\left( x\right) \) converges uniformly on \( R \) to a bounded function \( f \) . \( f \) is continuous at any point where all of the terms are continuous, hence at each point of \( R - E \) . On the other hand, at each point of \( {F}_{n} - {F}_{n - 1} \) the oscillation of \( f \) is at least equal to \( {a}_{n} - \mathop{\sum }\limits_{{i > n}}{a}_{i} \) . Hence the set of points of discontinuity of \( f \) is exactly \( E \) .
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Yes
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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.
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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\[ \n{E}_{n} = \mathop{\bigcap }\limits_{{i, j \geqq n}}\left\{ {x : \left| {{f}_{i}\left( x\right) - {f}_{j}\left( x\right) }\right| \leqq \varepsilon }\right\} \;\left( {n = 1,2,\ldots }\right) .\n\]\n\nThen \( {E}_{n} \) is closed, \( {E}_{n} \subset {E}_{n + 1} \), and \( \bigcup {E}_{n} \) is the whole line. Consider any closed interval \( I \) . Since \( I = \bigcup \left( {{E}_{n} \cap I}\right) \), the sets \( {E}_{n} \cap I \) cannot all be nowhere dense. Hence, for some positive integer \( n,{E}_{n} \cap I \) contains an open interval \( J \) . We have \( \left| {{f}_{i}\left( x\right) - {f}_{j}\left( x\right) }\right| \leqq \varepsilon \) for all \( x \) in \( J;i, j \geqq n \) . Putting \( j = n \) and letting \( i \rightarrow \infty \), it follows that \( \left| {f\left( x\right) - {f}_{n}\left( x\right) }\right| \leqq \varepsilon \) for all \( x \) in \( J \) . For any \( {x}_{0} \) in \( J \) there is a neighborhood \( I\left( {x}_{0}\right) \subset J \) such that \( \left| {{f}_{n}\left( x\right) - {f}_{n}\left( {x}_{0}\right) }\right| \leqq \varepsilon \) for all \( x \) in \( I\left( {x}_{0}\right) \) . Hence \( \left| {f\left( x\right) - {f}_{n}\left( {x}_{0}\right) }\right| \leqq {2\varepsilon } \) for all \( x \) in \( I\left( {x}_{0}\right) \) . Therefore \( \omega \left( {x}_{0}\right) \leqq {4\varepsilon } \) , and so no point of \( J \) belongs to \( F \) . Thus for every closed interval \( I \) there is an open interval \( J \subset I - F \) . This shows that \( F \) is nowhere dense.
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Yes
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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.
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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.
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Yes
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Lemma 7.6. If \( \omega \left( x\right) < \varepsilon \) for each \( x \) in \( I \), then \( F\left( I\right) < \varepsilon \left| I\right| \) .
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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, \( F\left( {I}_{2}\right) \geqq \varepsilon \left| {I}_{1}\right| /2 \) for at least one of the intervals \( {I}_{2} \) obtained by bisecting \( {I}_{1} \) . By repeated bisection we obtain a nested sequence of closed intervals \( {I}_{n} \) such that \( F\left( {I}_{n}\right) \geqq \varepsilon \left| I\right| /{2}^{n}\left( {n = 1,2,\ldots }\right) \) . These intersect in a point \( x \) of \( I \) . By hypothesis, \( \omega \left( x\right) < \varepsilon \) and therefore \( \omega \left( J\right) < \varepsilon \) for some open interval \( J \) containing \( x \) . Choose \( n \) so that \( {I}_{n} \subset J \) . Then\n\n\[ F\left( {I}_{n}\right) \leqq \omega \left( {I}_{n}\right) \left| {I}_{n}\right| \leqq \omega \left( J\right) \left| I\right| /{2}^{n} < \varepsilon \left| I\right| /{2}^{n} \leqq F\left( {I}_{n}\right) ,\]\n\n a contradiction.
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Yes
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Any continuous function on a closed interval is integrable.
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It may be noted that the above proof of this fact did not involve the notion of uniform continuity.
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No
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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.
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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 \( \left\{ {x}_{i}\right\} \) be any countable set, and let \( \sum {\varepsilon }_{i} \) be a convergent series of positive real numbers. The function \( f\left( x\right) \) \( = \mathop{\sum }\limits_{{{x}_{i} \leqq x}}{\varepsilon }_{i} \) is a monotone bounded function. It has the property that \( \omega \left( {x}_{i}\right) = {\varepsilon }_{i} \) for each \( i \), and \( \omega \left( x\right) = 0 \) for all \( x \) not in the sequence \( {x}_{i} \) .
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Yes
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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.
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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 category. Put \( P = \mathop{\bigcup }\limits_{1}^{\infty }{P}_{i} \) . Then \( P \) is of first category. The restriction \( g \) of \( f \) to \( R - P \) is continuous, since \( {g}^{-1}\left( {U}_{i}\right) = {f}^{-1}\left( {U}_{i}\right) - P = \left( {{G}_{i}\bigtriangleup {P}_{i}}\right) - P \) \( = {G}_{i} - P \) is open relative to \( R - P \) for each \( i \), and therefore so is \( {g}^{-1}\left( U\right) \) , for every open set \( U \) .\n\nConversely, if the restriction \( g \) of \( f \) to the complement of some set \( P \) of first category is continuous, then for any open set \( U,{g}^{-1}\left( U\right) = G - P \) for some open set \( G \) . Since\n\n\[ \n{g}^{-1}\left( U\right) \subset {f}^{-1}\left( U\right) \subset {g}^{-1}\left( U\right) \cup P, \n\]\n\nwe have\n\n\[ \nG - P \subset {f}^{-1}\left( U\right) \subset G \cup P. \n\]\n\nTherefore \( {f}^{-1}\left( U\right) = G\bigtriangleup Q \) for some set \( Q \subset P \) . Thus \( f \) has the property of Baire. \( ▱ \)
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Yes
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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.
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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\[ \n{F}_{i} \subset {f}^{-1}\left( {U}_{i}\right) \subset {G}_{i}\;\text{ and }\;m\left( {{G}_{i} - {F}_{i}}\right) < \varepsilon /{2}^{i}.\n\]\n\nPut \( E = \mathop{\bigcup }\limits_{1}^{\infty }\left( {{G}_{i} - {F}_{i}}\right) \) . Then \( m\left( E\right) < \varepsilon \) . If \( g \) denotes the restriction of \( f \) to\n\n\( R - E \), then\n\[ \n{g}^{-1}\left( {U}_{i}\right) = {f}^{-1}\left( {U}_{i}\right) - E = {F}_{i} - E = {G}_{i} - E.\n\]\n\nHence \( {g}^{-1}\left( {U}_{i}\right) \) is both closed and open relative to \( R - E \), and it follows that \( g \) is continuous.\n\nConversely, if \( f \) has the stated property there is a sequence of sets \( {E}_{i} \) with \( m\left( {E}_{i}\right) < 1/i \) such that the restriction \( {f}_{i} \) of \( f \) to \( R - {E}_{i} \) is continuous. For any open set \( U \) there are open sets \( {G}_{i} \) such that \( {f}_{i}^{-1}\left( U\right) = {G}_{i} - {E}_{i} \) \( \left( {i = 1,2,\ldots }\right) \) . Putting \( E = \mathop{\bigcap }\limits_{1}^{\infty }{E}_{i} \), we have\n\n\[ \n{f}^{-1}\left( U\right) - E = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{f}^{-1}\left( U\right) - {E}_{i}}\right) = \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{f}_{i}^{-1}\left( U\right) .\n\]\n\nConsequently,\n\n\[ \n{f}^{-1}\left( U\right) = \left\lbrack {{f}^{-1}\left( U\right) \cap E}\right\rbrack \cup \mathop{\bigcup }\limits_{{i = 1}}^{\infty }\left( {{G}_{i} - {E}_{i}}\right) .\n\]\n\nAll of these sets are measurable, since \( m\left( E\right) = 0 \), and therefore \( f \) is a measurable function.
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Yes
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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 \( E - F \) .
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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 \( {\bigcap }_{n = 1}^{\infty }{E}_{n, k} = \varnothing \), for each \( k \) . Given \( \varepsilon > 0 \), for each \( k \) there is an integer \( n\left( k\right) \) such that \( m\left( {E}_{n\left( k\right), k}\right) < \varepsilon /{2}^{k} \) . Put \( F = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{E}_{n\left( k\right), k} \) . Then \( m\left( F\right) < \varepsilon \) . For each \( k \) we have \( E - F \subset E - {E}_{n\left( k\right), k} \) . Therefore \( \left| {{f}_{i}\left( x\right) - f\left( x\right) }\right| < 1/k \) for all \( i \geqq n\left( k\right) \) and all \( x \in E - F \) . Thus \( {f}_{n} \) converges to \( f \) uniformly on \( E - F \) .
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Yes
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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 \( X \) .
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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 \( \overline{{S}_{n}} \subset {S}_{n - 1} \) \( - {A}_{n}\left( {n \geqq 1}\right) \) . This can be done step by step, taking for \( {S}_{n} \) a ball with center \( {x}_{n} \) in \( {S}_{n - 1} - {\bar{A}}_{n} \) (which is non-empty because \( {\overline{\bar{A}}}_{n} \) is nowhere dense) and with sufficiently small radius. Then \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence, since\n\n\[ \varrho \left( {{x}_{i},{x}_{j}}\right) \leqq \varrho \left( {{x}_{i},{x}_{n}}\right) + \varrho \left( {{x}_{n},{x}_{j}}\right) < 2{r}_{n}\;\text{ for }\;i, j \geqq n. \]\n\nHence \( {x}_{n} \rightarrow x \) for some \( x \) in \( X \) . Since \( {x}_{i} \in {\bar{S}}_{n} \) for \( i \geqq n \), it follows that \( x \in \bigcap {\bar{S}}_{n} \) \( \subset {S}_{0} - A \) . This shows that \( X - A \) is dense in \( X \) .
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Yes
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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 \( X \) .
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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 nowhere dense, let \( B = \bigcap \left( {X - {\bar{A}}_{n}}\right) \) . Then \( B \) is a \( {G}_{\delta } \) set contained in \( E \) . Its complement \( X - B = \bigcup {\bar{A}}_{n} \) is of first category. Since \( X \) is a Baire space, it follows that \( B \) is dense in \( X \) .
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Yes
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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.
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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\[ d\left( {x,{F}_{i}}\right) = \inf \left\{ {\varrho \left( {x, y}\right) : y \in {F}_{i}}\right\} .\n\]\n\nWe may assume that each of the sets \( {F}_{i} \) is non-empty. Then \( d\left( {x,{F}_{i}}\right) \) is a real-valued continuous function on \( Y \), positive on \( X \) . The functions \( {f}_{i}\left( x\right) = 1/d\left( {x,{F}_{i}}\right) \;\left( {i = 1,2,\ldots }\right) \) satisfy the hypotheses of Lemma 12.2. For suppose that \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence of points of \( X \), and that for each \( i \) the sequence \( \left\{ {{f}_{i}\left( {x}_{n}\right) }\right\} \) is bounded. Then \( {x}_{n} \) converges in \( Y \) to some point \( y \), since \( Y \) is complete. The point \( y \) cannot belong to
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No
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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 \( \left\{ {{f}_{1}\left( {x}_{n}\right) }\right\} ,\left\{ {{f}_{2}\left( {x}_{n}\right) }\right\} ,\ldots \) is bounded. Then \( X \) can be remetrized so as to be complete.
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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 suffices to observe that it is satisfied by each term. The other axioms are clearly satisfied.\n\nFor any \( \varepsilon > 0 \) and \( x \in X \) there is an integer \( N \) such that \( {2}^{-N} < \varepsilon \) and a positive number \( \delta < \varepsilon \) such that\n\n\[ \varrho \left( {x, y}\right) < \delta \;\text{ implies }\;\left| {{f}_{i}\left( x\right) - {f}_{i}\left( y\right) }\right| < \varepsilon \;\left( {i = 1,2,\ldots, N}\right) .\n\]\n\nIf \( \varrho \left( {x, y}\right) < \delta \), then\n\n\[ \sigma \left( {x, y}\right) < \varepsilon + \mathop{\sum }\limits_{{i = 1}}^{N}\left( {1/{2}^{i}}\right) \left| {{f}_{i}\left( x\right) - {f}_{i}\left( y\right) }\right| + 1/{2}^{N} < {3\varepsilon }.\n\]\n\nTherefore \( \sigma \left( {x,{x}_{n}}\right) \rightarrow 0 \) whenever \( \varrho \left( {x,{x}_{n}}\right) \rightarrow 0 \) . The converse follows from the inequality \( \varrho \left( {x, y}\right) \leqq \sigma \left( {x, y}\right) \) . Thus \( \sigma \) and \( \varrho \) are equivalent metrics.\n\nTo show that \( \left( {X,\sigma }\right) \) is complete, let \( \left\{ {x}_{n}\right\} \) be Cauchy relative to \( \sigma \) . Then for any positive integer \( i \) there is an integer \( N \) such that \( \sigma \left( {{x}_{n},{x}_{m}}\right) < 1/{2}^{i} \) for all \( n, m \geqq N \) . For all \( n, m \geqq N \), we have\n\n\[ 1 > {2}^{i}\sigma \left( {{x}_{n},{x}_{m}}\right) \geqq \min \left( {1,\left| {{f}_{i}\left( {x}_{n}\right) - {f}_{i}\left( {x}_{m}\right) }\right| }\right) ,\n\]\n\nand therefore\n\n\[ \left| {{f}_{i}\left( {x}_{n}\right) - {f}_{i}\left( {x}_{m}\right) }\right| < 1 \]\n\nHence the sequence \( \left\{ {{f}_{i}\left( {x}_{n}\right) }\right\} \) is bounded, for each \( i \) . Since \( \varrho \left( {x, y}\right) \leqq \sigma \left( {x, y}\right) \) , the sequence \( \left\{ {x}_{n}\right\} \) is also Cauchy relative to \( \varrho \) . Therefore, by hypothesis, the sequence \( \left\{ {x}_{n}\right\} \) is convergent.
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Yes
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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 \( Z \) .
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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 \( \varrho \left( {x,{x}^{\prime }}\right) < \delta \left( {x, n}\right) \) and \( {x}^{\prime } \in X \) . We may assume that \( \delta \left( {x, n}\right) < 1/n \) . Let \( {G}_{n} \) be the union of the balls in \( Z \) with center \( x \) and radius \( \delta \left( {x, n}\right) /2 \) , the union being taken as \( x \) ranges over \( X \) . Then \( {G}_{n} \) is open in \( Z \) . Let \( z \in \bigcap {G}_{n} \) . For each \( n \) there is a point \( {x}_{n} \in X \) such that \( \varrho \left( {z,{x}_{n}}\right) < \delta \left( {{x}_{n}, n}\right) /2 \) . Since \( \delta \left( {{x}_{n}, n}\right) < 1/n \) it follows that \( {x}_{n} \rightarrow z \) . Also, for any \( m > n \), we have\n\n\( \varrho \left( {{x}_{n},{x}_{m}}\right) \leqq \varrho \left( {z,{x}_{n}}\right) + \varrho \left( {z,{x}_{m}}\right) < \delta \left( {{x}_{n}, n}\right) /2 + \delta \left( {{x}_{m}, m}\right) /2 \leqq \delta \left( {{x}_{n}, n}\right) \) or \( \delta \left( {{x}_{m}, m}\right) .\)\n\nTherefore \( \sigma \left( {f\left( {x}_{n}\right), f\left( {x}_{m}\right) }\right) < 1/n \) for all \( m > n \) . The sequence \( {y}_{n} = f\left( {x}_{n}\right) \) is therefore Cauchy in \( \left( {Y,\sigma }\right) \) . Hence it is convergent, say \( {y}_{n} \rightarrow y \) . Put \( x = {f}^{-1}\left( y\right) \) . Then \( x \in X \) and \( {x}_{n} \rightarrow x \), because \( {f}^{-1} \) is continuous. Since \( {x}_{n} \) has already been shown to converge to \( z \), it follows that \( z = x \), and therefore \( z \in X \) . This proves that \( \bigcap {G}_{n} \subset X \) . The opposite inclusion is obvious from the definition of \( {G}_{n} \) . Hence \( X \) is a \( {G}_{\delta } \) subset of \( Z \) .
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Yes
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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.
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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 strictly increasing continuous map of \( \left\lbrack {0,1}\right\rbrack \) onto itself. On each of the open intervals that compose \( \left( {0,1}\right) - F \), we have \( {h}^{\prime }\left( x\right) = 1/2 \) . Hence \( m\left( {h\left( {I - F}\right) }\right) = 1/2 \), and therefore \( m\left( {h\left( A\right) }\right) \geqq m\left( {h\left( F\right) }\right) = 1/2 \) .
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Yes
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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 \( x \) except a set of first category in \( X \) .
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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 \( {\left( \bar{E}\right) }_{x} \) is of first category. Hence the second statement follows from the first. It is therefore sufficient to prove the second statement for any nowhere dense closed set \( E \) .\n\nLet \( \left\{ {V}_{n}\right\} \) be a countable base for \( Y \), and put \( G = \left( {X \times Y}\right) - E \) . Then \( G \) is a dense open subset of the plane. For each positive integer \( n \), let \( {G}_{n} \) be the projection of \( G \cap \left( {X \times {V}_{n}}\right) \) in \( X \), that is,\n\n\[ \n{G}_{n} = \left\{ {x : \left( {x, y}\right) \in G\text{ for some }y \in {V}_{n}}\right\} .\n\]\n\nLet \( x \in {G}_{n} \) and \( y \in {V}_{n} \) be such that \( \left( {x, y}\right) \in G \) . Since \( G \) is open, there exist open intervals \( U \) and \( V \) such that \( x \in U, y \in V \subset {V}_{n} \), and \( U \times V \subset G \) . It follows that \( U \subset {G}_{n} \) . Hence \( {G}_{n} \) is an open subset of \( X \) . For any nonempty open set \( U \), the set \( G \cap \left( {U \times {V}_{n}}\right) \) is non-empty, since \( G \) is dense in the plane. Hence \( {G}_{n} \) contains points of \( U \) . Therefore \( {G}_{n} \) is a dense open subset of \( X \), for each \( n \) . Consequently, the set \( \bigcap {G}_{n} \) is the complement of a set of first category in \( X \) . For any \( x \in \bigcap {G}_{n} \), the section \( {G}_{x} \) contains points of \( {V}_{n} \) for every \( n \) . Hence \( {G}_{x} \) is a dense open subset of \( Y \), and therefore \( {E}_{x} = Y - {G}_{x} \) is nowhere dense. This shows that for all \( x \) except a set of first category, \( {E}_{x} \) is nowhere dense.
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Yes
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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 \( X \) .
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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 the case for all \( x \) except a set of first category.
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Yes
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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.
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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 \( \left( {\bigcup {A}_{i}}\right) \times B = \bigcup \left( {{A}_{i} \times B}\right) \), it follows that \( A \times B \) is of first category whenever \( A \) is of first category. Similar reasoning applies to \( B \) .\n\nConversely, if \( A \times B \) is of first category and \( A \) is not, then by Theorem 15.1 there exists a point \( x \) in \( A \) such that \( {\left( A \times B\right) }_{x} \) is of first category. Since \( {\left( A \times B\right) }_{x} = B \) for all \( x \) in \( A \), it follows that \( B \) is of first category.
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Yes
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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.
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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 it follows from the Banach category theorem, which will be discussed in the next chapter.) By Theorem 15.3, both \( U \) and \( V \) are of second category. For all \( x \) in \( U,{E}_{x} \supset V - {P}_{x} \) . By Theorem 15.1, \( {P}_{x} \) is of first category for all \( x \) except a set of first category. Therefore \( {E}_{x} \) is of second category for all \( x \) in \( U \) except a set of first category. This implies that \( {E}_{x} \) is of second category for all \( x \) in a set of second category, contrary to hypothesis.
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Yes
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Theorem 15.5. There exists a plane set \( E \) of second category such that no three points of \( E \) are collinear.
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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 collinear and with \( {p}_{\beta } \in {F}_{\beta } \), have been chosen for all \( \beta < \alpha \) . Since the set of all lines joining pairs of points \( {p}_{\beta } \) with \( \beta < \alpha \) has power less than \( c \), we can find a direction not parallel to any of these lines. By Theorem 15.4, some line in this direction meets \( {E}_{\alpha } \) in a set of second category, and therefore, by Lemma 5.1, in a set of power \( c \) . We can therefore choose \( {p}_{\alpha } \) in \( {E}_{\alpha } \) in such a way that \( {p}_{\alpha } \) is not collinear with any two points \( {p}_{\beta } \) with \( \beta < \alpha \) . The set of all points \( {p}_{\alpha } \) so chosen contains no three collinear points. It is of second category because its complement contains no \( {G}_{\delta } \) set of second category.
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Yes
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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.
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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 set \( \bar{G} - \bigcup \mathcal{F} \) is nowhere dense. (Otherwise \( \mathcal{F} \) would not be maximal.) Each set \( {U}_{\alpha } \) can be represented as a countable union of nowhere dense sets, say \( {U}_{\alpha } = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{N}_{\alpha, n} \) . Put \( {N}_{n} = \mathop{\bigcup }\limits_{{\alpha \in A}}{N}_{\alpha, n} \) . If an open set \( U \) meets \( {N}_{n} \), then it meets some \( {N}_{\alpha, n} \) and there exists a non-empty open set \( V \subset \left( {U \cap {U}_{\alpha }}\right) - {N}_{\alpha, n} \) . Hence \( V \subset U - {N}_{n} \), and so \( {N}_{n} \) is nowhere dense. Therefore\n\n\[ G \subset \left( {\bar{G}-\bigcup \mathcal{F}}\right) \cup \mathop{\bigcup }\limits_{{\alpha \in A}}{U}_{\alpha } = \left( {\bar{G}-\bigcup \mathcal{F}}\right) \cup \mathop{\bigcup }\limits_{1}^{\infty }{N}_{n} \]\n\nis of first category.
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Yes
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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\[ \n{H}_{\\alpha } = {G}_{\\alpha } - \\mathop{\\bigcup }\\limits_{{\\beta < \\alpha }}{G}_{\\beta }\n\]\n\nThen the set \( E = \\mathop{\\bigcup }\\limits_{{\\alpha \\in A}}{F}_{\\alpha } \) is an \( {F}_{\\sigma } \) .
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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 \( {F}_{\\alpha } = \\left( {\\mathop{\\bigcup }\\limits_{{n = 1}}^{\\infty }{F}_{\\alpha, n}}\\right. \) . If \( \\alpha \\neq \\beta \), we have \( \\varrho \\left( {x, y}\\right) \\geqq 1/n \) for every \( x \\in {F}_{\\alpha, n} \) and \( y \\in {F}_{\\beta, n} \) . Hence any convergent sequence contained in the set \( {F}_{n} = \\mathop{\\bigcup }\\limits_{{\\alpha \\in A}}{F}_{\\alpha, n} \) must, except for a finite number of terms, be contained in a single set \( {F}_{\\alpha, n} \) . It follows that \( {F}_{n} \) is closed, and that\n\n\[ \nE = \\mathop{\\bigcup }\\limits_{{\\alpha \\in A}}{F}_{\\alpha } = \\mathop{\\bigcup }\\limits_{{n = 1}}^{\\infty }{F}_{n}\n\]\n\nis an \( {F}_{\\sigma } \) .
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Yes
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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 \( \mu \left( G\right) = 0 \) .
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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, therefore an \( {F}_{\sigma } \), say \( {H}_{\alpha } = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{\alpha, n} \), where \( {F}_{\alpha, n} \) is closed. For any set \( E \subset A \) we have\n\n\[ \n\mathop{\bigcup }\limits_{{\alpha \in E}}{H}_{\alpha } = \mathop{\bigcup }\limits_{{\alpha \in E}}\mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{\alpha, n} = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }\left\lbrack {\mathop{\bigcup }\limits_{{\alpha \in E}}{F}_{\alpha, n}}\right\rbrack .\n\]\n\nBy Montgomery’s lemma (taking \( {F}_{\alpha } = \varnothing \) when \( \alpha \in A - E \) ), the set in brackets is an \( {F}_{\sigma } \), for each \( n \), and therefore the union is too. Hence the\n\nset function\n\[ \nv\left( E\right) = \mu \left( {\mathop{\bigcup }\limits_{{\alpha \in E}}{H}_{\alpha }}\right) \n\]\n\nis defined for all subsets of \( A \) . It is evidently a finite measure, and nonatomic. Since card \( A \) has measure zero it follows that \( \mu \left( G\right) = \mu \left( {\mathop{\bigcup }\limits_{{\alpha \in A}}{H}_{\alpha }}\right) \) \( = v\left( A\right) = 0 \) .
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Yes
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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.
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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 \( \mu \left( {\bigcup \mathcal{G}}\right) = \mu \left( {\bigcup {\mathcal{B}}_{0}}\right) = 0 \) , by Theorem 16.3.
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Yes
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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 measure zero.\n\nThen \( X \) can be represented as the union of a \( {G}_{\delta } \) set of measure zero and a set of first category.
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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 \( {F}_{n} \) . Put \( D = \mathop{\bigcup }\limits_{1}^{\infty }{F}_{n} \) . Then \( D \) is dense in \( X \), and since every subset of \( {F}_{n} \) is closed, every subset of \( D \) is an \( {F}_{\sigma } \) . Hence \( \mu \) is defined for all subsets of \( D \) . Because \( \mu \) is zero for points and card \( D \) has measure zero, it follows that no subset of \( D \) has positive finite measure. Therefore, by (i) and (ii), \( \mu \left( D\right) = 0 \) and \( D \) is contained in a \( {G}_{\delta } \) set \( E \) with \( \mu \left( E\right) = 0 \) . The complement of \( E \) is a set of first category.
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Yes
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Theorem 17.2. An S-measurable mapping \( T \) of \( X \) into \( X \) has the recurrence property if and only if \( T \) is nondissipative.
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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\[ \n{T}^{-j}F \cap {T}^{-i}F \subset {T}^{-j}E - \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{T}^{-i - k}E = \varnothing .\n\]\n\nThis shows that \( F \), and each of the sets \( {T}^{-k}F\left( {k = 1,2,\ldots }\right) \), is a wandering set. Since all of these sets belong to \( S \), and \( T \) is nondissipative, it follows that \( {T}^{-k}F \in I \) for all \( k \geqq 0 \) . Since \( I \) is a \( \sigma \) -ideal, the union \( \mathop{\bigcup }\limits_{0}^{\infty }{T}^{-k}F \) belongs to \( I \), and so does the set \( H = E \cap \left( {\mathop{\bigcup }\limits_{0}^{\infty }{T}^{-k}F}\right. \) . But \( {T}^{-k}F \) consists of all points \( x \) such that \( {T}^{k}x \in E \) and such that \( {T}^{i}x \in X - E \) for all \( i > k \) . Hence \( H = D\left( E\right) \) . Thus we have shown that \( D\left( E\right) \in I \), for every \( E \in S \) ; that is, \( T \) has the recurrence property.\n\nConversely, if \( T \) is dissipative there exists a wandering set \( E \) that belongs to \( S - I \) . Then \( D\left( E\right) = E \), and we have \( E \in S \) but \( D\left( E\right) \notin I \) . This shows that \( T \) lacks the recurrence property.
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Yes
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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 that each member of \( K \) is contained in some member of \( G \),\n\n(d) the complement of each member of \( K \) contains a set of power \( {\aleph }_{1} \) that belongs to \( K \).\n\nThen \( X \) can be decomposed into \( {\aleph }_{1} \) disjoint sets \( {X}_{\alpha } \), each of power \( {\aleph }_{1} \), such that a subset \( E \) of \( X \) belongs to \( K \) if and only if \( E \) is contained in a countable union of the sets \( {X}_{\alpha } \).
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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 \( \alpha \rightarrow {G}_{\alpha } \) of \( A \) onto \( G \) . For each \( \alpha \in A \) define\n\n\[ \n{H}_{\alpha } = \mathop{\bigcup }\limits_{{\beta \leqq \alpha }}{G}_{\beta }\;\text{ and }\;{K}_{\alpha } = {H}_{\alpha } - \mathop{\bigcup }\limits_{{\beta < \alpha }}{H}_{\beta }.\n\]\n\nPut \( B = \left\{ {\alpha \in A : {K}_{\alpha }}\right. \) is uncountable \( \} \) . Properties (a),(c) and (d) imply that \( B \) has no upper bound in \( A \) . Therefore there exists a one-to-one order-preserving map \( \phi \) of \( A \) onto \( B \) . For each \( \alpha \) in \( A \), define\n\n\[ \n{X}_{\alpha } = {H}_{\phi \left( \alpha \right) } - \mathop{\bigcup }\limits_{{\beta < \alpha }}{H}_{\phi \left( \beta \right) }.\n\]\n\nBy construction and property (a), the sets \( {X}_{\alpha } \) are disjoint and belong to \( K \) . Since \( {X}_{\alpha } \supset {K}_{\phi \left( \alpha \right) } \), each of the sets \( {X}_{\alpha } \) has power \( {\aleph }_{1} \) . For any \( \beta \in A \), we have \( \beta < \phi \left( \alpha \right) \) for some \( \alpha \in A \), and therefore\n\n\[ \n{G}_{\beta } \subset {H}_{\beta } \subset {H}_{\phi \left( \alpha \right) } = \mathop{\bigcup }\limits_{{\gamma \leqq \alpha }}{X}_{\gamma }.\n\]\n\nHence, by (c), each member of \( K \) is contained in a countable union of the sets \( {X}_{\alpha } \) . Using (b), it follows that\n\n\[ \nX = \bigcup K \subset \mathop{\bigcup }\limits_{{\alpha \in A}}{X}_{\alpha }\n\]\n\nThus \( \left\{ {{X}_{\alpha } : \alpha \in A}\right\} \) is a decomposition of \( X \) with the required properties.
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Yes
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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 there exists a one-to-one mapping \( f \) of \( X \) onto itself such that \( f = {f}^{-1} \) and such that \( f\left( E\right) \in L \) if and only if \( E \in K \) .
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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 \( {X}_{0} = M \), because \( M \) cannot be countable. Similarly, let \( {Y}_{\alpha }\left( {0 \leqq \alpha < \Omega }\right) \) be a decomposition of \( X \) corresponding to \( L \), with \( {Y}_{0} = N \) . Then\n\n\[ M = \mathop{\bigcup }\limits_{{0 < \alpha < \Omega }}{Y}_{\alpha }\;\text{ and }\;N = \mathop{\bigcup }\limits_{{0 < \alpha < \Omega }}{X}_{\alpha }.\]\n\nThe sets \( {X}_{\alpha } \) and \( {Y}_{\alpha } \), for \( 0 < \alpha < \Omega \), constitute a decomposition of \( X \) into sets of power \( {\aleph }_{1} \) . For each \( 0 < \alpha < \Omega \), let \( {f}_{\alpha } \) be a one-to-one mapping of \( {X}_{\alpha } \) onto \( {Y}_{\alpha } \) . Define \( f \) equal to \( {f}_{\alpha } \) on \( {X}_{\alpha } \), and equal to \( {f}_{\alpha }^{-1} \) on \( {Y}_{\alpha } \), for \( 0 < \alpha < \Omega \) . Then \( f \) is a one-to-one mapping of \( X \) onto itself, \( f \) is equal to \( {f}^{-1} \), and \( f\left( {X}_{\alpha }\right) = {Y}_{\alpha } \) for all \( 0 < \alpha < \Omega \) . Since\n\n\[ {X}_{0} = \mathop{\bigcup }\limits_{{0 < \alpha < \Omega }}{Y}_{\alpha }\;\text{ and }\;{Y}_{0} = \mathop{\bigcup }\limits_{{0 < \alpha < \Omega }}{X}_{\alpha },\]\n\nwe have also \( f\left( {X}_{0}\right) = {Y}_{0} \) . Thus \( f\left( {X}_{\alpha }\right) = {Y}_{\alpha } \) for all \( 0 \leqq \alpha < \Omega \) . From the properties of \( {X}_{\alpha } \) and \( {Y}_{\alpha } \) stated in Theorem 19.5 it follows that \( f\left( E\right) \in L \) if and only if \( E \in K \) .
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Yes
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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.
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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 \( E \) is of second category, \( N \) is uncountable and therefore of power \( c \) . No uncountable subset of \( N \) can be covered by countably many of the sets \( {X}_{\alpha } \) . Hence no uncountable subset of \( N \) is of first category.
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Yes
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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.
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Proof. Let \( f \) be any one-to-one mapping of the line onto a Lusin set. \( ▱ \)
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No
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Proposition 20.3. Any linear set \( E \) of second category contains \( c \) disjoint subsets of second category.
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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.
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No
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Theorem 21.3. If \( E \) is a measurable tail set in \( X \), then either \( \mu \left( E\right) = 0 \) or \( \mu \left( E\right) = 1 \) .
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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, \( {X}^{n} \) is a finite set. If \( {A}_{n} \) has \( k \) points, then \( \mu \left( F\right) = k/{2}^{n} \) and \( \mu \left( {{A}_{n} \times {B}_{n}}\right) = \left( {k/{2}^{n}}\right) \mu \left( {{X}^{n} \times {B}_{n}}\right) \). Hence \( \mu \left( {E \cap F}\right) = \mu \left( E\right) \mu \left( F\right) \). Every measurable set can be approximated in the space of measurable sets (Chapter 10) by a set of the form \( F \), since \( g\left( F\right) \) can be any finite union of dyadic subintervals of \( \left\lbrack {0,1}\right\rbrack \). It follows that the equation \( \mu \left( {E \cap F}\right) = \mu \left( E\right) \mu \left( F\right) \) holds for every measurable set \( F \). In particular, it holds when \( F = E \). Hence \( \mu \left( E\right) = 0 \) or 1 .
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No
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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.
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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 the Cantor set). Hence \( G \) contains a set \( U \) of the form \( U = {A}_{n} \times {Y}^{n} \), where \( {A}_{n} \) is non-empty. By hypothesis, \( E \) can be written in the form \( E = {X}^{n} \times {B}_{n} \) . Hence \( U \cap E \) \( = {A}_{n} \times {B}_{n} \) . But also,\n\n\[ \n{A}_{n} \times {B}_{n} \subset G \subset \left( {X - E}\right) \cup P.\n\]\n\nHence\n\n\[ \n{A}_{n} \times {B}_{n} \subset E \cap \left\lbrack {\left( {X - E}\right) \cup P}\right\rbrack \subset P.\n\]\n\nTherefore \( {A}_{n} \times {B}_{n} \) is of first category. Since \( {A}_{n} \) is a non-empty subset of the finite discrete space \( {X}^{n} \), it is of second category. It follows from Theorem 15.3 that \( {B}_{n} \) is of first category in \( {Y}^{n} \) . Hence \( E = {X}^{n} \times {B}_{n} \) is of first category in \( X \) .
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Yes
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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 \( G \) such that \( F \subset G \) and \( \mu \left( G\right) < \mu \left( F\right) + \varepsilon \) .
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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 \( U \subset G \) . The maximality of \( \mathcal{F} \) implies that \( G \subset \bar{U} \) . Hence \( G - U \), which is contained in \( \bar{U} - U \), is nowhere dense, and so \( \mu \left( G\right) = \sum \mu \left( {U}_{i}\right) \) . Choose \( n \) so that \( \mathop{\sum }\limits_{1}^{n}\mu \left( {U}_{i}\right) \) \( > \mu \left( G\right) - \varepsilon \) . Then \( F = \mathop{\bigcup }\limits_{1}^{n}{\bar{U}}_{i} \) is a closed subset of \( G \), and \( \mu \left( F\right) > \mu \left( G\right) - \varepsilon \) . This proves the first assertion; the second follows by complementation. []
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Yes
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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.
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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 \( \mu \left( {G}_{ij}\right) < 1/{2}^{i + j} \) . Put \( {H}_{j} = \left( {\mathop{\bigcup }\limits_{{i = 1}}^{\infty }{G}_{ij}}\right. \) . Then \( {H}_{j} \) is open, \( P \subset {H}_{j} \), and \( \mu \left( {\bar{H}}_{j}\right) = \mu \left( {H}_{j}\right) < 1/{2}^{j} \) . Put \( F = \mathop{\bigcap }\limits_{{j = 1}}^{\infty }{\bar{H}}_{j} \) . Then \( F \) is closed and \( P \subset F \) . Since \( \mu \left( F\right) = 0 \), the interior of \( F \) must be empty. Hence \( F \), and therefore \( P \), is nowhere dense.
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Yes
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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\[ \n\mu \left( E\right) = \left\{ \begin{array}{l} \inf \{ \mu \left( G\right) : E \subset G, G\text{ open }\} \\ \sup \{ \mu \left( F\right) : E \supset F, F\text{ closed }\} . \end{array}\right.\n\]
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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 and last of these sets differ by a nowhere dense set, hence all have equal measure. This proves the first assertion; the second then follows from Theorem 22.1.
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Yes
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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.
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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 \( \left. {{N}_{1} \in \mathcal{N}\text{, then}\phi \left( {A}_{1}\right) \in \mathcal{N}\text{and so}\phi \left( {A}_{1}\right) = \varnothing \text{, by 2}}\right) \) and 3 \( ) \). Hence \( \phi \left( {A}_{1}\right) \) \( - {N}_{1} = \varnothing \), and therefore \( N \) is nowhere dense. Conversely, if \( F \) is closed and nowhere dense, then \( X - F = \phi \left( A\right) - N \) for some \( A \in S \) and \( N \in \mathcal{N} \) , hence \( F \) belongs to \( S \). Since\n\n\[ F \supset \phi \left( F\right) - \left\lbrack {\phi \left( F\right) - F}\right\rbrack \in \mathcal{T}, \]\n\nthe nowhere denseness of \( F \) implies that \( \phi \left( F\right) \subset \phi \left( F\right) - F \). Hence \( \phi \left( F\right) = \varnothing \) , by 1),2), and 3). Therefore \( F \sim \varnothing \), that is, \( F \in \mathcal{N} \). Thus, \( \mathcal{N} \) is identical with the class of closed nowhere dense sets. Since every nowhere dense set is contained in a closed nowhere dense set, and every subset of a member of \( \mathcal{N} \) belongs to \( \mathcal{N} \), it follows that every nowhere dense set is closed.
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Yes
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Theorem 22.7. A set \( A \subset X \) has the property of Baire if and only if \( A \in S \) .
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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. Conversely, if \( A \) has the property of Baire, then \( A = \left\lbrack {\phi \left( B\right) - N}\right\rbrack \vartriangle M \) for some \( B \in S \), some \( N \in \mathcal{N} \), and some set \( M \) of first category. By Theorem 22.6, \( M \) belongs to \( \mathcal{N} \), and therefore \( A \in S \) .
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Yes
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Theorem 22.8. A set \( G \subset X \) is regular open if and only if \( G = \phi \left( A\right) \) for some \( A \in S \) .
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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 \( \phi \left( A\right) \cup N \) . Then\n\n\[ \phi \left( {A}_{1}\right) - {N}_{1} \subset \phi \left( {A}_{1}\right) = \phi \left( {\phi \left( {A}_{1}\right) - {N}_{1}}\right) \subset \phi \left( {\phi \left( A\right) \cup N}\right) = \phi \left( A\right) . \]\n\nThus \( \phi \left( A\right) \) is the largest open subset of \( \phi \left( A\right) \cup N \) . This shows that \( \phi \left( A\right) \) is equal to the interior of its closure, that is, \( \phi \left( A\right) \) is regular open. Conversely, if \( G \) is regular open, then \( G = \phi \left( A\right) - N \) for some \( A \in S \) and \( N \in \mathcal{N} \) . Since \( \phi \left( A\right) \bigtriangleup \left\lbrack {\phi \left( A\right) - N}\right\rbrack \) is contained in \( N \), we have \( \phi \left( A\right) \sim \phi \left( A\right) - N = G \) . Since \( G \) and \( \phi \left( A\right) \) differ by a nowhere dense set, and both are regular open, it follows that \( G = \phi \left( A\right) \) .
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Yes
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