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Theorem 14.15.4 For a connected graph \( X \) on \( n \) vertices, the following assertions are equivalent:\n\n(a) \( X \) is pedestrian.\n\n(b) The Laplacian \( Q \) of \( X \) has binary rank \( n - 1 \) .\n\n(c) The number of spanning trees of \( X \) is odd.
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Proof. It follows at once from the previous lemma that (a) and (b) are equivalent.\n\nWe will prove that (b) and (c) are equivalent. If \( {\operatorname{rk}}_{2}\left( Q\right) = n - 1 \), then by Theorem 8.9.1, \( Q \) has a principal \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) submatrix of full rank. If we denote this submatrix by \( Q\left\lbrack u\right\rbrack \), then \( \det Q\left\lbrack u\right\rbrack ≢ 0{\;\operatorname{mod}\;2} \), and so over the integers, \( \det Q\left\lbrack u\right\rbrack \) is odd. By Theorem 13.2.1, we conclude that the number of spanning trees of \( X \) is odd.\n\nConversely, if the number of spanning trees of \( X \) is odd, then by Theorem 13.2.1, any principal \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) submatrix of \( Q \) has odd determinant over \( \mathbb{Z} \) . Therefore, it has nonzero determinant over \( {GF}\left( 2\right) \) , which implies that \( {\operatorname{rk}}_{2}\left( Q\right) \geq n - 1 \) .
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Yes
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Theorem 14.16.2 Let \( y \) be the column of the incidence matrix \( B \) corresponding to the edge \( e \), and consider the system of linear equations \( {Qx} = y \) . Then:\n\n(a) If \( {Qx} = y \) has no solution, then \( e \) is of bicycle-type.\n\n(b) If \( {Qx} = y \) has a solution \( x \), where \( {x}^{T}{Qx} \neq 0 \), then \( e \) is of cut-type.\n\n(c) If \( {Qx} = y \) has a solution \( x \), where \( {x}^{T}{Qx} = 0 \), then \( e \) is of flow-type.
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Proof. Identify \( e \) with its characteristic vector in \( {GF}{\left( 2\right) }^{E} \), and suppose that there is a vector \( x \) such that \( {Qx} = y \) . Then \( B{B}^{T}x = y = {Be} \), and as \( \ker B = F \), we have\n\n\[ \n{B}^{T}x \in F + e.\n\]\n\nNow, \( {B}^{T}x \) is a cut, and so this provides us with a representation of \( e \) as the symmetric difference of a cut and an even subgraph. Therefore, if \( e \) is of bicycle-type, the system of linear equations has no solution. If there is a solution, then the additional condition merely checks whether \( e \) is in the cut or not.
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No
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Theorem 14.16.3 In a pedestrian graph \( X \), the union of the cut-type edges is a cut, and the union of the flow-type edges is a flow.
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Proof. Since \( X \) is pedestrian, any edge \( e \) has a unique expression of the form\n\n\[ e = c\left( e\right) + f\left( e\right) \]\n\nwhere \( c\left( e\right) \in C \) and \( f\left( e\right) \in F \) . Define \( {c}^{ * } \) and \( {f}^{ * } \) by\n\n\[ {c}^{ * } \mathrel{\text{:=}} \mathop{\sum }\limits_{e}c\left( e\right) ,\;{f}^{ * } \mathrel{\text{:=}} \mathop{\sum }\limits_{e}f\left( e\right) .\n\nThen \( {c}^{ * } \) is the characteristic vector of a cut, \( {f}^{ * } \) is the characteristic vector of a flow, and\n\n\[ 1 = \mathop{\sum }\limits_{e}e = {c}^{ * } + {f}^{ * } \]\n\nUsing the fact that the characteristic vector of any cut is orthogonal to the characteristic vector of any flow, we obtain the following string of equalities:\n\n\[ c{\left( e\right) }^{T}e = c{\left( e\right) }^{T}\left( {c\left( e\right) + f\left( e\right) }\right) = c{\left( e\right) }^{T}c\left( e\right) \]\n\n\[ = c{\left( e\right) }^{T}\mathbf{1} \]\n\n\[ = c{\left( e\right) }^{T}\left( {{c}^{ * } + {f}^{ * }}\right) \]\n\n\[ = c{\left( e\right) }^{T}{c}^{ * } \]\n\n\[ = {\left( e + f\left( e\right) \right) }^{T}{c}^{ * } \]\n\n\[ = {e}^{T}{c}^{ * }\text{.} \]\n\nNow, \( c{\left( e\right) }^{T}e = 1 \) if and only if \( e \) is of cut-type, and therefore \( {e}^{T}{c}^{ * } = 1 \) if and only if \( e \) is of cut-type.
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Yes
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Lemma 15.2.1 If \( A \subseteq \Omega \), then the rank of \( A \) is the size of any maximal (with respect to inclusion) independent set contained in \( A \) .
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Proof. If \( A \) is itself independent, then the result follows immediately. So let \( {A}^{\prime } \) be any maximal independent set properly contained in \( A \) . Then for every element \( x \in A \smallsetminus {A}^{\prime } \) we have \( \operatorname{rk}\left( {{A}^{\prime }\cup \{ x\} }\right) = \operatorname{rk}\left( A\right) \) . Now, consider any subset \( B \) such that \( {A}^{\prime } \subset B \subseteq A \), and let \( x \in B \smallsetminus {A}^{\prime } \) . Then by applying the submodularity property to \( B \smallsetminus \{ x\} \) and \( {A}^{\prime } \cup \{ x\} \) we get\n\n\[ \operatorname{rk}\left( {A}^{\prime }\right) + \operatorname{rk}\left( B\right) \leq \operatorname{rk}\left( {B\smallsetminus \{ x\} }\right) + \operatorname{rk}\left( {{A}^{\prime }\cup \{ x\} }\right) ,\]\n\nand so \( \operatorname{rk}\left( B\right) = \operatorname{rk}\left( {B\smallsetminus \{ x\} }\right) \), which in turn implies that \( \operatorname{rk}\left( B\right) = \operatorname{rk}\left( {A}^{\prime }\right) \) . By taking \( B = A \) this yields that \( \operatorname{rk}\left( A\right) = \operatorname{rk}\left( {A}^{\prime }\right) = \left| {A}^{\prime }\right| \) .
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Yes
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Theorem 15.3.1 If \( \mathrm{{rk}} \) is a rank function on \( \Omega \), then the function \( {\mathrm{{rk}}}^{ \bot } \) given by\n\n\[ \n{\mathrm{{rk}}}^{ \bot }\left( A\right) \mathrel{\text{:=}} \left| A\right| + \mathrm{{rk}}\left( \bar{A}\right) - \mathrm{{rk}}\left( \Omega \right) \n\]\n\nis also a rank function on \( \Omega \) .
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Proof. If \( f \) is a function on the subsets of \( \Omega \), let \( \bar{f} \) be the function (on subsets of \( \Omega \) ) defined by\n\n\[ \n\bar{f}\left( A\right) \mathrel{\text{:=}} f\left( \bar{A}\right) \n\]\nIf \( A \) and \( B \) are subsets of \( \Omega \) and \( f \) is submodular, we have\n\n\[ \nf\left( \bar{A}\right) + f\left( \bar{B}\right) \geq f\left( {\bar{A} \cap \bar{B}}\right) + f\left( {\bar{A} \cup \bar{B}}\right) = f\left( \overline{A \cup B}\right) + f\left( \overline{A \cap B}\right) , \n\]\n\nwhich implies that \( \bar{f} \) is submodular. It is immediate that the sum of two submodular functions is submodular. Since the size of a set is a submodular function, it follows that, if \( f \) is submodular, then so is \( \bar{f} + \left| \cdot \right| \) .\n\nNext we show that if \( f \) is a rank function on \( \Omega \), then the function mapping a subset \( A \) to \( \bar{f}\left( A\right) + \left| A\right| \) is monotone. If \( A \subseteq B \), then\n\n\[ \nf\left( \bar{A}\right) = f\left( {\bar{B} \cup \left( {B \smallsetminus A}\right) }\right) \leq f\left( \bar{B}\right) + f\left( {B \smallsetminus A}\right) \leq f\left( \bar{B}\right) + \left| {B \smallsetminus A}\right| . \n\]\n\nHence\n\n\[ \n\bar{f}\left( A\right) + \left| A\right| \leq f\left( \bar{B}\right) + \left| {B \smallsetminus A}\right| + \left| A\right| = \bar{f}\left( B\right) + \left| B\right| . \n\]\n\nThus we have established that \( {\mathrm{{rk}}}^{ \bot } \) is monotone and submodular, and it remains only to show that \( {\mathrm{{rk}}}^{ \bot } \) is nonnegative and satisfies \( {\mathrm{{rk}}}^{ \bot }\left( A\right) \leq \left| A\right| \) . Because \( \mathrm{{rk}} \) is submodular and monotone,\n\n\[ \n\operatorname{rk}\left( \Omega \right) \leq \operatorname{rk}\left( \bar{A}\right) + \operatorname{rk}\left( A\right) \n\]\n\nand so\n\n\[ \n0 \leq \operatorname{rk}\left( \Omega \right) - \operatorname{rk}\left( \bar{A}\right) \leq \operatorname{rk}\left( A\right) \leq \left| A\right| \n\]\n\nTherefore, if we rewrite \( {\mathrm{{rk}}}^{ \bot } \) as\n\n\[ \n{\mathrm{{rk}}}^{ \bot }\left( A\right) = \left| A\right| - \left( {\mathrm{{rk}}\left( \Omega \right) - \mathrm{{rk}}\left( \bar{A}\right) }\right) \n\]\n\nwe see that\n\n\[ \n0 \leq {\operatorname{rk}}^{ \bot }\left( A\right) \leq \left| A\right| \n\]\n\nTherefore, \( {\mathrm{{rk}}}^{ \bot } \) is a rank function.
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Yes
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Lemma 15.3.2 The bases of \( {M}^{ \bot } \) are the complements of the bases of \( M \) .
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Proof. If \( A \) is a subset of \( \Omega \), then\n\n\[ \n{\mathrm{{rk}}}^{ \bot }\left( \bar{A}\right) = \left| \bar{A}\right| + \mathrm{{rk}}\left( A\right) - \mathrm{{rk}}\left( \Omega \right)\n\]\n\nand so if \( A \) is an independent set in \( M \), then \( \bar{A} \) is a spanning set in \( {M}^{ \bot } \) . Conversely, if \( \bar{A} \) is a spanning set in \( {M}^{ \bot } \), then \( A \) is independent in \( M \) . By duality, \( A \) is a spanning set in \( M \) if and only if \( \bar{A} \) is independent in \( {M}^{ \bot } \) . Therefore, \( A \) is a base of \( M \) if and only if \( \bar{A} \) is a base of \( {M}^{ \bot } \) .
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Yes
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Theorem 15.4.1 Let \( X \) be a graph, and let \( e \in E\left( X\right) \). Then \( M\left( X\right) \smallsetminus e = \) \( M\left( {X \smallsetminus e}\right) \) and \( M\left( X\right) /e = M\left( {X/e}\right) \).
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Proof. The result for deletion is a direct consequence of the definitions of deletion, so we need only consider the result for contraction. If \( e \) is a loop, then \( X/e = X \smallsetminus e \), and it is easy to see that \( M\left( X\right) /e = M\left( X\right) \smallsetminus e \), and so the result follows from the result for deletion. So suppose that \( e \) is not a loop. From the definition of \( \rho \), a set \( A \subseteq E\left( X\right) \smallsetminus e \) is independent in \( M\left( X\right) /e \) if and only if \( A \cup e \) is independent in \( M\left( X\right) \). It is not hard to see that \( A \cup e \) contains no cycles of \( X \) if and only if \( A \) contains no cycles of \( X/e \). Therefore, \( M\left( X\right) /e \) and \( M\left( {X/e}\right) \) have the same independent sets. \( ▱ \)
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Yes
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Lemma 15.5.2 If \( C \) is a code of dimension \( k \) in \( {\mathbb{F}}^{n} \), then \( M\left( {C}^{ \bot }\right) = \) \( M{\left( C\right) }^{ \bot } \) .
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Proof. Suppose that \( A \subseteq \Omega \) is a base of \( M\left( C\right) \) . Then there is a generator matrix for \( C \) such that the matrix formed by the columns corresponding to \( A \) is \( {I}_{k} \) . Hence by Lemma 15.5.1, there is a generator matrix \( H \) for \( {C}^{ \bot } \) where the matrix formed by the columns corresponding to \( \bar{A} \) is \( {I}_{n - k} \) . Therefore, \( \bar{A} \) is a base of \( M\left( {C}^{ \bot }\right) \) . By duality, if \( \bar{A} \) is a base of \( M\left( {C}^{ \bot }\right) \) , then \( A \) is a base of \( M\left( C\right) \) . Hence the bases of \( M\left( {C}^{ \bot }\right) \) are precisely the complements of the bases of \( M\left( C\right) \), and the result follows.
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Yes
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Theorem 15.5.3 If \( C/i \) is the code obtained by shortening \( C \) at coordinate position \( i \), then \( M\left( {C/i}\right) = M\left( C\right) /i \) .
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Proof. Let \( G \) be the generator matrix for \( C \) . If \( i \) is a loop, then the \( i \) th column of \( G \) is zero, and puncturing and shortening the code are the same thing, and so \( M\left( {C/i}\right) = M\left( C\right) \smallsetminus i = M\left( C\right) /i \) . So suppose that \( i \) is not a loop. Using row operations if necessary, we can assume that the \( i \) th column of \( G \) contains a single nonzero entry in the first row and that the remaining \( k - 1 \) rows of \( G \) form a generator matrix for \( C/i \) . Recall that if \( \mathrm{{rk}} \) is the rank function for \( M\left( C\right) \), then \( \operatorname{rk}\left( {A \cup i}\right) - 1 \) is the rank function for \( M\left( {C/i}\right) \) . Then it is clear that \( A \) is independent in \( M\left( C\right) /i \) if and only if \( A \cup i \) is independent in \( M\left( C\right) \) if and only if \( A \) is independent in \( M\left( {C/i}\right) \) . Hence \( M\left( {C/i}\right) = M\left( C\right) /i \) .
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Yes
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Theorem 15.6.1 If \( X \) is a graph, then the number \( \kappa \left( X\right) \) of acyclic orientations of \( X \) is given by\n\n\[ \kappa \left( X\right) = \left\{ \begin{array}{ll} 0, & \text{ if }X\text{ contains a loop; } \\ {2\kappa }\left( {X/e}\right) , & \text{ if }e\text{ is a cut-edge; } \\ \kappa \left( {X \smallsetminus e}\right) + \kappa \left( {X/e}\right) ; & \text{ if }e\text{ is not a loop or a cut-edge. } \end{array}\right. \]
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Proof. If \( X \) contains a loop, then this loop forms a directed cycle in any orientation. If \( e \) is a cut-edge in \( X \), then any acyclic orientation of \( X \) can be formed by taking an acyclic orientation of \( X/e \) and orienting \( e \) in either direction. Suppose, then, that \( e = {uv} \) is neither a loop nor a cut-edge. If \( \sigma \) is an acyclic orientation of \( X \), then \( \tau = \sigma \upharpoonright \left( {V\left( X\right) \smallsetminus e}\right) \) is an acyclic orientation of \( X \smallsetminus e \) . We partition the orientations of \( X \smallsetminus e \) according to whether or not there is a path respecting the orientation joining the end-vertices of \( e \) (either from \( u \) to \( v \), or \( v \) to \( u \) ; clearly, there cannot be both). If there is no path joining the end-vertices of \( e \), then \( \tau \) is an acyclic orientation of \( X/e \) , and so there are \( \kappa \left( {X/e}\right) \) such acyclic orientations, leaving \( \kappa \left( {X \smallsetminus e}\right) - \kappa \left( {X/e}\right) \) remaining acyclic orientations. If \( \tau \) is in the former category, then it is the restriction of two acyclic orientations of \( X \) (because \( e \) may be oriented in either direction), and if it is in the latter category, then it is the restriction of one acyclic orientation of \( X \) (since \( e \) is oriented in the same direction as the directed path joining its end-vertices). Therefore,\n\n\[ \kappa \left( X\right) = {2\kappa }\left( {X/e}\right) + \left( {\kappa \left( {X \smallsetminus e}\right) - \kappa \left( {X/e}\right) }\right) = \kappa \left( {X/e}\right) + \kappa \left( {X \smallsetminus e}\right) ,\] \n\nas claimed.
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Yes
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Theorem 15.6.2 If \( X \) is a graph, then the number \( \kappa \left( {X, v}\right) \) of acyclic orientations of \( X \) where \( v \) is the unique source is given by\n\n\[ \kappa \left( {X, v}\right) = \left\{ \begin{array}{ll} 0, & \text{ if }X\text{ contains a loop; } \\ \kappa \left( {X/e, v}\right) , & \text{ if }e\text{ is a cut-edge; } \\ \kappa \left( {X \smallsetminus e, v}\right) + \kappa \left( {X/e, v}\right) ; & \text{ if }e\text{ is not a loop or a cut-edge. } \end{array}\right. \]
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Proof. This is left as Exercise 8.
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No
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Theorem 15.7.1 Assume that \( C \) is a binary code of length \( n \) and let \( {e}_{i} \) denote the ith standard basis vector of \( {GF}{\left( 2\right) }^{n} \) . Then exactly one of the following holds:\n\n(a) \( i \) lies in the support of a bicycle,\n\n(b) \( i \) lies in the support of a codeword \( \gamma \) such that \( \gamma + {e}_{i} \in {C}^{ \bot } \),\n\n(c) \( i \) lies in the support of a word \( \varphi \) in \( {C}^{ \bot } \) such that \( \varphi + {e}_{i} \in C \) .
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Proof. If \( i \) does not lie in the support of a bicycle, then \( \left\langle {{e}_{i},\beta }\right\rangle = 0 \) for each element \( \beta \) of \( C \cap {C}^{ \bot } \), and consequently\n\n\[ {e}_{i} \in {\left( C \cap {C}^{ \bot }\right) }^{ \bot } = {C}^{ \bot } + C.\]\n\nThus \( {e}_{i} = \gamma + \varphi \), where \( \gamma \in C \) and \( \varphi \in {C}^{ \bot } \) .\n\nSuppose we also have \( {e}_{i} = {\gamma }^{\prime } + {\varphi }^{\prime } \), where \( {\gamma }^{\prime } \in C \) and \( {\varphi }^{\prime } \in {C}^{ \bot } \) . Then\n\n\[ \gamma + {\gamma }^{\prime } = \varphi + {\varphi }^{\prime } \in C \cap {C}^{ \bot }.\]\n\nTherefore, \( i \) is in the support of both or neither of \( \gamma \) and \( {\gamma }^{\prime } \), and similarly for \( \varphi \) and \( {\varphi }^{\prime } \) . Hence precisely one of the last two conditions holds.
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Yes
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Lemma 15.7.3 Let \( i \) be an element of the code \( C \) . Then\n\n\[ \n\operatorname{bike}\left( C\right) = \left\{ \begin{array}{ll} \left( {-1}\right) \operatorname{bike}\left( {C \smallsetminus i}\right) , & \text{ if }i\text{ is a loop; } \\ \left( {-1}\right) \operatorname{bike}\left( {C/i}\right) , & \text{ if }i\text{ is a coloop; } \\ \operatorname{bike}\left( {C \smallsetminus i}\right) + \operatorname{bike}\left( {C/i}\right) , & \text{ otherwise. } \end{array}\right.\n\]
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Proof. If \( i \) is a loop or a coloop, then \( b\left( {C \smallsetminus i}\right) = b\left( {C/i}\right) = b\left( C\right) \), whence the first two claims follow. So suppose that \( i \) is neither a loop nor coloop, and assume that \( \ell \left( C\right) = n \) and \( b\left( C\right) = d \) . If \( i \) is of cut-type, then\n\n\[ \n\operatorname{bike}\left( {C \smallsetminus i}\right) + \operatorname{bike}\left( {C/i}\right) = {\left( -1\right) }^{n - 1}{\left( -2\right) }^{d + 1} + {\left( -1\right) }^{n - 1}{\left( -2\right) }^{d}\n\]\n\n\[ \n= {\left( -1\right) }^{n}{\left( -2\right) }^{d}.\n\]\n\nIf \( i \) is of flow-type, then\n\n\[ \n\text{bike}\left( {C \smallsetminus i}\right) + \text{bike}\left( {C/i}\right) = {\left( -1\right) }^{n - 1}{\left( -2\right) }^{d} + {\left( -1\right) }^{n - 1}{\left( -2\right) }^{d + 1}\n\]\n\n\[ \n= {\left( -1\right) }^{n}{\left( -2\right) }^{d}.\n\]\n\nFinally, if \( i \) lies in the support of a bicycle, then\n\n\[ \n\operatorname{bike}\left( {C \smallsetminus i}\right) + \operatorname{bike}\left( {C/i}\right) = 2{\left( -1\right) }^{n - 1}{\left( -2\right) }^{d - 1}\n\]\n\n\[ \n= {\left( -1\right) }^{n}{\left( -2\right) }^{d}.\n\]\n\nThus in all cases, \( \operatorname{bike}\left( {C \smallsetminus i}\right) + \operatorname{bike}\left( {C/i}\right) = \operatorname{bike}\left( C\right) \) .
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Yes
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Theorem 15.8.1 Let \( P\left( {X, t}\right) \) denote the number of proper colourings of \( X \) with \( t \) colours. Then\n\n\[ P\left( {X, t}\right) = \left\{ \begin{array}{ll} 0, & \text{ if }X\text{ contains a loop } \\ P\left( {X \smallsetminus e, t}\right) - P\left( {X/e, t}\right) , & \text{ if }e\text{ is not a loop. } \end{array}\right. \]
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Proof. If \( X \) is empty, then \( P\left( {X, t}\right) = {t}^{n} \) . If \( X \) contains a loop, then it has no proper colourings. Otherwise, suppose that \( u \) and \( v \) are the ends of the edge \( e \) . Any \( t \) -colouring of \( X \smallsetminus e \) where \( u \) and \( v \) have different colours is a proper \( t \) -colouring of \( X \), while colourings where \( u \) and \( v \) are assigned the same colour are in one-to-one correspondence with colourings of \( X/e \) . Therefore,\n\n\[ P\left( {X \smallsetminus e, t}\right) = P\left( {X, t}\right) + P\left( {X/e, t}\right) \]\n\nand the result follows.
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Yes
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Theorem 15.8.2 Let \( C\left( {X, p}\right) \) denote the probability that no component of \( X \) is disconnected when each edge of \( X \) is deleted independently with probability p. Then\n\n\[ C\left( {X, p}\right) = \left\{ \begin{array}{ll} C\left( {X \smallsetminus e, p}\right) , & \text{ if }e\text{ is a loop; } \\ \left( {1 - p}\right) C\left( {X/e, p}\right) , & \text{ if }e\text{ is a cut-edge; } \\ {pC}\left( {X \smallsetminus e, p}\right) + \left( {1 - p}\right) C\left( {X/e, p}\right) , & \text{ otherwise. } \end{array}\right. \]
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Proof. In each case we consider the two possibilities that \( e \) is deleted or not deleted and sum the conditional probabilities that no component of \( X \) is disconnected given the fate of \( e \) . For example, if \( e \) is a cut-edge, then the number of components of \( X \) remains the same if and only if \( e \) survives (probability \( \left( {1 - p}\right) \) ) and the deleted edges do not form a cut in the rest of the graph (probability \( C\left( {X/e, p}\right) \) ). Arguments for the other cases are broadly similar.
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Yes
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Theorem 15.9.1 If \( M \) is a matroid on \( \Omega \) and \( e \in \Omega \), then\n\n\[ R\left( {M;x, y}\right) = \left\{ \begin{array}{ll} \left( {1 + y}\right) R\left( {M \smallsetminus e;x, y}\right) , & \text{ if }e\text{ is a loop; } \\ \left( {1 + x}\right) R\left( {M/e;x, y}\right) , & \text{ if }e\text{ is a coloop; } \\ R\left( {M \smallsetminus e;x, y}\right) + R\left( {M/e;x, y}\right) , & \text{ otherwise. } \end{array}\right. \]
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Proof. If \( e \) is a loop or coloop, then this follows directly from the definition of the rank polynomial. If \( e \) is neither a loop nor a coloop, then the subsets of \( \Omega \) that do not contain \( e \) contribute \( R\left( {M \smallsetminus e;x, y}\right) \) to the rank polynomial, while the subsets that do contain \( e \) contribute \( R\left( {M/e;x, y}\right) \) .\n\nTherefore, the rank polynomial of any matroid can be calculated using the deletion-contraction algorithm.
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Yes
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Lemma 15.10.1 The number of spanning trees in a connected graph \( X \) is
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\[ \tau \left( X\right) = R\left( {M\left( X\right) ;0,0}\right) . \]
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No
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Lemma 15.10.2 The number of acyclic orientations of a graph \( X \) is\n\n\[ \kappa \left( X\right) = R\left( {M\left( X\right) ;1, - 1}\right) . \]
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Proof. Using Theorem 15.9.3 and the deletion-contraction expression given in Theorem 15.6.1, we see that\n\n\[ a\left( {1 + y}\right) = 0 \]\n\n\[ b\left( {1 + x}\right) = 2 \]\n\n\[ a = 1 \]\n\n\[ b = 1 \]\n\nand so \( x = 1 \) and \( y = - 1 \) .
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No
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Theorem 15.10.3 Let \( C \) be a binary code of length \( n \), with a bicycle space of dimension \( d \) . Then\n\n\[{\left( -1\right) }^{n}{\left( -2\right) }^{d} = R\left( {M\left( C\right) ; - 2, - 2}\right) .
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Proof. This follows directly from Theorem 15.9.3 and Lemma 15.7.3.
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No
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Theorem 15.10.4 If \( X \) is a graph with \( n \) vertices, \( c \) components, and chromatic polynomial \( P\left( {X, t}\right) \), then\n\n\[ P\left( {X, t}\right) = {\left( -1\right) }^{n - c}{t}^{c}R\left( {M\left( X\right) ; - t, - 1}\right) . \]
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Proof. We cannot work directly with \( P\left( {X, t}\right) \) because it does not satisfy the conditions of Theorem 15.9.3. Instead, consider the function\n\n\[ \widehat{P}\left( {X, t}\right) \mathrel{\text{:=}} {t}^{-c}P\left( {X, t}\right) \]\n\nwhich takes the value 1 on the empty matroid, that is, on a graph with no edges. If \( e \) is a coloop, then some thought shows that \( P\left( {X \smallsetminus e, t}\right) = \) \( {tP}\left( {X/e, t}\right) \), and therefore putting \( \widehat{P} \) into Theorem 15.9.3, we get\n\n\[ a\left( {1 + y}\right) = 0 \]\n\n\[ b\left( {1 + x}\right) = \left( {t - 1}\right) \]\n\n\[ a = 1 \]\n\n\[ b = - 1 \]\n\nand therefore \( y = - 1 \) and \( x = - t \) . Multiplying by \( {t}^{c} \) to recover \( P\left( {X, t}\right) \) gives the stated result.
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Yes
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Theorem 15.11.1 Let \( C \) be a linear code over the finite field \( {GF}\left( q\right) \) with weight enumerator \( W\left( {C, t}\right) \) . Then\n\n\[ W\left( {C, t}\right) = \left\{ \begin{array}{ll} W\left( {C \smallsetminus i, t}\right) , & \text{ if }i\text{ is a loop in }M\left( C\right) ; \\ \left( {1 + \left( {q - 1}\right) t}\right) W\left( {C/i, t}\right) , & \text{ if }i\text{ is a coloop in }M\left( C\right) ; \\ {tW}\left( {C \smallsetminus i, t}\right) + \left( {1 - t}\right) W\left( {C/i, t}\right) , & \text{ otherwise. } \end{array}\right. \]
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Proof. The result is clear if \( i \) is a loop. If \( i \) is a coloop, then consider a generator matrix \( G \) in the form of (15.1). Any codeword of \( C \) is either zero or nonzero in coordinate position \( i \) . The codewords that are zero in this position contribute \( W\left( {C/i, t}\right) \) to the weight enumerator. All other codewords consist of a codeword in \( C/i \) plus a nonzero multiple of the first row of \( G \) . These codewords contribute \( \left( {q - 1}\right) {tW}\left( {C/i, t}\right) \) to the weight enumerator. Now, suppose that \( i \) is neither a loop nor a coloop, and so there is a one-toone correspondence between codewords of \( C \smallsetminus i \) and \( C \) . If every codeword of \( C \) had nonzero \( i \) th coordinate, then the weight enumerator of \( C \) would be \( {tW}\left( {C \smallsetminus i, t}\right) \) . However, every codeword does not have nonzero \( i \) th coordinate, and those with \( i \) th coordinate zero actually contribute \( W\left( {C/i, t}\right) \) to the weight enumerator rather than \( {tW}\left( {C/i, t}\right) \) . Subtracting the overcount gives the stated result.
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Yes
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If \( M \) is the matroid of a linear code \( C \) of length \( n \) and dimension \( k \) over the field \( {GF}\left( q\right) \), then the number of codewords of weight \( n \) in \( C \) is \[ {\left( -1\right) }^{k}R\left( {M; - q, - 1}\right) \text{.} \]
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Proof. Let \( \sigma \left( C\right) \) denote the number of codewords with no zero entries in a code \( C \) . It is straightforward to see from Theorem 15.11.1 that \[ \sigma \left( C\right) = \left\{ \begin{array}{ll} 0, & \text{ if }i\text{ is a loop in }M\left( C\right) ; \\ \left( {q - 1}\right) \sigma \left( {C/i}\right) , & \text{ if }i\text{ is a coloop in }M\left( C\right) ; \\ \sigma \left( {C \smallsetminus i}\right) - \sigma \left( {C/i}\right) , & \text{ otherwise. } \end{array}\right. \] The result then follows immediately from Theorem 15.9.3.
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No
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Lemma 15.12.2 Let \( C \) be a code of length \( n \) over a field \( q \) and let \( M \) be the matroid of \( C \) . Then the number of ordered \( r \) -tuples of codewords in \( C \) such that the union of the supports of the codewords in the \( r \) -tuple has size \( n \) is
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\[ {\left( -1\right) }^{k}R\left( {M; - {q}^{r}, - 1}\right) \text{.} \]
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Yes
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Theorem 15.13.3 Let \( \widehat{R}\left( X\right) \) be the modified rank polynomial of the signed graph \( X \). Then\n\n\[ \widehat{R}\left( X\right) = \left\{ \begin{array}{ll} \widehat{R}\left( e\right) \widehat{R}\left( {X \smallsetminus e}\right) , & \text{ if }e\text{ is a loop; } \\ \alpha \widehat{R}\left( {X \smallsetminus e}\right) + \beta \widehat{R}\left( {X/e}\right) , & \text{ if }e\text{ is positive and not a loop; } \\ \beta \widehat{R}\left( {X \smallsetminus e}\right) + \alpha \widehat{R}\left( {X/e}\right) , & \text{ if }e\text{ is negative and not a loop. } \end{array}\right. \]
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Proof. If \( e \) is a coloop in \( X \), then\n\n\[ {x}^{-c}\widehat{R}\left( {X \smallsetminus e}\right) = R\left( {X \smallsetminus e}\right) = R\left( {X/e}\right) = {x}^{-c + 1}\widehat{R}\left( {X/e}\right) ,\]\n\nand so \( \widehat{R}\left( {X \smallsetminus e}\right) = x\widehat{R}\left( {X/e}\right) \). Since \( R\left( X\right) = R\left( e\right) R\left( {X/e}\right) \), it follows that if \( e \) is a positive coloop, then \( \widehat{R}\left( X\right) = \left( {{\alpha x} + \beta }\right) \widehat{R}\left( {X/e}\right) \), and so\n\n\[ \widehat{R}\left( X\right) = \alpha \widehat{R}\left( {X \smallsetminus e}\right) + \beta \widehat{R}\left( {X/e}\right) \]\n\nSimilarly, if \( e \) is a negative coloop, then\n\n\[ \widehat{R}\left( X\right) = \beta \widehat{R}\left( {X \smallsetminus e}\right) + \alpha \widehat{R}\left( {X/e}\right) \]\n\nIt is easy to verify that the last two recurrences hold true for positive and negative edges that are neither loops nor coloops.
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Yes
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If \( R \) is a rotor of order less than six, then the cycle matroids of \( X \) and \( {X}^{\varphi } \) have the same rank polynomial.
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We describe a bijection between subsets of edges of \( X \) and subsets of edges of \( {X}^{\varphi } \) that preserves both size and rank.\n\nLet \( A = {A}_{R} \cup {A}_{S} \) be a subset of \( E\left( R\right) \cup E\left( S\right) \), where \( {A}_{R} \subseteq E\left( R\right) \) and \( {A}_{S} \subseteq E\left( S\right) \) . The connected components of the subgraph of \( R \) with edge-set\n\n\n\nFigure 15.5. The graph \( X \) obtained from \( R \) and \( S \)\n\n\( {A}_{R} \) induce a partition \( \rho \) on the vertices \( \left\{ {{x}_{0},\ldots ,{x}_{n - 1}}\right\} \), and the connected components of the subgraph of \( S \) with edge-set \( {A}_{S} \) induce a partition \( \sigma \) on the vertices \( \left\{ {{y}_{0},\ldots ,{y}_{n - 1}}\right\} \) .\n\nThe number of connected components of \( A \) depends on the number of components that do not contain any vertices of \( N \), and the number of cells in the join \( \rho \vee \sigma \) .\n\nIt is straightforward to check that any partition of a set of size five or less is invariant under at least one of the \
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Yes
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Theorem 15.15.1 Let \( f \) be an integer-valued monotone submodular function on the subsets of \( \Omega \) such that \( f\left( \varnothing \right) = 0 \) . The function \( \widehat{f} \) on the subsets of \( \Omega \) defined by\n\n\[ \widehat{f}\left( A\right) \mathrel{\text{:=}} \min \{ f\left( {A \smallsetminus B}\right) + \left| B\right| : B \subseteq A\} \]\n\nis a rank function.
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Proof. Clearly, \( \widehat{f} \) is nonnegative. We show that it is monotone, submodular and that \( \widehat{f}\left( A\right) \leq \left| A\right| \) . First, suppose that \( A \subseteq {A}_{1} \) . If \( B \subseteq {A}_{1} \), then \( A \smallsetminus B \subseteq \) \( {A}_{1} \smallsetminus B \), and since \( f \) is monotone,\n\n\[ f\left( {{A}_{1} \smallsetminus B}\right) + \left| B\right| \geq f\left( {A \smallsetminus B}\right) + \left| {A \cap B}\right| .\n\nTherefore, \( \widehat{f}\left( {A}_{1}\right) \geq \widehat{f}\left( A\right) \), which shows that \( \widehat{f} \) is monotone.\n\nNow, let \( A \) and \( B \) be subsets of \( \Omega \) . Suppose that \( C \subseteq A \) and \( D \subseteq B \) are the subsets that provide the values for \( \widehat{f}\left( A\right) \) and \( \widehat{f}\left( B\right) \), respectively. Let \( W = \left( {C \smallsetminus B}\right) \cup \left( {D \smallsetminus A}\right) \cup \left( {C \cap D}\right) \) and \( X = \left( {C \cup D}\right) \cap \left( {A \cap B}\right) \), and observe that \( \left| C\right| + \left| D\right| = \left| W\right| + \left| X\right| \) . Then\n\n\[ \widehat{f}\left( A\right) + \widehat{f}\left( B\right) = f\left( {A \smallsetminus C}\right) + f\left( {B \smallsetminus D}\right) + \left| C\right| + \left| D\right| \]\n\n\[ \geq f\left( {\left( {A \backslash C}\right) \cup \left( {B \backslash D}\right) }\right) + f\left( {\left( {A \backslash C}\right) \cap \left( {B \backslash D}\right) }\right) + \left| C\right| + \left| D\right| \]\n\n\[ = f\left( {\left( {A \cup B}\right) \smallsetminus W}\right) + f\left( {\left( {A \cap B}\right) \smallsetminus X}\right) + \left| W\right| + \left| X\right| \]\n\n\[ \geq \widehat{f}\left( {A \cup B}\right) + \widehat{f}\left( {A \cap B}\right) \]\n\nand so \( \widehat{f} \) is submodular.\n\nFinally, by taking \( B \) equal to \( A \) in the definition of \( \widehat{f} \), we see that \( \widehat{f}\left( A\right) \leq \) \( \left| A\right| \), and therefore \( \widehat{f} \) is a rank function.\n\nTherefore, any integer-valued monotone submodular function \( f \) such that \( f\left( \varnothing \right) = 0 \) determines a matroid \( M \) whose rank function is \( \widehat{f} \) .
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Yes
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Corollary 15.15.2 Let \( f \) be an integer-valued monotone submodular function on the subsets of \( \Omega \) such that \( f\left( \varnothing \right) = 0 \), and let \( M \) be the matroid determined by the rank function \( \widehat{f} \) . A subset \( A \) of \( \Omega \) is independent in \( M \) if and only if \( f\left( B\right) \geq \left| B\right| \) for all subsets \( B \) of \( A \) .
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Proof. A subset \( A \) of \( \Omega \) is independent in \( M \) if and only if \( f\left( {A \smallsetminus B}\right) + \left| B\right| \geq \) \( \left| A\right| \) for all subsets \( B \) of \( A \), or, equivalently, if and only if \( f\left( {A \smallsetminus B}\right) \geq \left| {A \smallsetminus B}\right| \) for all subsets \( B \) .
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Yes
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Lemma 15.15.4 Let \( M \) be the union of the matroids \( {M}_{1} \) and \( {M}_{2} \) on \( \Omega \) . \( A \) subset \( A \) of \( \Omega \) is independent in \( M \) if and only if \( A = {A}_{1} \cup {A}_{2} \), where \( {A}_{i} \) is independent in \( {M}_{i} \) .
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Proof. Let \( {\mathrm{{rk}}}_{1} \) and \( {\mathrm{{rk}}}_{2} \) be the rank functions of \( {M}_{1} \) and \( {M}_{2} \) respectively. Suppose that \( A \) is the union of the sets \( {A}_{1} \) and \( {A}_{2} \), where \( {A}_{i} \) is independent in \( {M}_{i} \) . For any \( B \subseteq A \), set \( {B}_{i} \mathrel{\text{:=}} B \cap {A}_{i} \) . Then\n\n\[ \left| B\right| \leq \left| {B}_{1}\right| + \left| {B}_{2}\right| = {\operatorname{rk}}_{1}\left( {B}_{1}\right) + {\operatorname{rk}}_{2}\left( {B}_{2}\right) \leq {\operatorname{rk}}_{1}\left( B\right) + {\operatorname{rk}}_{2}\left( B\right) ,\]\n\nand by Corollary 15.15.2, \( A \) is independent in \( {M}_{1} \vee {M}_{2} \) .\n\nThe converse takes more work. Let \( N \) denote the direct sum \( {M}_{1} \oplus {M}_{2} \) , with ground set \( \Phi \) consisting of two disjoint copies of \( \Omega \) . Let \( \Gamma \) denote a third copy of \( \Omega \) and let \( X \) be the bipartite graph with bipartition \( \left( {\Gamma ,\Phi }\right) \) , where the \( i \) th element of \( \Gamma \) is joined to the \( i \) th elements in each of the two copies of \( \Omega \) in \( \Phi \) . This is not a very interesting graph, but the matroid induced from \( {M}_{1} \oplus {M}_{2} \) by \( X \) is \( {M}_{1} \vee {M}_{2} \) . From this it follows that a subset \( A \) of \( \Gamma \) is independent in \( {M}_{1} \vee {M}_{2} \) if and only if there is a matching in \( X \) that pairs the vertices of \( A \) with an independent set in \( \Phi \), whence we infer that \( A \) is independent if and only if it is the union of two sets, one independent in \( {M}_{1} \) and the other in \( {M}_{2} \) .
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Yes
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Theorem 15.15.5 A connected graph \( X \) has \( k \) edge-disjoint spanning trees if and only if for every partition \( \pi \) of the vertices, \[ e\left( {X,\pi }\right) \geq k\left( {\left| \pi \right| - 1}\right) \]
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Proof. For any partition \( \pi \), it is easy to see that a spanning tree contributes at least \( \left| \pi \right| - 1 \) edges to \( e\left( {X,\pi }\right) \), and so if \( X \) has \( k \) edge-disjoint spanning trees, the inequality holds.\n\nFor the converse, suppose that \( X \) has \( n \) vertices, and that \( \mathrm{{rk}} \) is the rank function of the cycle matroid \( M\left( X\right) \) . Let \( {kM} = M\left( X\right) \vee \cdots \vee M\left( X\right) \) denote the \( k \) -fold union of \( M\left( X\right) \) with itself. By the previous result, \( X \) has \( k \) edge-disjoint spanning trees if and only if the rank of \( {kM} \) is \( k\left( {n - 1}\right) \) . By the definition of the rank function of \( {kM} \), this is true if and only if for all \( A \subseteq E\left( X\right) \)\n\n\[ k\operatorname{rk}\left( A\right) + \left| \bar{A}\right| \geq k\left( {n - 1}\right) . \]\n\nIf \( A \) is a subset of \( E\left( X\right) \), then let \( \pi \) be the partition whose cells are the connected components of the graph with vertex set \( V\left( X\right) \) and edge set \( A \) . Then \( \operatorname{rk}\left( A\right) = n - \left| \pi \right| \) and \( e\left( {X,\pi }\right) = \left| \bar{A}\right| \), and so\n\n\[ k\operatorname{rk}\left( A\right) + \left| \bar{A}\right| \geq k\left( {n - \left| \pi \right| }\right) + k\left( {\left| \pi \right| - 1}\right) = k\left( {n - 1}\right) ,\]\n\nand thus the result follows.
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Yes
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Theorem 16.2.1 Two link diagrams determine the same link if and only if one can be obtained from the other by a sequence of Reidemeister moves and planar isotopies.
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We omit the proof of this because it is entirely topological; nonetheless, it is neither long nor especially difficult. This result shows that equivalence of links can be determined entirely from consideration of link diagrams. We regard this as justification for our focus on link diagrams, which are 2-dimensional combinatorial objects, rather than links themselves, which are 3-dimensional topological objects. Therefore, to show that two links are equivalent it is sufficient to present a sequence of Reidemeister moves leading from a diagram of one link to a diagram of the other.
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No
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Lemma 16.5.1 Let \( Y \) be a signed graph and let \( e \) and \( f \) be parallel edges of opposite sign, not loops. If \( {\alpha \beta } = 1 \) and \( y = - \left( {{\alpha }^{2} + {\alpha }^{-2}}\right) \), then \( \widehat{R}\left( Y\right) = \widehat{R}\left( {Y\smallsetminus \{ e, f\} }\right) \) .
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Proof. Without loss of generality, suppose that \( e \) is positive. Then by Theorem 15.13.3,\n\n\[ \widehat{R}\left( Y\right) = \alpha \widehat{R}\left( {Y \smallsetminus e}\right) + \beta \widehat{R}\left( {Y/e}\right) \]\n\nSince \( f \) has the opposite sign to \( e \) and is an edge in \( Y \smallsetminus e \) and a loop in \( Y/e \) , we have\n\n\[ \widehat{R}\left( Y\right) = \alpha (\beta \widehat{R}\left( {Y\smallsetminus \{ e, f\} }\right) + \alpha \widehat{R}\left( {\left( {Y \smallsetminus e}\right) /f}\right) + \beta \left( {{\alpha y} + \beta }\right) \widehat{R}\left( {\left( {Y/e}\right) \smallsetminus f}\right) .\n\nSince \( \left( {Y \smallsetminus e}\right) /f = \left( {Y/e}\right) \smallsetminus f \), the lemma follows, provided that \( {\alpha \beta } = 1 \) and\n\n\[ {\alpha }^{2} + {\beta }^{2} + {\alpha \beta y} = 0 \]\n\nwhich follow from the given conditions.
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Yes
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Lemma 16.5.3 If \( Y \) is a signed graph, \( {\alpha \beta } = 1 \), and \( x = y = - \left( {{\alpha }^{2} + {\alpha }^{-2}}\right) \) , then \( \widehat{R}\left( {Y;\alpha ,\beta, x, y}\right) \) is invariant under Reidemeister moves of type III.
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Proof. Let \( Y \) be a graph containing a star with two negative legs and a positive leg, and \( {Y}^{\prime } \) the graph obtained by performing the star-triangle move on \( Y \) as shown in Figure 16.13. Let \( Z \) be the graph we get from \( Y \smallsetminus e \) by contracting \( f \) and \( g \) . Alternatively, it is the graph obtained from \( {Y}^{\prime } \) by contracting \( \widehat{e} \) and deleting \( \widehat{f} \) and \( \widehat{g} \) .\n\nBy applying Theorem 15.13.3 to the edge \( e \) in \( Y \) we find that\n\n\[ \widehat{R}\left( Y\right) = \beta \widehat{R}\left( {Y \smallsetminus e}\right) + \alpha \widehat{R}\left( {Y/e}\right) \]\n\nApplying Lemma 16.5.2 to the edges \( f \) and \( g \) in \( Y \backslash e \), we see that \( \widehat{R}\left( {Y \backslash e}\right) = \) \( \widehat{R}\left( Z\right) \), and therefore\n\n\[ \widehat{R}\left( Y\right) = \beta \widehat{R}\left( Z\right) + \alpha \widehat{R}\left( {Y/e}\right) \]\n\nBy applying Theorem 15.13.3 to the edge \( \widehat{e} \) in \( {Y}^{\prime } \) we find that\n\n\[ \widehat{R}\left( {Y}^{\prime }\right) = \alpha \widehat{R}\left( {{Y}^{\prime } \smallsetminus \widehat{e}}\right) + \beta \widehat{R}\left( {{Y}^{\prime }/\widehat{e}}\right) . \]\n\nIn \( {Y}^{\prime }/\widehat{e} \), the edges \( \widehat{f} \) and \( \widehat{g} \) are parallel with opposite sign. Hence Lemma 16.5.1 yields that \( \widehat{R}\left( {{Y}^{\prime }/\widehat{e}}\right) = \widehat{R}\left( Z\right) \) . Since \( {Y}^{\prime } \smallsetminus \widehat{e} = Y/e \), we conclude that \( \widehat{R}\left( Y\right) = \widehat{R}\left( {Y}^{\prime }\right) \) .
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Yes
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Theorem 16.7.1 Let \( L \) be a link diagram of an oriented link. Then\n\n\[ \n{\left( -{\alpha }^{3}\right) }^{\mathrm{{wr}}\left( L\right) }\left\lbrack L\right\rbrack \n\]\n\nis invariant under all Reidemeister moves, and hence is an invariant of the oriented link.
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Proof. The expressions \( \left\lbrack L\right\rbrack \) and \( {\left( -{\alpha }^{-3}\right) }^{\operatorname{wr}\left( L\right) } \) are both regular isotopy invariants. A Reidemeister move of type I has the same effect on both expressions, and hence their ratio is invariant under all three Reidemeister moves.
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Yes
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Theorem 16.8.3 Let \( {Y}_{1} \) and \( {Y}_{2} \) be signed graphs that are related by a Whitney flip. Then their rank polynomials are equal.
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Proof. The graphs \( {Y}_{1} \) and \( {Y}_{2} \) have the same edge set, and it is clear that a set \( S \subseteq E\left( {Y}_{1}\right) \) is independent in \( M\left( {Y}_{1}\right) \) if and only if it is independent in \( M\left( {Y}_{2}\right) \) . Therefore, the two graphs have the same cycle matroid.
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Yes
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Lemma 17.1.1 Let \( \mathcal{T} \) be an eulerian tour of the 4-valent graph \( X \) . If \( u \in \) \( V\left( X\right) \), then there is unique eulerian tour \( {\mathcal{T}}^{\prime } \) in \( X \) that differs from \( \mathcal{T} \) at \( u \) and agrees with \( \mathcal{T} \) at all other vertices of \( X \) .
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Proof. There are three eulerian partitions that agree with \( \mathcal{T} \) at the vertices in \( V\left( X\right) \smallsetminus u \) . We show that one of these is an eulerian tour and that the other has two eulerian cycles.\n\nSuppose that \( u \in V\left( X\right) \) and that \( a, b, c \), and \( d \) are the edges on \( u \) and that \( \mathcal{T} \) contains the subsequence \( \left( {a, u, b, S, c, u, d}\right) \), where \( S \) is an alternating sequence of incident edges and vertices. Thus \( \mathcal{T} \) partitions the edges on \( u \) into the pairs \( \{ a, b\} \) and \( \{ c, d\} \) . Let \( {S}^{\prime } \) be the reverse of the eulerian walk \( S \) . If we replace the subsequence \( \left( {b, S, c}\right) \) of \( \mathcal{T} \) by its reversal \( \left( {c,{S}^{\prime }, b}\right) \), the result is a new eulerian tour that partitions the edges on \( u \) into the pairs \( \{ a, c\} \) and \( \{ b, d\} \) .\n\nThere is a unique eulerian partition that agrees with \( \mathcal{T} \) at each vertex other than \( u \), and partitions the edges at \( u \) into the pairs \( \{ a, d\} \) and \( \{ b, c\} \) . But this eulerian partition has two components, one of which is the eulerian cycle \( \left( {u, b, S, c, u}\right) \) .\n\nIf \( {\mathcal{T}}^{\prime } \) is the unique eulerian tour that differs from \( \mathcal{T} \) only at \( u \), then we say that \( {\mathcal{T}}^{\prime } \) is obtained from \( \mathcal{T} \) by flipping at \( u \) .
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Yes
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Lemma 17.1.2 Let \( \mathcal{S} \) and \( \mathcal{T} \) be two eulerian tours in the 4-valent graph \( X \) . Then there is a sequence of vertices such that \( \mathcal{T} \) can be obtained from \( \mathcal{S} \) by flipping at each vertex of the sequence in turn.
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Proof. Suppose \( \mathcal{S} \) and \( \mathcal{T} \) are two eulerian tours that do not agree at a vertex \( u \) . Let \( {\mathcal{S}}^{\prime } \) and \( {\mathcal{T}}^{\prime } \) denote the tours obtained from \( \mathcal{S} \) and \( \mathcal{T} \), respectively, by flipping at \( u \) . Since there are only three partitions of the edges on \( u \) into two pairs, and since \( \mathcal{S} \) and \( \mathcal{T} \) do not agree at \( u \), one of the three pairs of tours\n\n\[ \left\{ {{\mathcal{S}}^{\prime },\mathcal{T}}\right\} ,\;\left\{ {\mathcal{S},{\mathcal{T}}^{\prime }}\right\} ,\;\left\{ {{\mathcal{S}}^{\prime },{\mathcal{T}}^{\prime }}\right\} \]\n\nmust agree at \( u \), and therefore differ at one fewer vertex than \( \{ \mathcal{S},\mathcal{T}\} \) . A straightforward induction on the number of vertices at which two tours differ yields the result.
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Yes
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Lemma 17.3.1 The core of a left-right cycle in a plane graph \( Y \) is a bicycle of \( Y \) .
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Proof. Let \( P \) be a left-right cycle and let \( Q \) be its core. Let \( u \) be a vertex of \( Y \), and consider the edges on \( u \) . The total number of occurrences of these edges in \( P \) is even, and so the total number of these edges in \( Q \) is also even. Therefore, \( Q \) determines an even subgraph of \( Y \) . Since \( P \) is also a left-right cycle in \( {Y}^{ * } \), it follows that \( Q \) is also an even subgraph of \( {Y}^{ * } \) . Therefore, \( Q \) is a bicycle of \( Y \) .
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Yes
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Lemma 17.3.2 Let \( Y \) be a connected plane graph and let \( \mathcal{P} \) be the set of all the left-right cycles of \( Y \) . Each edge of \( Y \) occurs in an even number of members of \( \mathcal{P} \), but no proper subset of \( \mathcal{P} \) covers every edge an even number of times.
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Proof. The first claim follows directly from the comments preceding this result. For the second claim, suppose that \( {\mathcal{P}}^{\prime } \) is a proper subset of \( \mathcal{P} \) containing every edge an even number of times. Then there is at least one edge that does not occur at all in the walks in \( {\mathcal{P}}^{\prime } \) . Since \( Y \) is connected, we can find a face of \( Y \) containing two consecutive edges \( e \) and \( f \), where \( e \) does not occur in the walks in \( {\mathcal{P}}^{\prime } \) but \( f \) occurs twice. This, however, is impossible, because one of the left-right cycles through \( f \) must use the neighbouring edge \( e \) .
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Yes
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Lemma 17.3.3 Let \( Y \) be a connected plane graph with exactly c left-right cycles. Then the subspace of \( {GF}{\left( 2\right) }^{E\left( Y\right) } \) spanned by the characteristic vectors of the cores has dimension \( c - 1 \) .
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Proof. Let \( \mathcal{P} = \left\{ {{P}_{1},\ldots ,{P}_{c}}\right\} \) be the set of left-right cycles in \( Y \), and let \( \mathcal{Q} = \left\{ {{Q}_{1},\ldots ,{Q}_{c}}\right\} \) be the corresponding cores. Identify a core with its characteristic vector, and suppose that there is some linear combination of the cores equal to the zero vector. Let \( {\mathcal{Q}}^{\prime } \) be the set of cores with nonzero coefficients in this linear combination, and let \( {\mathcal{P}}^{\prime } \) be the corresponding left-right cycles. Then \( {\mathcal{Q}}^{\prime } \) covers every edge an even number of times, and so \( {\mathcal{P}}^{\prime } \) covers every edge an even number of times. By the previous result \( {\mathcal{P}}^{\prime } \) is either empty or \( {\mathcal{P}}^{\prime } = \mathcal{P} \) . Therefore, there is a unique nontrivial linear combination of the cores equal to the zero vector, and so the subspace they span has dimension \( c - 1 \) .
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Yes
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Lemma 17.3.4 If \( Y \) is a plane graph, and \( e \) is an edge of \( Y \), then\n\n(a) If \( e \) is a loop or coloop, then \( c\left( {Y \smallsetminus e}\right) = c\left( {Y/e}\right) = c\left( Y\right) \).\n\n(b) If \( e \) is a parallel edge, then \( c\left( {Y \smallsetminus e}\right) = c\left( Y\right) = c\left( {Y/e}\right) - 1 \).\n\n(c) If \( e \) is a skew edge, then \( c\left( {Y/e}\right) = c\left( Y\right) = c\left( {Y \smallsetminus e}\right) - 1 \).\n\n(d) If \( e \) is a crossing edge, then \( c\left( {Y \smallsetminus e}\right) = c\left( {Y/e}\right) = c\left( Y\right) - 1 \).
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Proof. We leave (a) as an exercise.\n\nIf \( e = {uv} \) is a parallel edge, then we can assume that the left-right cycle containing \( e \) has the form \( P = \left( {u, e, v, S, u, e, v, T}\right) \) . The graph \( Y \smallsetminus e \) contains all the left-right cycles of \( Y \) except for \( P \), and in addition the left-right cycle \( \left( {u,{S}^{\prime }, v, T}\right) \) where \( {S}^{\prime } \) is the reverse of \( S \) . The graph \( Y/e \) has also lost \( P \), but it has gained two new left-right cycles, namely \( \left( {x, S}\right) \) and \( \left( {x, T}\right) \) , where \( x \) is the vertex that resulted from merging \( u \) and \( v \) .\n\nIf \( e \) is a skew edge, then it is parallel in \( {Y}^{ * } \), and the result follows directly from (b).\n\nFinally, if \( e \) is a crossing edge, then the two left-right cycles through \( e \) are merged into a single walk in both \( Y/e \) and \( Y \smallsetminus e \) . Therefore, both graphs have one fewer left-right cycle than \( Y \) .
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No
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Theorem 17.3.5 If \( Y \) is a connected plane graph with exactly \( c \) left-right cycles, then the bicycle space has dimension \( c - 1 \) and is spanned by the cores of \( Y \) .
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Proof. We prove this by induction on the number of left-right cycles in \( Y \) . First we consider the case where \( Y \) has a single left-right cycle. Assume by way of contradiction that \( Y \) has a nonempty bicycle \( B \) . Then \( B \) is both an even subgraph and an edge cutset whose shores we denote by \( L \) and \( R \) . Since \( B \) is even, it divides the plane into regions that can be coloured black and white so that every edge in \( B \) has a black side and a white side, and every other edge of \( Y \) has two sides of the same colour. Consider a left-right cycle starting from a vertex in \( L \) on the black side of an edge in \( B \) . After it uses this edge, it is on the white side of the edge at a vertex in \( R \) . Every time the walk returns to \( L \) it uses an edge in \( B \), and therefore returns to the black side of that edge. Therefore, the edges in \( B \) are used at most once by this walk, contradicting the assumption that \( Y \) has only one left-right cycle.\n\nNow, let \( Y \) be a plane graph with \( c > 1 \) left-right cycles. The dimension of its bicycle space is at least \( c - 1 \) by Lemma 17.3.3. If \( e \) is a crossing edge of \( Y \), then \( Y \smallsetminus e \) has \( c - 1 \) left-right cycles and by the inductive hypothesis, a bicycle space of dimension \( c - 2 \) . Since deleting an edge cannot reduce the dimension of the bicycle space by more than one, the bicycle space of \( Y \) has dimension at most \( c - 1 \), and the result follows.
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Yes
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Lemma 17.4.1 If \( \omega \) is the Gauss code of the shadow of a knot diagram, then\n\n(a) Each symbol occurs twice in \( \omega \) .\n\n(b) The two occurrences of each symbol are separated by an even number of other symbols.
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Proof. The Gauss code gives the unique straight eulerian tour of the knot shadow, and since an eulerian tour of a 4-valent graph visits each vertex twice, the first part of the claim follows immediately.\n\nThe two occurrences of the symbol \( v \) partition the Gauss code into two sections. The edges of the shadow of the knot diagram determined by one of these sections form a closed curve \( C \) in the plane starting and ending at the crossing labelled \( v \) . The edges determined by the other section form another closed curve starting and ending at \( v \), and the first and last edges of this curve are both inside or both outside \( C \) . Therefore, by the Jordan curve theorem these two curves intersect in an even number of points other than \( v \) .
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Yes
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Theorem 17.4.2 Let \( X \) be a 4-valent graph embedded in a surface and suppose \( \omega \) is the double occurrence word corresponding to a straight eulerian tour of \( X \) . If the surface is orientable and \( {X}^{ * } \) is bipartite, then \( \omega \) is even.
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Proof. Assume that the surface is orientable. Since \( {X}^{ * } \) is bipartite, we may partition the faces of \( X \) into two classes, so that two faces in the same class do not have an edge in common. Choose one of these classes and call the faces in it white; call the other faces black. The straight eulerian tour given by \( \omega \) has the property that as we go along it, the faces on the left alternate in colour. Since the embedding is orientable, we can orient the white faces. This gives an orientation of the edges of \( X \) so that at each vertex there are two edges pointing in and two pointing out. This orientation has the further property that any two consecutive edges in \( \omega \) point in opposite directions, which implies that \( \omega \) is even.
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Yes
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Lemma 17.5.1 A chord diagram is a planar graph if and only if its circle graph is bipartite.
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Proof. Let \( Z \) be a chord diagram with a bipartite circle graph. We can map the rim of \( Z \) onto a circle in the plane. The chords in one colour class of the circle graph can be embedded inside the circle without intersecting, and the chords in the other colour class similarly embedded outside the circle.\n\nConversely, suppose we have an embedding of a chord diagram in the plane. The rim of the chord diagram is a continuous closed curve in the plane, which divides the plane into two disjoint parts: inside and outside. Thus there are two classes of chords, those embedded on the inside and those on the outside, and no two chords in the same class are adjacent in the circle graph. Hence the circle graph of \( Z \) is bipartite.
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Yes
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Lemma 17.8.1 Let \( X \) be a 4-valent plane graph with face graph \( Y \) . Then there is a bijection between the bent eulerian tours of \( X \) and the spanning trees of \( Y \) .
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Proof. Suppose that \( Y \) is the graph on the white faces, and that \( {Y}^{ * } \) is the graph on the black faces. Let \( T \) be a spanning tree of \( Y \), and then \( \bar{T} = E\left( Y\right) \smallsetminus T \) is a spanning tree of \( {Y}^{ * } \) . We can identify \( E\left( Y\right) \) and \( E\left( {Y}^{ * }\right) \) with \( V\left( X\right) \), and so \( T \cup \bar{T} \) is a partition of \( V\left( X\right) \) . Now, define an eulerian partition of \( X \) by taking the white partition at every vertex in \( T \) and the black partition at every vertex in \( \bar{T} \) . (That is, take the white partition for every edge in the white spanning tree, and the black partition for every edge in the black spanning tree.) From our earlier remarks this is a bent eulerian partition, so it remains to show that it has just one component.\n\nConsider an adjacency relation on the faces of \( X \), with two white faces being adjacent if they meet at a vertex where the induced partition is white, and two black faces being adjacent if they meet at a vertex where the induced partition is black. It follows immediately that this adjacency relation is described by \( T \) and \( \bar{T} \), and so has two components, being all the white faces and all the black faces. An eulerian cycle of \( X \) determines a closed curve \( C \) in the plane, and so has an inside and an outside. The adjacency relation cannot connect a face inside \( C \) to a face outside \( C \) . If the bent eulerian partition has more than one component, then there are either two white or two black faces on opposite sides of an eulerian cycle, which is a contradiction.\n\nConversely, given a bent eulerian tour, partition the vertices into two sets \( T \) and \( \bar{T} \), according to whether the bent eulerian partition induces the white partition or the black partition, respectively, at that vertex. Then \( T \) is a spanning tree of \( Y \), and \( \bar{T} \) is a spanning tree of \( {Y}^{ * } \) .
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Yes
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Lemma 17.8.2 Suppose that \( X \) is a 4-valent plane graph with a straight eulerian tour \( \mathcal{S} \). For every bent eulerian tour \( \mathcal{T} \) of \( X \), there is a sequence \( \sigma \) containing each vertex of \( X \) once such that \( \mathcal{T} \) is obtained from \( \mathcal{S} \) by flipping on the elements of \( \sigma \) in turn.
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Proof. See Exercise 7.
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No
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Lemma 17.9.1 Let \( X \) be a 4-valent plane graph with face graphs \( Y \) and \( {Y}^{ * } \), and let \( \mathcal{T} \) be a bent eulerian partition of \( X \) with \( c \) components. Then \( \mathcal{T} \) determines a set of edges \( S \) in \( Y \) and the complementary set of edges \( \bar{S} \) in \( {Y}^{ * } \). If \( {c}_{w} \) is the number of components in the subgraph of \( Y \) with edge set \( S \), and \( {c}_{b} \) the number of components in the subgraph of \( {Y}^{ * } \) with edge set \( \bar{S} \), then\n\n\[ c = {c}_{b} + {c}_{w} - 1 \]
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Proof. The eulerian cycles of the bent eulerian partition \( \mathcal{T} \) form a collection of disjoint closed cycles in the plane, and hence divide the plane into \( c + 1 \) regions. Every face of \( X \) lies completely within one of these regions, and each region contains faces of only one colour. Two white faces of \( X \) lie in the same region if and only if there is a path in \( Y \) connecting them using only edges of \( S \), and two black faces lie in the same region if and only if there is a path joining them in \( {Y}^{ * } \) using only edges of \( \bar{S} \). Therefore, the partition into regions is the same as the partition into components of \( Y \) and \( {Y}^{ * } \). Hence \( c + 1 = {c}_{b} + {c}_{w} \), and the result follows.
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Yes
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Corollary 17.9.2 Let \( X \) be a 4-valent plane graph with face graph \( Y \) . Let \( R\left( {Y;x, y}\right) \) be the rank polynomial of \( Y \) . Then the number of bent eulerian partitions of \( X \) with \( c \) components is the coefficient of \( {x}^{c - 1} \) in \( R\left( {Y;x, x}\right) \) .
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Proof. Let \( E \) denote the edge set of \( Y \), and identify it with the edge set of \( {Y}^{ * } \) . By the definition of the rank polynomial we see that\n\n\[ R\left( {Y;x, x}\right) = \mathop{\sum }\limits_{{S \subseteq E}}{x}^{\mathrm{{rk}}\left( E\right) - \mathrm{{rk}}\left( S\right) }{x}^{{\mathrm{{rk}}}^{ \bot }\left( E\right) - {\mathrm{{rk}}}^{ \bot }\left( {E \smallsetminus S}\right) }.\]\n\nFrom the results of Section 15.2, the expression \( \operatorname{rk}\left( E\right) - \operatorname{rk}\left( S\right) \) is equal to the number of components of the subgraph of \( Y \) with edge set \( S \), and \( {\mathrm{{rk}}}^{ \bot }\left( E\right) - {\mathrm{{rk}}}^{ \bot }\left( {E \smallsetminus S}\right) \) is the number of components of the subgraph of \( {Y}^{ * } \) with edge set \( \bar{S} \) .
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Yes
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Lemma 17.12.1 If \( X \) is a 4-valent plane graph and \( \rho \) is an orientation such that every vertex is the head of two edges and the tail of two edges, then \( {X}^{\rho } \) has a unique straight oriented eulerian partition \( \mathcal{S} \) and a unique bent oriented eulerian partition \( {\mathcal{S}}^{ * } \) .
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The conditions of this lemma are satisfied if \( X \) is the shadow of the link diagram of an oriented link and \( \rho \) is the orientation of \( X \) inherited from the orientation of the link. Since \( {\mathcal{S}}^{ * } \) is bent, it forms a collection of noncrossing closed curves in the plane, which are known as the Seifert circles of the link diagram. The minimum number of Seifert circles in any diagram of an oriented link is an important invariant in knot theory.
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No
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Lemma 17.13.1 Let \( Z \) be a chord diagram with associated circle graph \( Y \) , and let \( A\left( Y\right) \) be the adjacency matrix of \( Y \) . Then the characteristic vector of the core of any alternating cycle is in the kernel of \( A\left( Y\right) \) over \( {GF}\left( 2\right) \) . If \( Z \) has \( s \) alternating cycles, then the cores of these cycles span a subspace of dimension \( s - 1 \) in \( \ker \left( {A\left( Y\right) }\right) \) .
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Proof. Let \( C \) be an alternating cycle of \( Z \), and let \( \alpha {\alpha }^{\prime } \) be an arbitrary chord of \( Z \) . Since \( C \) is a cycle, it starts and ends on the same side of \( \alpha {\alpha }^{\prime } \) , and so crosses \( \alpha {\alpha }^{\prime } \) an even number of times in total. A chord not in the core of \( C \) is used exactly twice, and so the number of chords in the core of \( C \) that cross \( \alpha {\alpha }^{\prime } \) is even. Hence the core of \( C \) contains an even number of neighbours of any vertex \( \alpha \), and so its characteristic vector is in \( \ker \left( {A\left( Y\right) }\right) \) .\n\nLet \( {C}_{1},\ldots ,{C}_{s} \) be the collection of alternating cycles of \( Z \), and let \( {c}_{1},\ldots ,{c}_{s} \) be the characteristic vectors of the cores of \( {C}_{1},\ldots ,{C}_{s} \) , respectively. Clearly,\n\n\[ {c}_{1} + {c}_{2} + \cdots + {c}_{s} = 0 \]\n\nand so we must show that no other linear combination of the vectors is equal to zero. So suppose instead for a contradiction that there is some set of indices \( I \subset \{ 1,\ldots, s\} \) such that\n\n\[ \mathop{\sum }\limits_{{i \in I}}{c}_{i} = 0 \]\n\nThen the corresponding set of alternating cycles \( {\mathcal{C}}_{I} = \left\{ {{C}_{i} : i \in I}\right\} \) contains every chord an even number of times. As this collection of alternating cycles does not contain every chord twice, there is a pair of chords \( \alpha {\alpha }^{\prime } \) and \( \beta {\beta }^{\prime } \) with \( \alpha \) and \( \beta \) successive vertices of the rim of \( Z \), and such that \( \alpha {\alpha }^{\prime } \) occurs twice in \( {\mathcal{C}}_{I} \) and \( \beta {\beta }^{\prime } \) does not occur at all. This is clearly impossible, because \( \beta {\beta }^{\prime } \) must be the chord used immediately after \( {\alpha }^{\prime }\alpha \) in an alternating cycle, and so we have the desired contradiction.
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Yes
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Lemma 17.13.2 Let \( Z \) be a chord diagram with associated circle graph \( Y \) . If \( Z \) has only one alternating cycle, then \( A\left( Y\right) \) is invertible over \( {GF}\left( 2\right) \) .
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Proof. If \( C \) is an alternating cycle, then form a word \( {\omega }^{\prime } \) by listing the vertex at the end of each \
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No
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Theorem 17.13.3 Let \( Z \) be a chord diagram with \( n \) chords, and with associated circle graph \( Y \) . If \( Z \) has \( s \) alternating cycles, then \( {\operatorname{rk}}_{2}\left( {A\left( Y\right) }\right) = \) \( n + 1 - s \) .
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Proof. We will prove this by induction; our previous result shows that it is true when \( s = 1 \) .\n\nSo suppose that \( Z \) has \( s \geq 2 \) alternating cycles \( {C}_{1},\ldots ,{C}_{s} \) . Select a chord \( \alpha {\alpha }^{\prime } \) that is in two distinct cores, say \( {C}_{s - 1} \) and \( {C}_{s} \), and consider the chord diagram \( Z \smallsetminus \alpha {\alpha }^{\prime } \) obtained by deleting \( \alpha \) and \( {\alpha }^{\prime } \) from the rim of \( Z \) . The alternating cycles of this chord diagram are \( {C}_{1},\ldots ,{C}_{s - 2} \) and an alternating cycle obtained by merging \( {C}_{s - 1} \) and \( {C}_{s} \), whose core has characteristic vector \( {c}_{s - 1} + {c}_{s} \) . Therefore, \( Z \smallsetminus \alpha {\alpha }^{\prime } \) has \( s - 1 \) alternating cycles, and by induction\n\n\[ \n{\operatorname{rk}}_{2}\left( {A\left( {Y \smallsetminus \alpha }\right) }\right) = \left( {n - 1}\right) + 1 - \left( {s - 1}\right) = n + 1 - s. \n\]\n\nSince \( A\left( {Y \smallsetminus \alpha }\right) \) is a principal submatrix of \( A\left( Y\right) \), we see that \( {\operatorname{rk}}_{2}\left( {A\left( Y\right) }\right) \) is at least \( n + 1 - s \) . By Lemma 17.13.1, \( {\operatorname{rk}}_{2}(A\left( Y\right) \) is at most \( n + 1 - s \), and so the result follows.
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Yes
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Example 4. Let \( X = {\mathbb{R}}^{n} \) . Here the elements of \( X \) are \( n \) -tuples of real numbers that we can display in the form \( x = \left\lbrack {x\left( 1\right), x\left( 2\right) ,\ldots, x\left( n\right) }\right\rbrack \) or \( x = \left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) . A useful norm is defined by the equation\n\n\[ \parallel x{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {x\left( i\right) }\right| \]
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Note that an \( n \) -tuple is a function on the set \( \{ 1,2,\ldots, n\} \), and so the notation \( x\left( i\right) \) is consistent with that interpretation. (This is the \
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No
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Theorem 1. Let \( K \) be a compact set in a normed linear space \( X \) . To each \( x \) in \( X \) there corresponds at least one point in \( K \) of minimum distance from \( x \) .
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Proof. Let \( x \) be any member of \( X \) . The distance from \( x \) to \( K \) is defined to be the number\n\n\[ \operatorname{dist}\left( {x, K}\right) = \mathop{\inf }\limits_{{z \in K}}\parallel x - z\parallel \]\n\nBy the definition of an infimum (Problem 12 in Section 1.1, page 6), there exists a sequence \( \left\lbrack {y}_{n}\right\rbrack \) in \( K \) such that \( \begin{Vmatrix}{x - {y}_{n}}\end{Vmatrix} \rightarrow \operatorname{dist}\left( {x, K}\right) \) . Since \( K \) is compact, there is a subsequence converging to a point in \( K \), say \( {y}_{{n}_{i}} \rightarrow y \in K \) . Since\n\n\[ \parallel x - y\parallel \leq \begin{Vmatrix}{x - {y}_{{n}_{i}}}\end{Vmatrix} + \begin{Vmatrix}{{y}_{{n}_{i}} - y}\end{Vmatrix} \]\n\nwe have in the limit \( \parallel x - y\parallel \leq \operatorname{dist}\left( {x, K}\right) \leq \parallel x - y\parallel \) . (The final inequality follows from the definition of the distance function.)
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Yes
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On the real line, an open interval \( \left( {a, b}\right) \) is not compact, for we can take a sequence in the interval that converges to the endpoint \( b \), say. Then every subsequence will also converge to \( b \) . Since \( b \) is not in the interval, the interval cannot be compact. On the other hand, a closed and bounded interval, say \( \left\lbrack {a, b}\right\rbrack \), is compact.
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This is a special case of the Heine-Borel theorem. See the discussion before Lemma 1 in Section 1.4, page 20.
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No
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Theorem 2. The space \( C\left\lbrack {a, b}\right\rbrack \) with norm \( \parallel x\parallel = \mathop{\max }\limits_{s}\left| {x\left( s\right) }\right| \) is a Banach space.
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Proof. Let \( \left\lbrack {x}_{n}\right\rbrack \) be a Cauchy sequence in \( C\left\lbrack {a, b}\right\rbrack \) . (This space is described in Example 6, page 3.) Then for each \( s,\left\lbrack {{x}_{n}\left( s\right) }\right\rbrack \) is a Cauchy sequence in \( \mathbb{R} \) . Since \( \mathbb{R} \) is complete, this latter sequence converges to a real number that we may denote by \( x\left( s\right) \) . The function \( x \) thus defined must now be shown to be continuous, and we must also show that \( \begin{Vmatrix}{{x}_{n} - x}\end{Vmatrix} \rightarrow 0 \) . Let \( t \) be fixed as the point at which continuity is to be proved. We write (1) \[ \left| {x\left( s\right) - x\left( t\right) }\right| \leq \left| {x\left( s\right) - {x}_{n}\left( s\right) }\right| + \left| {{x}_{n}\left( s\right) - {x}_{n}\left( t\right) }\right| + \left| {{x}_{n}\left( t\right) - x\left( t\right) }\right| \] This inequality should suggest to the reader how the proof must proceed. Let \( \varepsilon > 0 \) . Select \( N \) so that \( \begin{Vmatrix}{{x}_{n} - {x}_{m}}\end{Vmatrix} \leq \varepsilon /3 \) whenever \( m \geq n \geq N \) (Cauchy property). Then for \( m \geq n \geq N,\left| {{x}_{n}\left( s\right) - {x}_{m}\left( s\right) }\right| \leq \varepsilon /3 \) . By letting \( m \rightarrow \infty \) we get \( \left| {{x}_{n}\left( s\right) - x\left( s\right) }\right| \leq \varepsilon /3 \) for all \( s \) . This shows that \( \begin{Vmatrix}{{x}_{n} - x}\end{Vmatrix} \leq \varepsilon /3 \) and that the sequence \( \begin{Vmatrix}{{x}_{n} - x}\end{Vmatrix} \) converges to 0 . By the continuity of \( {x}_{n} \) there exists a \( \delta > 0 \) such that \( \left| {{x}_{n}\left( s\right) - {x}_{n}\left( t\right) }\right| < \varepsilon /3 \) whenever \( \left| {t - s}\right| < \delta \) . Inequality (1) now shows that \( \left| {x\left( s\right) - x\left( t\right) }\right| < \varepsilon \) when \( \left| {t - s}\right| < \delta \) . (This proof illustrates what is sometimes called \
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Yes
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Theorem 1. Let \( f \) be a continuous mapping whose domain \( D \) is a compact set in a normed linear space and whose range is contained in another normed linear space. Then \( f\left( D\right) \) is compact.
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Proof. To show that \( f\left( D\right) \) is compact, we let \( \left\lbrack {y}_{n}\right\rbrack \) be any sequence in \( f\left( D\right) \) , and prove that this sequence has a convergent subsequence whose limit is in \( f\left( D\right) \) . There exist points \( {x}_{n} \in D \) such that \( f\left( {x}_{n}\right) = {y}_{n} \) . Since \( D \) is compact, the sequence \( \left\lbrack {x}_{n}\right\rbrack \) has a subsequence \( \left\lbrack {x}_{{n}_{i}}\right\rbrack \) that converges to a point \( x \in D \) . Since \( f \) is continuous,\n\n\[ f\left( x\right) = f\left( {\mathop{\lim }\limits_{i}{x}_{{n}_{i}}}\right) = \mathop{\lim }\limits_{i}f\left( {x}_{{n}_{i}}\right) = \mathop{\lim }\limits_{i}{y}_{{n}_{i}} \]\n\nThus the subsequence \( \left\lbrack {y}_{{n}_{i}}\right\rbrack \) converges to a point in \( f\left( D\right) \) .
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Yes
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Theorem 2. A continuous real-valued function whose domain is a compact set in a normed linear space attains its supremum and infimum; both of these are therefore finite.
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Proof. Let \( f \) be a continuous real-valued function whose domain is a compact set \( D \) in a normed linear space. Let \( M = \sup \{ f\left( x\right) : x \in D\} \) . Then there is a sequence \( \left\lbrack {x}_{n}\right\rbrack \) in \( D \) for which \( f\left( {x}_{n}\right) \rightarrow M \) . (At this stage, we admit the possibility that \( M \) may be \( + \infty \) .) By compactness, there is a subsequence \( \left\lbrack {x}_{{n}_{i}}\right\rbrack \) converging to a point \( x \in D \) . By continuity, \( f\left( {x}_{{n}_{i}}\right) \rightarrow f\left( x\right) \) . Hence \( f\left( x\right) = M \) , and of course \( M < \infty \) . The proof for the infimum is similar.
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Yes
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Theorem 3. A continuous function whose domain is a compact subset of a normed space and whose values lie in another normed space is uniformly continuous.
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Proof. Let \( f \) be a function (defined on a compact set) that is not uniformly continuous. We shall show that \( f \) is not continuous. There exists an \( \varepsilon > 0 \) for which there is no corresponding \( \delta \) to fulfill the condition of uniform continuity. That implies that for each \( n \) there is a pair of points \( \left( {{x}_{n},{y}_{n}}\right) \) satisfying the condition \( \begin{Vmatrix}{{x}_{n} - {y}_{n}}\end{Vmatrix} < 1/n \) and \( \begin{Vmatrix}{f\left( {x}_{n}\right) - f\left( {y}_{n}\right) }\end{Vmatrix} \geq \varepsilon \) . By compactness the sequence \( \left\lbrack {x}_{n}\right\rbrack \) has a subsequence \( \left\lbrack {x}_{{n}_{i}}\right\rbrack \) that converges to a point \( x \) in the domain of \( f \) . Then \( {y}_{{n}_{i}} \rightarrow x \) also because \( \begin{Vmatrix}{{y}_{{n}_{i}} - x}\end{Vmatrix} \leq \begin{Vmatrix}{{y}_{{n}_{i}} - {x}_{{n}_{i}}}\end{Vmatrix} + \begin{Vmatrix}{{x}_{{n}_{i}} - x}\end{Vmatrix} \) . Now the continuity of \( f \) at \( x \) fails because\n\n\[ \varepsilon \leq \begin{Vmatrix}{f\left( {x}_{{n}_{i}}\right) - f\left( {y}_{{n}_{i}}\right) }\end{Vmatrix} \leq \begin{Vmatrix}{f\left( {x}_{{n}_{i}}\right) - f\left( x\right) }\end{Vmatrix} + \begin{Vmatrix}{f\left( x\right) - f\left( {y}_{{n}_{i}}\right) }\end{Vmatrix} \]
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Yes
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Theorem 4. The inverse image of a closed set by a continuous map is closed.
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Proof. Recall that the inverse image of a set \( A \) by a map \( f \) is defined to be \( {f}^{-1}\left( A\right) = \{ x : f\left( x\right) \in A\} \) . Let \( f : X \rightarrow Y \), where \( X \) and \( Y \) are normed spaces and \( f \) is continuous. Let \( K \) be a closed set in \( Y \) . To show that \( {f}^{-1}\left( K\right) \) is closed, we start by letting \( \left\lbrack {x}_{n}\right\rbrack \) be a convergent sequence in \( {f}^{-1}\left( K\right) \) . Thus \( {x}_{n} \rightarrow x \) and \( f\left( {x}_{n}\right) \in K \) . By continuity, \( f\left( {x}_{n}\right) \rightarrow f\left( x\right) \) . Since \( K \) is closed, \( f\left( x\right) \in K \) . Hence \( x \in {f}^{-1}\left( K\right) \) .
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Yes
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Theorem 5. If a series in a Banach space is absolutely convergent, then all rearrangements of the series converge to a common value.
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Proof. Let \( \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}_{i} \) be such a series and \( \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}_{{k}_{i}} \) a rearrangement of it. Put \( x = \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}_{i},{S}_{n} = \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i},{s}_{n} = \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{{k}_{i}} \), and \( M = \mathop{\sum }\limits_{{i = 1}}^{\infty }\begin{Vmatrix}{x}_{i}\end{Vmatrix} \) . Then \( \mathop{\sum }\limits_{{i = 1}}^{n}\begin{Vmatrix}{x}_{{k}_{i}}\end{Vmatrix} \leq M \) . This proves that \( \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}_{{k}_{i}} \) is absolutely convergent and hence convergent. (Here we require the completeness of the space.) Put \( y = \mathop{\sum }\limits_{{i = 1}}^{\infty }{x}_{{k}_{i}} \) . Let \( \varepsilon > 0 \) . Select \( n \) such that \( \mathop{\sum }\limits_{{i \geq n}}\begin{Vmatrix}{x}_{i}\end{Vmatrix} < \varepsilon \) and such that \( \begin{Vmatrix}{{S}_{m} - x}\end{Vmatrix} < \varepsilon \) when \( m \geq n \) . Select \( r \) so that \( \begin{Vmatrix}{{s}_{r} - y}\end{Vmatrix} < \varepsilon \) and so that \( \{ 1,\ldots, n\} \subset \left\{ {{k}_{1},\ldots ,{k}_{r}}\right\} \) . Select \( m \) such that \( \left\{ {{k}_{1},\ldots ,{k}_{r}}\right\} \subset \{ 1,\ldots, m\} \) . Then \( m \geq n \) and\n\n\[ \begin{Vmatrix}{{S}_{m} - {s}_{r}}\end{Vmatrix} = \begin{Vmatrix}{\left( {{x}_{1} + \cdots + {x}_{m}}\right) - \left( {{x}_{{k}_{1}} + \cdots + {x}_{{k}_{r}}}\right) }\end{Vmatrix} \leq \mathop{\sum }\limits_{{i = n + 1}}^{m}\begin{Vmatrix}{x}_{i}\end{Vmatrix} < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel \leq \begin{Vmatrix}{x - {S}_{m}}\end{Vmatrix} + \begin{Vmatrix}{{S}_{m} - {s}_{r}}\end{Vmatrix} + \begin{Vmatrix}{{s}_{r} - y}\end{Vmatrix} < {3\varepsilon } \]
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Yes
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Lemma 1. In the space \( {\mathbb{R}}^{n} \) with norm \( \parallel x{\parallel }_{\infty } = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {x\left( i\right) }\right| \)\neach ball \( \left\{ {x : \parallel x{\parallel }_{\infty } \leq c}\right\} \) is compact.
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Proof. Let \( \left\lbrack {x}_{k}\right\rbrack \) be a sequence of points in \( {\mathbb{R}}^{n} \) satisfying \( {\begin{Vmatrix}{x}_{k}\end{Vmatrix}}_{\infty } \leq c \) . Then the components obey the inequality \( - c \leq {x}_{k}\left( i\right) \leq c \) . By the compactness of the interval \( \left\lbrack {-c, c}\right\rbrack \), there exists an increasing sequence \( {I}_{1} \subset \mathbb{N} \) having the property that \( \lim \left\lbrack {{x}_{k}\left( 1\right) : k \in {I}_{1}}\right\rbrack \) exists. Next, there exists another increasing sequence \( {I}_{2} \subset {I}_{1} \) such that \( \lim \left\lbrack {{x}_{k}\left( 2\right) : k \in {I}_{2}}\right\rbrack \) exists. Then \( \lim \left\lbrack {{x}_{k}\left( 1\right) : k \in {I}_{2}}\right\rbrack \) exists also, because \( {I}_{2} \subset {I}_{1} \) . Continuing in this way, we obtain at the \( n \) th step an increasing sequence \( {I}_{n} \) such that \( \lim \left\lbrack {{x}_{k}\left( i\right) : k \in {I}_{n}}\right\rbrack \) exists for each \( i = 1,\ldots, n \) . Denoting that limit by \( {x}^{ * }\left( i\right) \), we have defined a vector \( {x}^{ * } \) such that \( {\begin{Vmatrix}{x}_{k} - {x}^{ * }\end{Vmatrix}}_{\infty } \rightarrow 0 \) as \( k \) runs through the sequence of integers \( {I}_{n} \) .
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Yes
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Lemma 2. A closed subset of a compact set is compact.
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Proof. If \( F \) is a closed subset of a compact set \( K \), and if \( \left\lbrack {x}_{n}\right\rbrack \) is a sequence in \( F \), then by the compactness of \( K \) a subsequence converges to a point of \( K \) . The limit point must be in \( F \), since \( F \) is closed.
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Yes
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Corollary 1. Every finite-dimensional normed linear space is complete.
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Proof. Let \( \left\lbrack {x}_{n}\right\rbrack \) be a Cauchy sequence in such a space. Let us prove that the sequence is bounded. Select an index \( m \) such that \( \begin{Vmatrix}{{x}_{i} - {x}_{j}}\end{Vmatrix} < 1 \) whenever \( i, j \geq m \) . Then we have\n\n\[ \begin{Vmatrix}{x}_{i}\end{Vmatrix} \leq \begin{Vmatrix}{{x}_{i} - {x}_{m}}\end{Vmatrix} + \begin{Vmatrix}{x}_{m}\end{Vmatrix} \leq 1 + \begin{Vmatrix}{x}_{m}\end{Vmatrix}\;\left( {i \geq m}\right) \]\n\nHence for all \( i \) ,\n\n\[ \begin{Vmatrix}{x}_{i}\end{Vmatrix} \leq 1 + \begin{Vmatrix}{x}_{1}\end{Vmatrix} + \cdots + \begin{Vmatrix}{x}_{m}\end{Vmatrix} \equiv c \]\n\nSince the ball of radius \( c \) is compact, our sequence must have a convergent subsequence, say \( {x}_{{n}_{i}} \rightarrow {x}^{ * } \) . Given \( \varepsilon > 0 \), select \( N \) so that \( \begin{Vmatrix}{{x}_{i} - {x}_{j}}\end{Vmatrix} < \varepsilon \) when \( i, j \geq N \) . Then \( \begin{Vmatrix}{{x}_{j} - {x}_{{n}_{i}}}\end{Vmatrix} < \varepsilon \) when \( i, j \geq N \), because \( {n}_{i} > i \) . By taking the limit as \( i \rightarrow \infty \), we conclude that \( \begin{Vmatrix}{{x}_{j} - {x}^{ * }}\end{Vmatrix} \leq \varepsilon \) when \( j \geq N \) . This shows that \( {x}_{j} \rightarrow x \) .
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Yes
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Corollary 2. Every finite-dimensional subspace in a normed linear space is closed.
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Proof. Recall that a subset \( Y \) in a linear space is a subspace if it is a linear space in its own right. (The only axioms that require verification are the ones concerned with algebraic closure of \( Y \) under addition and scalar multiplication.) Let \( Y \) be a finite-dimensional subspace in a normed space. To show that \( Y \) is closed, let \( {y}_{n} \in Y \) and \( {y}_{n} \rightarrow y \) . We want to know that \( y \in Y \) . The preceding corollary establishes this: The convergent sequence has the Cauchy property and hence converges to a point in \( Y \), because \( Y \) is complete.
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Yes
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Theorem 2. If the unit ball in a normed linear space is compact, then the space has finite dimension.
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Proof. If the space is not finite dimensional, then a sequence \( \left\lbrack {x}_{n}\right\rbrack \) can be defined inductively as follows. Let \( {x}_{1} \) be any point such that \( \begin{Vmatrix}{x}_{1}\end{Vmatrix} = 1 \) . If \( {x}_{1},\ldots ,{x}_{n - 1} \) have been defined, let \( {U}_{n - 1} \) be the subspace that they span. By Corollary 2, above, \( {U}_{n - 1} \) is closed. Use Riesz’s Lemma to select \( {x}_{n} \) so that \( \begin{Vmatrix}{x}_{n}\end{Vmatrix} = 1 \) and \( \operatorname{dist}\left( {{x}_{n},{U}_{n - 1}}\right) > \frac{1}{2} \) . Then \( \begin{Vmatrix}{{x}_{n} - {x}_{i}}\end{Vmatrix} > \frac{1}{2} \) whenever \( i < n \) . This sequence cannot have any convergent subsequence.
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Yes
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If \( X = {\mathbb{R}}^{n} \) and \( Y = {\mathbb{R}}^{m} \), then each linear map of \( X \) into \( Y \) is of the form \( f\left( x\right) = y \) ,
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\[ y\left( i\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}x\left( j\right) \;\left( {1 \leq i \leq m}\right) \] where the \( {a}_{ij} \) are certain real numbers that form an \( m \times n \) matrix.
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero.
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Proof. Let \( T : X \rightarrow Y \) be such a linear transformation. If it is continuous, then of course it is continuous at 0 . For the converse, suppose that \( T \) is continuous at 0 . For each \( \varepsilon > 0 \) there is a \( \delta > 0 \) such that for all \( x \) ,\n\n\[ \parallel x\parallel < \delta \; \Rightarrow \;\parallel {Tx}\parallel < \varepsilon \]\n\nHence\n\n\[ \parallel x - y\parallel < \delta \; \Rightarrow \;\parallel {Tx} - {Ty}\parallel = \parallel T\left( {x - y}\right) \parallel < \varepsilon \]
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Yes
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Theorem 2. A linear transformation acting between normed linear spaces is continuous if and only if it is bounded.
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Proof. Let \( T : X \rightarrow Y \) be such a map. If it is continuous, then there is a \( \delta > 0 \) such that\n\n\[ \parallel x\parallel \leq \delta \Rightarrow \parallel {Tx}\parallel \leq 1 \]\n\nIf \( \parallel x\parallel \leq 1 \), then \( {\delta x} \) is a vector of norm at most \( \delta \). Consequently, \( \parallel T\left( {\delta x}\right) \parallel \leq 1 \), whence \( \parallel {Tx}\parallel \leq 1/\delta \). Conversely, if \( \parallel {Tx}\parallel \leq M \) whenever \( \parallel x\parallel \leq 1 \), then\n\n\[ \parallel x\parallel \leq \frac{\varepsilon }{M} \Rightarrow \begin{Vmatrix}\frac{Mx}{\varepsilon }\end{Vmatrix} \leq 1 \Rightarrow \begin{Vmatrix}{T\left( \frac{Mx}{\varepsilon }\right) }\end{Vmatrix} \leq M \Rightarrow \parallel {Tx}\parallel \leq \varepsilon \]\n\nThis proves continuity at 0, which suffices, by the preceding theorem.\n\nIf \( T : X \rightarrow Y \) is a bounded linear transformation, we define\n\n\[ \parallel T\parallel = \sup \{ \parallel {Tx}\parallel : \parallel x\parallel \leq 1\} \]\n\nIt can be shown that this defines a norm on the family of all bounded linear transformations from \( X \) into \( Y \) ; this family is a vector space, and it now becomes a normed linear space, denoted by \( \mathcal{L}\left( {X, Y}\right) \).\n\nThe definition of \( \parallel T\parallel \) leads at once to the important inequality\n\n\[ \parallel {Tx}\parallel \leq \parallel T\parallel \parallel x\parallel \]\n\nTo prove this, notice first that it is correct for \( x = 0 \), since \( {T0} = 0 \). On the other hand, if \( x \neq 0 \), then \( x/\parallel x\parallel \) is a vector of norm 1. By the definition of \( \parallel T\parallel \), we have \( \parallel T\left( {x/\parallel x\parallel }\right) \parallel \leq \parallel T\parallel \), which is equivalent to the inequality displayed above. That inequality contains three distinct norms: the ones defined on \( X, Y \), and \( \mathcal{L}\left( {X, Y}\right) \).
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Yes
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Theorem 2. A linear transformation acting between normed linear spaces is continuous if and only if it is bounded.
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Proof. Let \( T : X \rightarrow Y \) be such a map. If it is continuous, then there is a \( \delta > 0 \) such that\n\n\[ \parallel x\parallel \leq \delta \Rightarrow \parallel {Tx}\parallel \leq 1 \]\n\nIf \( \parallel x\parallel \leq 1 \), then \( {\delta x} \) is a vector of norm at most \( \delta \) . Consequently, \( \parallel T\left( {\delta x}\right) \parallel \leq 1 \), whence \( \parallel {Tx}\parallel \leq 1/\delta \) . Conversely, if \( \parallel {Tx}\parallel \leq M \) whenever \( \parallel x\parallel \leq 1 \), then\n\n\[ \parallel x\parallel \leq \frac{\varepsilon }{M} \Rightarrow \begin{Vmatrix}\frac{Mx}{\varepsilon }\end{Vmatrix} \leq 1 \Rightarrow \begin{Vmatrix}{T\left( \frac{Mx}{\varepsilon }\right) }\end{Vmatrix} \leq M \Rightarrow \parallel {Tx}\parallel \leq \varepsilon \]\n\nThis proves continuity at 0, which suffices, by the preceding theorem.\n\nIf \( T : X \rightarrow Y \) is a bounded linear transformation, we define\n\n\[ \parallel T\parallel = \sup \{ \parallel {Tx}\parallel : \parallel x\parallel \leq 1\}\]\n\nIt can be shown that this defines a norm on the family of all bounded linear transformations from \( X \) into \( Y \) ; this family is a vector space, and it now becomes a normed linear space, denoted by \( \mathcal{L}\left( {X, Y}\right) \) .\n\nThe definition of \( \parallel T\parallel \) leads at once to the important inequality\n\n\[ \parallel {Tx}\parallel \leq \parallel T\parallel \parallel x\parallel \]\n\nTo prove this, notice first that it is correct for \( x = 0 \), since \( {T0} = 0 \) . On the other hand, if \( x \neq 0 \), then \( x/\parallel x\parallel \) is a vector of norm 1 . By the definition of \( \parallel T\parallel \), we have \( \parallel T\left( {x/\parallel x\parallel }\right) \parallel \leq \parallel T\parallel \), which is equivalent to the inequality displayed above. That inequality contains three distinct norms: the ones defined on \( X, Y \), and \( \mathcal{L}\left( {X, Y}\right) \) .
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Yes
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Corollary 2. Every linear transformation from a finite-dimensional normed space to another normed space is continuous.
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Proof. Let \( T : X \rightarrow Y \) be such a transformation. Let \( \left\{ {{b}_{1},\ldots ,{b}_{n}}\right\} \) be a basis for \( X \) . Then each \( x \in X \) has a unique expression as a linear combination of basis elements. The coefficients depend on \( x \), and so we write \( x = \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}\left( x\right) {b}_{i} \) . These functionals \( {\lambda }_{i} \) are in fact linear. Indeed, from the previous equation and the equation \( u = \sum {\lambda }_{i}\left( u\right) {b}_{i} \) we conclude that\n\n\[ \n{\alpha x} + {\beta u} = \mathop{\sum }\limits_{{i = 1}}^{n}\left\lbrack {\alpha {\lambda }_{i}\left( x\right) + \beta {\lambda }_{i}\left( u\right) }\right\rbrack {b}_{i} \n\]\n\nSince we have also\n\n\[ \n{\alpha x} + {\beta u} = \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}\left( {{\alpha x} + {\beta u}}\right) {b}_{i} \n\]\n\nwe may conclude (by the uniqueness of the representations) that\n\n\[ \n{\lambda }_{i}\left( {{\alpha x} + {\beta u}}\right) = \alpha {\lambda }_{i}\left( x\right) + \beta {\lambda }_{i}\left( u\right) \n\]\n\nNow use the preceding corollary to infer that the functionals \( {\lambda }_{i} \) are continuous. Getting back to \( T \), we have\n\n\[ \n{Tx} = T\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}\left( x\right) {b}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}{\lambda }_{i}\left( x\right) T{b}_{i} \n\]\n\nand this is obviously continuous.
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Yes
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Corollary 3. All norms on a finite-dimensional vector space are equivalent, as defined in Problem 3, page 23.
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Proof. Let \( X \) be a finite-dimensional vector space having two norms \( \parallel {\parallel }_{1} \) and \( \parallel {\parallel }_{2} \) . The identity map \( I \) from \( \left( {X,\parallel {\parallel }_{1}}\right) \) to \( \left( {X,\parallel {\parallel }_{2}}\right) \) is continuous by the preceding result. Hence it is bounded. This implies that\n\n\[ \parallel x{\parallel }_{2} = \parallel {Ix}{\parallel }_{2} \leq \alpha \parallel x{\parallel }_{1} \]\n\nBy the symmetry in the hypotheses, there is a \( \beta \) such that \( \parallel x{\parallel }_{1} \leq \beta \parallel x{\parallel }_{2} \) .
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Yes
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Theorem 4. If \( X \) is a normed linear space and \( Y \) is a Banach space, then \( \mathcal{L}\left( {X, Y}\right) \) is a Banach space.
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Proof. The only issue is the completeness of \( \mathcal{L}\left( {X, Y}\right) \) . Let \( \left\lbrack {A}_{n}\right\rbrack \) be a Cauchy sequence in \( \mathcal{L}\left( {X, Y}\right) \) . For each \( x \in X \), we have \[ \begin{Vmatrix}{{A}_{n}x - {A}_{m}x}\end{Vmatrix} = \begin{Vmatrix}{\left( {{A}_{n} - {A}_{m}}\right) x}\end{Vmatrix} \leq \begin{Vmatrix}{{A}_{n} - {A}_{m}}\end{Vmatrix}\parallel x\parallel \] This shows that \( \left\lbrack {{A}_{n}x}\right\rbrack \) is a Cauchy sequence in \( Y \) . By the completeness of \( Y \) we can define \( {Ax} = \lim {A}_{n}x \) . The linearity of \( A \) follows by letting \( n \rightarrow \infty \) in the equation \[ {A}_{n}\left( {{\alpha x} + {\beta u}}\right) = \alpha {A}_{n}x + \beta {A}_{n}u \] The boundedness of \( A \) follows from the boundedness of the Cauchy sequence \( \left\lbrack {A}_{n}\right\rbrack \) . If \( \begin{Vmatrix}{A}_{n}\end{Vmatrix} \leq M \) then \( \begin{Vmatrix}{{A}_{n}x}\end{Vmatrix} \leq M\parallel x\parallel \) for all \( x \), and in the limit we have \( \parallel {Ax}\parallel \leq M\parallel x\parallel \) . Finally, we have \( \begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} \rightarrow 0 \) because if \( \begin{Vmatrix}{{A}_{n} - {A}_{m}}\end{Vmatrix} \leq \varepsilon \) when \( m, n \geq N \), then for all \( x \) of norm 1 we have \( \begin{Vmatrix}{{A}_{n}x - {A}_{m}x}\end{Vmatrix} \leq \varepsilon \) when \( m, n \geq N \) . Then we can let \( m \rightarrow \infty \) to get \( \begin{Vmatrix}{{A}_{n}x - {Ax}}\end{Vmatrix} \leq \varepsilon \) and \( \begin{Vmatrix}{{A}_{n} - A}\end{Vmatrix} \leq \varepsilon \) .
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Yes
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Theorem 5. The Neumann Theorem. Let \( A \) be a bounded linear operator on a Banach space \( X \) (and taking values in \( X \) ). If \( \parallel A\parallel < 1 \), then \( I - A \) is invertible, and\n\n\[{\left( I - A\right) }^{-1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{A}^{k}\]
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Proof. Put \( {B}_{n} = \mathop{\sum }\limits_{{k = 0}}^{n}{A}^{k} \) . The sequence \( \left\lbrack {B}_{n}\right\rbrack \) has the Cauchy property, for if \( n > m \), then\n\n\[\\begin{Vmatrix}{{B}_{n} - {B}_{m}}\\end{Vmatrix} = \\begin{Vmatrix}{\\mathop{\\sum }\\limits_{{k = m + 1}}^{n}{A}^{k}}\\end{Vmatrix} \\leq \\mathop{\\sum }\\limits_{{k = m + 1}}^{n}\\begin{Vmatrix}{A}^{k}\\end{Vmatrix} \\leq \\mathop{\\sum }\\limits_{{k = m}}^{\\infty }\\parallel A{\\parallel }^{k}\]\n\n\[= \\parallel A{\\parallel }^{m}\\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }\\parallel A{\\parallel }^{k} = \\parallel A{\\parallel }^{m}/\\left( {1 - \\parallel A\\parallel }\\right)\]\n\n(In this calculation we used Problem 20.) Since the space of all bounded linear operators on \( X \) into \( X \) is complete (Theorem 4), the sequence \( \left\lbrack {B}_{n}\right\rbrack \) converges to a bounded linear operator \( B \) . We have\n\n\[\\left( {I - A}\\right) {B}_{n} = {B}_{n} - A{B}_{n} = \\mathop{\\sum }\\limits_{{k = 0}}^{n}{A}^{k} - \\mathop{\\sum }\\limits_{{k = 1}}^{{n + 1}}{A}^{k} = I - {A}^{n + 1}\]\n\nTaking a limit, we obtain \( \\left( {I - A}\\right) B = I \) . Similarly, \( B\\left( {I - A}\\right) = I \) . Hence \( B = {\\left( I - A\\right) }^{-1} \).
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Yes
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Theorem 1. If a contradiction can be derived from the Zermelo-Fraenkel axioms of set theory (which include the Axiom of Choice), then a contradiction can be derived within the restricted set theory based on the Zermelo-Fraenkel axioms without the Axiom of Choice.
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In 1963, Paul Cohen [Coh] proved that the Axiom of Choice is independent of the remaining axioms in the Zermelo-Fraenkel system. Thus it cannot be proved from them.
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No
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Theorem 2. Every nontrivial vector space has a Hamel base.
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Proof. Let \( X \) be a nontrivial vector space. To show that \( X \) has a Hamel base we first prove that \( X \) has a maximal linearly independent set, and then we show that any such set is necessarily a Hamel base. Consider the collection of all linearly independent subsets of \( X \), and partially order this collection by inclusion, C. In order to use Zorn's Lemma, we verify that every chain in this partially ordered set has an upper bound. Let \( C \) be a chain. Consider \( {S}^{ * } = \bigcup \{ S : S \in C\} \) . This certainly satisfies \( S \subset {S}^{ * } \) for all \( S \in C \) . But is \( {S}^{ * } \) linearly independent? Suppose that \( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{s}_{i} = 0 \) for some scalars \( {\alpha }_{i} \) and for some distinct points \( {s}_{i} \) in \( {S}^{ * } \) . Each \( {s}_{i} \) belongs to some \( {S}_{i} \in C \) . Since \( C \) is a chain (and since there are only finitely many \( {s}_{i} \) ), one of these sets (say \( {S}_{j} \) ) contains all the others. Since \( {S}_{j} \) is linearly independent, we conclude that \( \sum \left| {\alpha }_{i}\right| = 0 \) . This establishes the linear independence of \( {S}^{ * } \) and the fact that every chain in our partially ordered set has an upper bound. Now by Zorn's Lemma, the collection of all linearly independent sets in \( X \) has a maximal element, \( H \) . To see that \( H \) is a Hamel base, let \( x \) be any element of \( X \) . By the maximality of \( H \), either \( H \cup \{ x\} \) is linearly dependent or \( H \cup \{ x\} \subset H \) (and then \( x \in H \) ). In either case, \( x \) is a linear combination of elements of \( H \) . If \( x \) can be represented in two different ways as a linear combination of members of \( H \), then by subtraction, we obtain 0 as a nontrivial linear combination of elements of \( H \), contradicting the linear independence of \( H \) .
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Yes
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Theorem 2. Every nontrivial vector space has a Hamel base.
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Proof. Let \( X \) be a nontrivial vector space. To show that \( X \) has a Hamel base we first prove that \( X \) has a maximal linearly independent set, and then we show that any such set is necessarily a Hamel base. Consider the collection of all linearly independent subsets of \( X \), and partially order this collection by inclusion, C. In order to use Zorn's Lemma, we verify that every chain in this partially ordered set has an upper bound. Let \( C \) be a chain. Consider \( {S}^{ * } = \bigcup \{ S : S \in C\} \) . This certainly satisfies \( S \subset {S}^{ * } \) for all \( S \in C \) . But is \( {S}^{ * } \) linearly independent? Suppose that \( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{s}_{i} = 0 \) for some scalars \( {\alpha }_{i} \) and for some distinct points \( {s}_{i} \) in \( {S}^{ * } \) . Each \( {s}_{i} \) belongs to some \( {S}_{i} \in C \) . Since \( C \) is a chain (and since there are only finitely many \( {s}_{i} \) ), one of these sets (say \( {S}_{j} \) ) contains all the others. Since \( {S}_{j} \) is linearly independent, we conclude that \( \sum \left| {\alpha }_{i}\right| = 0 \) . This establishes the linear independence of \( {S}^{ * } \) and the fact that every chain in our partially ordered set has an upper bound. Now by Zorn's Lemma, the collection of all linearly independent sets in \( X \) has a maximal element, \( H \) . To see that \( H \) is a Hamel base, let \( x \) be any element of \( X \) . By the maximality of \( H \), either \( H \cup \{ x\} \) is linearly dependent or \( H \cup \{ x\} \subset H \) (and then \( x \in H \) ). In either case, \( x \) is a linear combination of elements of \( H \) . If \( x \) can be represented in two different ways as a linear combination of members of \( H \), then by subtraction, we obtain 0 as a nontrivial linear combination of elements of \( H \), contradicting the linear independence of \( H \) .
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Yes
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Theorem 2. Every nontrivial vector space has a Hamel base.
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Proof. Let \( X \) be a nontrivial vector space. To show that \( X \) has a Hamel base we first prove that \( X \) has a maximal linearly independent set, and then we show that any such set is necessarily a Hamel base. Consider the collection of all linearly independent subsets of \( X \), and partially order this collection by inclusion, C. In order to use Zorn's Lemma, we verify that every chain in this partially ordered set has an upper bound. Let \( C \) be a chain. Consider \( {S}^{ * } = \bigcup \{ S : S \in C\} \) . This certainly satisfies \( S \subset {S}^{ * } \) for all \( S \in C \) . But is \( {S}^{ * } \) linearly independent? Suppose that \( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{s}_{i} = 0 \) for some scalars \( {\alpha }_{i} \) and for some distinct points \( {s}_{i} \) in \( {S}^{ * } \) . Each \( {s}_{i} \) belongs to some \( {S}_{i} \in C \) . Since \( C \) is a chain (and since there are only finitely many \( {s}_{i} \) ), one of these sets (say \( {S}_{j} \) ) contains all the others. Since \( {S}_{j} \) is linearly independent, we conclude that \( \sum \left| {\alpha }_{i}\right| = 0 \) . This establishes the linear independence of \( {S}^{ * } \) and the fact that every chain in our partially ordered set has an upper bound. Now by Zorn's Lemma, the collection of all linearly independent sets in \( X \) has a maximal element, \( H \) . To see that \( H \) is a Hamel base, let \( x \) be any element of \( X \) . By the maximality of \( H \), either \( H \cup \{ x\} \) is linearly dependent or \( H \cup \{ x\} \subset H \) (and then \( x \in H \) ). In either case, \( x \) is a linear combination of elements of \( H \) . If \( x \) can be represented in two different ways as a linear combination of members of \( H \), then by subtraction, we obtain 0 as a nontrivial linear combination of elements of \( H \), contradicting the linear independence of \( H \) .
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Yes
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Corollary 1. Let \( \phi \) be a linear functional defined on a subspace \( Y \) in a normed linear space \( X \) and satisfying\n\n\[ \left| {\phi \left( y\right) }\right| \leq M\parallel y\parallel \;\left( {y \in Y}\right) \]\n\nThen \( \phi \) has a linear extension defined on all of \( X \) and satisfying the above inequality on \( X \) .
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Proof. Use the Hahn-Banach Theorem with \( p\left( x\right) = M\parallel x\parallel \) .
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Yes
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Corollary 2. Let \( Y \) be a subspace in a normed linear space \( X \) . If \( w \in X \) and \( \operatorname{dist}\left( {w, Y}\right) > 0 \), then there exists a continuous linear functional \( \phi \) defined on \( X \) such that \( \phi \left( y\right) = 0 \) for all \( y \in Y,\phi \left( w\right) = 1 \) , and \( \parallel \phi \parallel = 1/\operatorname{dist}\left( {w, Y}\right) \) .
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Proof. Let \( Z \) be the subspace generated by \( Y \) and \( w \) . Each element of \( Z \) has a unique representation as \( y + {\lambda w} \), where \( y \in Y \) and \( \lambda \in \mathbb{R} \) . It is clear that \( \phi \) must be defined on \( Z \) by writing \( \phi \left( {y + {\lambda w}}\right) = \lambda \) . The norm of \( \phi \) on \( Z \) is computed as follows, in which the supremum is over all nonzero vectors in \( Z \) :\n\n\[ \parallel \phi \parallel = \sup \;\left| {\phi \left( {y + {\lambda w}}\right) /\parallel y + {\lambda w}\parallel = \sup \;}\right| \lambda |/\parallel y + {\lambda w}\parallel = \sup \;1/\parallel y/\lambda + w\parallel \]\n\n\[ = 1/\inf \parallel y + w\parallel = 1/\operatorname{dist}\left( {w, Y}\right) \]\n\nBy Corollary 1, we can extend the functional \( \phi \) to all of \( X \) without increase of its norm.
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Yes
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Corollary 3. To each point \( w \) in a normed linear space there corresponds a continuous linear functional \( \phi \) such that \( \parallel \phi \parallel = 1 \) and \( \phi \left( w\right) = \parallel w\parallel \)
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Proof. In Corollary 2, take \( Y \) to be the 0 -subspace.
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No
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Example 4. Let \( X = {\mathbb{R}}^{n} \), endowed with the max-norm. Then \( {X}^{ * } \) is (or can be identified with) \( {\mathbb{R}}^{n} \) with the norm \( \parallel {\parallel }_{1} \) . To see that this is so, recall (Problem 1.5.25, page 30) that if \( \phi \in {X}^{ * } \), then \( \phi \left( x\right) = \mathop{\sum }\limits_{{i = 1}}^{n}u\left( i\right) x\left( i\right) \) for a suitable \( u \in {\mathbb{R}}^{n} \) .
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\[ \parallel \phi \parallel = \mathop{\sup }\limits_{{\parallel x{\parallel }_{\infty } \leq 1}}\left| {\mathop{\sum }\limits_{{i = 1}}^{n}u\left( i\right) x\left( i\right) }\right| = \mathop{\sum }\limits_{{i = 1}}^{n}\left| {u\left( i\right) }\right| = \parallel u{\parallel }_{1} \]
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Yes
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Corollary 4. For each \( x \) in a normed linear space \( X \), we have\n\n\[ \parallel x\parallel = \max \left\{ {\left| {\phi \left( x\right) }\right| : \phi \in {X}^{ * },\parallel \phi \parallel = 1}\right\} \]
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Proof. If \( \phi \in {X}^{ * } \) and \( \parallel \phi \parallel = 1 \), then\n\n\[ \left| {\phi \left( x\right) }\right| \leq \parallel \phi \parallel \parallel x\parallel = \parallel x\parallel \]\n\nTherefore,\n\n\[ \sup \left\{ {\left| {\phi \left( x\right) }\right| : \phi \in {X}^{ * },\parallel \phi \parallel = 1}\right\} \leq \parallel x\parallel \]\n\nFor the reverse inequality, note first that it is trivial if \( x = 0 \) . Otherwise, use Corollary 3. Then there is a functional \( \psi \in {X}^{ * } \) such that \( \psi \left( x\right) = \parallel x\parallel \) and \( \parallel \psi \parallel = 1 \) . Note that the supremum is attained.
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Yes
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Theorem 3. A subset in a normed space is fundamental if and only if its annihilator is \( \{ 0\} \) .
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Proof. Let \( X \) be the space and \( Z \) the subset in question. Let \( Y \) be the closure of the linear span of \( Z \) . If \( Y \neq X \), let \( x \in X \smallsetminus Y \) . Then by Corollary 2, there exists \( \phi \in {X}^{ * } \) such that \( \phi \left( x\right) = 1 \) and \( \phi \in {Y}^{ \bot } \) . Hence \( \phi \in {Z}^{ \bot } \) and \( {Z}^{ \bot } \neq 0 \) . If \( Y = X \), then any element of \( {Z}^{ \bot } \) annihilates the span of \( Z \) as well as \( Y \) and \( X \) . Thus it must be the zero functional; i.e., \( {Z}^{ \bot } = 0 \) .
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Yes
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Theorem 3. A subset in a normed space is fundamental if and only if its annihilator is \( \\{ 0\\} \) .
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Proof. Let \( X \) be the space and \( Z \) the subset in question. Let \( Y \) be the closure of the linear span of \( Z \) . If \( Y \\neq X \), let \( x \\in X \\smallsetminus Y \) . Then by Corollary 2, there exists \( \\phi \\in {X}^{ * } \) such that \( \\phi \\left( x\\right) = 1 \) and \( \\phi \\in {Y}^{ \\bot } \) . Hence \( \\phi \\in {Z}^{ \\bot } \) and \( {Z}^{ \\bot } \\neq 0 \) . If \( Y = X \), then any element of \( {Z}^{ \\bot } \) annihilates the span of \( Z \) as well as \( Y \) and \( X \) . Thus it must be the zero functional; i.e., \( {Z}^{ \\bot } = 0 \) .
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Yes
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Theorem 4. If \( X \) is a normed linear space (not necessarily complete)\nthen its conjugate space \( {X}^{ * } \) is complete.
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Proof. This follows from Theorem 4 in Section 1.5, page 27, by letting \( Y = \mathbb{R} \) in that theorem.
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Yes
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Theorem 1. Baire's Theorem. In a complete metric space, the intersection of a countable family of open dense sets is dense.
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Proof. (A set is \
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No
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Theorem 2. The Banach-Steinhaus Theorem. be a family of continuous linear transformations defined on a Banach space \( X \) and taking values in a normed linear space. In order that \( \mathop{\sup }\limits_{\alpha }\begin{Vmatrix}{A}_{\alpha }\end{Vmatrix} < \infty \), it is necessary and sufficient that the set \( \{ x \in X \) : \( \left. {\mathop{\sup }\limits_{\alpha }\begin{Vmatrix}{{A}_{\alpha }x}\end{Vmatrix} < \infty }\right\} \) be of the second category in \( X \) .
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Proof. Assume first that \( c = \mathop{\sup }\limits_{\alpha }\begin{Vmatrix}{A}_{\alpha }\end{Vmatrix} < \infty \) . Then every \( x \) satisfies \( \begin{Vmatrix}{{A}_{\alpha }x}\end{Vmatrix} \leq \) \( c\parallel x\parallel \), and every \( x \) belongs to the set \( F = \left\{ {x : \mathop{\sup }\limits_{\alpha }\begin{Vmatrix}{{A}_{\alpha }x}\end{Vmatrix} < \infty }\right\} \) . Since \( F = X \) , the preceding corollary implies that \( F \) is of the second category in \( X \) .\n\nFor the sufficiency, define\n\n\[ \n{F}_{n} = \left\{ {x \in X : \mathop{\sup }\limits_{\alpha }\begin{Vmatrix}{{A}_{\alpha }x}\end{Vmatrix} \leq n}\right\} \n\]\nand assume that \( F \) is of the second category in \( X \) . Notice that \( F = \mathop{\bigcup }\limits_{{n = 1}}^{\infty }{F}_{n} \) . Since \( F \) is of the second category, and each \( {F}_{n} \) is a closed subset of \( X \), the definition of second category implies that some \( {F}_{m} \) contains a ball. Suppose that\n\n\[ \nB \equiv \left\{ {x \in X : \begin{Vmatrix}{x - {x}_{0}}\end{Vmatrix} \leq r}\right\} \subset {F}_{m}\;\left( {r > 0}\right) \n\]\n\nFor any \( x \) satisfying \( \parallel x\parallel \leq 1 \) we have \( {x}_{0} + {rx} \in B \) . Hence\n\n\[ \n\begin{Vmatrix}{{A}_{\alpha }x}\end{Vmatrix} = \begin{Vmatrix}{{A}_{\alpha }\left\lbrack {{r}^{-1}\left( {{x}_{0} + {rx} - {x}_{0}}\right) }\right\rbrack }\end{Vmatrix} \n\]\n\n\[ \n\leq {r}^{-1}\begin{Vmatrix}{{A}_{\alpha }\left( {{x}_{0} + {rx}}\right) }\end{Vmatrix} + {r}^{-1}\begin{Vmatrix}{{A}_{\alpha }{x}_{0}}\end{Vmatrix} \leq 2{r}^{-1}m \n\]\n\nHence \( \begin{Vmatrix}{A}_{\alpha }\end{Vmatrix} \leq 2{r}^{-1}m \) for all \( \alpha \) .
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Yes
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Consider the familiar space \( C\left\lbrack {0,1}\right\rbrack \) . We are going to show that most members of \( C\left\lbrack {0,1}\right\rbrack \) are not differentiable.
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Select a point \( \xi \) in the open interval \( \left( {0,1}\right) \) . For small positive values of \( h \) we define a linear functional \( {\phi }_{h} \) by the equation\n\n\[ \n{\phi }_{h}\left( x\right) = \frac{x\left( {\xi + h}\right) - x\left( {\xi - h}\right) }{2h}\;\left( {x \in C\left\lbrack {0,1}\right\rbrack }\right) \n\]\n\nIt is elementary to prove that \( {\phi }_{h} \) is linear and that \( \begin{Vmatrix}{\phi }_{h}\end{Vmatrix} = {h}^{-1} \) . Consequently, by the Banach-Steinhaus Theorem, the set of \( x \) such that \( \mathop{\sup }\limits_{h}\left| {{\phi }_{h}\left( x\right) }\right| < \infty \) is of the first category. Hence the set of \( x \) for which \( \mathop{\sup }\limits_{h}\left| {{\phi }_{h}\left( x\right) }\right| = \infty \) is of the second category in \( C\left\lbrack {0,1}\right\rbrack \) . In other words, the set of functions in \( C\left\lbrack {0,1}\right\rbrack \) that are not differentiable at \( \xi \) is of the second category in \( C\left\lbrack {0,1}\right\rbrack \) .
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Yes
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Theorem 4. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a sequence of continuous linear transformations from a Banach space \( X \) into a normed linear space. In order that \( \mathop{\lim }\limits_{n}{A}_{n}x = 0 \) for all \( x \in X \) it is necessary and sufficient that \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{A}_{n}\end{Vmatrix} < \infty \) and that \( {A}_{n}u \rightarrow 0 \) for each \( u \) in some fundamental subset of \( X \) .
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Proof. If \( {A}_{n}x \rightarrow 0 \) for all \( x \), then obviously \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{{A}_{n}x}\end{Vmatrix} < \infty \) for all \( x \) . Hence \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{A}_{n}\end{Vmatrix} < \infty \), by the Principle of Uniform Boundedness.\n\nFor the other half of the theorem, assume that \( \begin{Vmatrix}{A}_{n}\end{Vmatrix} < M \) for all \( n \) and that \( {A}_{n}u \rightarrow 0 \) for all \( u \) in a fundamental set \( F \) . It is elementary to prove that \( {A}_{n}y \rightarrow 0 \) for all \( y \) in the linear span of \( F \) . Now let \( x \in X \) . Let \( \epsilon > 0 \) . Select \( y \) in the linear span of \( F \) so that \( \parallel x - y\parallel < \epsilon /{2M} \) . Select \( m \) so that \( \begin{Vmatrix}{{A}_{n}y}\end{Vmatrix} < \epsilon /2 \) whenever \( n \geq m \) . Then for \( n \geq m \) we have\n\n\[ \begin{Vmatrix}{{A}_{n}x}\end{Vmatrix} \leq \begin{Vmatrix}{{A}_{n}\left( {x - y}\right) }\end{Vmatrix} + \begin{Vmatrix}{{A}_{n}y}\end{Vmatrix} < M\parallel x - y\parallel + \epsilon /2 < \epsilon \]
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Yes
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Theorem 4. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a sequence of continuous linear transformations from a Banach space \( X \) into a normed linear space. In order that \( \mathop{\lim }\limits_{n}{A}_{n}x = 0 \) for all \( x \in X \) it is necessary and sufficient that \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{A}_{n}\end{Vmatrix} < \infty \) and that \( {A}_{n}u \rightarrow 0 \) for each \( u \) in some fundamental subset of \( X \) .
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Proof. If \( {A}_{n}x \rightarrow 0 \) for all \( x \), then obviously \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{{A}_{n}x}\end{Vmatrix} < \infty \) for all \( x \) . Hence \( \mathop{\sup }\limits_{n}\begin{Vmatrix}{A}_{n}\end{Vmatrix} < \infty \), by the Principle of Uniform Boundedness.\n\nFor the other half of the theorem, assume that \( \begin{Vmatrix}{A}_{n}\end{Vmatrix} < M \) for all \( n \) and that \( {A}_{n}u \rightarrow 0 \) for all \( u \) in a fundamental set \( F \) . It is elementary to prove that \( {A}_{n}y \rightarrow 0 \) for all \( y \) in the linear span of \( F \) . Now let \( x \in X \) . Let \( \epsilon > 0 \) . Select \( y \) in the linear span of \( F \) so that \( \parallel x - y\parallel < \epsilon /{2M} \) . Select \( m \) so that \( \begin{Vmatrix}{{A}_{n}y}\end{Vmatrix} < \epsilon /2 \) whenever \( n \geq m \) . Then for \( n \geq m \) we have\n\n\[ \begin{Vmatrix}{{A}_{n}x}\end{Vmatrix} \leq \begin{Vmatrix}{{A}_{n}\left( {x - y}\right) }\end{Vmatrix} + \begin{Vmatrix}{{A}_{n}y}\end{Vmatrix} < M\parallel x - y\parallel + \epsilon /2 < \epsilon \]
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Yes
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Theorem 5. Let \( \psi \) and \( {\phi }_{n} \) be as in Equations (1) and (2) above. In order that \( {\phi }_{n}\left( x\right) \rightarrow \psi \left( x\right) \) for each \( x \in C\left\lbrack {a, b}\right\rbrack \), it is necessary and sufficient that these two conditions be fulfilled:\n\n(i) \( \mathop{\sup }\limits_{n}\mathop{\sum }\limits_{{i = 1}}^{n}\left| {A}_{ni}\right| < \infty \)\n\n(ii) The convergence occurs for all the elementary monomial functions, \( s \mapsto {s}^{k}, k = 0,1,2,\ldots \) .
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Proof. Consider the sequence of functionals \( \left\lbrack {\psi - {\phi }_{n}}\right\rbrack \) . The norm of \( \psi \) is\n\n\[ \parallel \psi \parallel = \mathop{\sup }\limits_{{\parallel x\parallel \leq 1}}\left| {{\int }_{a}^{b}x\left( s\right) w\left( s\right) {dx}}\right| \leq {\int }_{a}^{b}\left| {w\left( s\right) }\right| {ds} \]\n\nConsequently, condition (i) is equivalent to the condition\n\n\[ \mathop{\sup }\limits_{n}\begin{Vmatrix}{\psi - {\phi }_{n}}\end{Vmatrix} < \infty \]\n\nNext observe that the functions \( {e}_{k} \) defined by the equation \( {e}_{k}\left( s\right) = {s}^{k} \), where \( k = 0,1,\ldots \), form a fundamental set in \( C\left\lbrack {a, b}\right\rbrack \), by the Weierstrass Polynomial Approximation Theorem. Now apply the preceding theorem.
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Yes
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Theorem 1. Let \( {x}_{n} \in {C}^{1}\left\lbrack {a, b}\right\rbrack ,{\begin{Vmatrix}{x}_{n} - x\end{Vmatrix}}_{\infty } \rightarrow 0 \), and \( {\begin{Vmatrix}{x}_{n}^{\prime } - y\end{Vmatrix}}_{\infty } \rightarrow 0 \) . Then \( y \in C\left\lbrack {a, b}\right\rbrack \) and \( {x}^{\prime } = y \) .
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Proof. Since \( {x}_{n} \in {C}^{1}\left\lbrack {a, b}\right\rbrack \), we have \( {x}_{n}^{\prime } \in C\left\lbrack {a, b}\right\rbrack \) . Thus \( y \in C\left\lbrack {a, b}\right\rbrack \), by Theorem 2 in Section 1.2, page 10. By the Fundamental Theorem of Calculus and the continuity of integration, \[ {\int }_{a}^{t}y\left( s\right) {ds} = {\int }_{a}^{t}\mathop{\lim }\limits_{n}{x}_{n}^{\prime }\left( s\right) {ds} = \mathop{\lim }\limits_{n}{\int }_{a}^{t}{x}_{n}^{\prime }\left( s\right) {ds} = \mathop{\lim }\limits_{n}\left\lbrack {{x}_{n}\left( t\right) - {x}_{n}\left( a\right) }\right\rbrack = x\left( t\right) - x\left( a\right) \] Differentiation with respect to \( t \) now yields \( y\left( t\right) = {x}^{\prime }\left( t\right) \) .
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Yes
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Corollary 1. If an algebraic isomorphism of one Banach space onto another is continuous, then its inverse is continuous.
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Proof. Let \( L : X \rightarrow Y \) be such a map. (The two-headed arrow denotes a surjective map. Thus \( L\left( X\right) = Y \) .) Being continuous, \( L \) is closed. By the Interior Mapping Theorem, \( L \) is an interior map. Hence \( {L}^{-1} \) is continuous. (Recall that a map \( f \) is continuous if \( {f}^{-1} \) carries open sets to open sets.)
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Yes
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Corollary 2. If a linear space can be made into a Banach space with two norms, one of which dominates the other, then these norms are equivalent.
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Proof. Let \( X \) be the space, and \( {N}_{1},{N}_{2} \) the two norms. The equivalence of two norms is explained in Problem 1.4.3, page 23. Let \( I \) denote the identity map acting from \( \left( {X,{N}_{2}}\right) \) to \( \left( {X,{N}_{1}}\right) \) . Assume that the norms bear the relationship \( {N}_{1} \leq {N}_{2} \) . Since \( {N}_{1}\left( {Ix}\right) \leq {N}_{2}\left( x\right) \), we see that \( I \) is continuous. By the preceding corollary, \( {I}^{-1} \) is continuous. Hence for some \( \alpha ,{N}_{2}\left( x\right) = {N}_{2}\left( {{I}^{-1}x}\right) \leq \alpha {N}_{1}\left( x\right) \) .
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Yes
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