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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. | 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... | Yes |
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}... | 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 wit... | No |
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. | 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\... | Yes |
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 \( A \) . | 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{... | Yes |
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 functio... | 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(... | Yes |
Lemma 15.3.2 The bases of \( {M}^{ \bot } \) are the complements of the bases of \( M \) . | 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 \( {M}^{ \bot } \) . Co... | Yes |
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 \( M\left( X\right) /e = M\left( {X/e}\right) \). | 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 ... | Yes |
Lemma 15.5.2 If \( C \) is a code of dimension \( k \) in \( {\mathbb{F}}^{n} \), then \( M\left( {C}^{ \bot }\right) = \) \( M{\left( C\right) }^{ \bot } \) . | 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... | Yes |
Theorem 15.5.3 If \( C/i \) is the code obtained by shortening \( C \) at coordinate position \( i \), then \( M\left( {C/i}\right) = M\left( C\right) /i \) . | 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. Us... | Yes |
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( ... | 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... | Yes |
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{ i... | Proof. This is left as Exercise 8. | No |
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 \( \gamma \) s... | 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 \( {e}_{i} = \gamma + \varphi \)... | Yes |
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... | 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 cu... | Yes |
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 ... | 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 p... | Yes |
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; ... | 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... | Yes |
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 ... | 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 d... | Yes |
Lemma 15.10.1 The number of spanning trees in a connected graph \( X \) is | \[ \tau \left( X\right) = R\left( {M\left( X\right) ;0,0}\right) . \] | No |
Lemma 15.10.2 The number of acyclic orientations of a graph \( X \) is\n\n\[ \kappa \left( X\right) = R\left( {M\left( X\right) ;1, - 1}\right) . \] | 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 \( y = - 1 \) . | No |
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) . | Proof. This follows directly from Theorem 15.9.3 and Lemma 15.7.3. | No |
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\[ P\left( {X, t}\right) = {\left( -1\right) }^{n - c}{t}^{c}R\left( {M\left( X\right) ; - t, - 1}\right) . \] | 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 wit... | Yes |
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 + \le... | 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 ... | Yes |
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 \[ {\left( -1\right) }^{k}R\left( {M; - q, - 1}\right) \text{.} \] | 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/... | No |
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 | \[ {\left( -1\right) }^{k}R\left( {M; - {q}^{r}, - 1}\right) \text{.} \] | Yes |
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 ... | 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 ... | Yes |
If \( R \) is a rotor of order less than six, then the cycle matroids of \( X \) and \( {X}^{\varphi } \) have the same rank polynomial. | 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... | Yes |
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... | 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 ... | Yes |
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 \( ... | 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 \( f\left( {A \smallsetminus B}\right) \geq \left| {A \smallsetminus B}\right| \) f... | Yes |
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 \( {M}_{i} \) . | 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} \math... | Yes |
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, \[ e\left( {X,\pi }\right) \geq k\left( {\left| \pi \right| - 1}\right) \] | 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 ... | Yes |
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. | 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 comb... | No |
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 \( \widehat{R}\left( Y\right) = \widehat{R}\left( {Y\smallsetminus \{ e, f\} }\right) \) . | 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... | Yes |
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. | 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... | Yes |
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. | 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. | Yes |
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. | 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. | Yes |
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 \( X \) . | 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 edg... | Yes |
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. | 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 th... | Yes |
Lemma 17.3.1 The core of a left-right cycle in a plane graph \( Y \) is a bicycle of \( Y \) . | 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 ... | Yes |
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. | 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... | Yes |
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 \( c - 1 \) . | 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 ... | Yes |
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... | 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 additio... | No |
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 \( Y \) . | 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 \... | Yes |
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. | 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 ... | Yes |
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. | 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 ... | Yes |
Lemma 17.5.1 A chord diagram is a planar graph if and only if its circle graph is bipartite. | 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\nCon... | Yes |
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 \( Y \) . | 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}^{... | Yes |
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 th... | Proof. See Exercise 7. | No |
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}^{ * }... | 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... | Yes |
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 \( R\left( {Y;x, x}\right) \) . | 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}}}^{ \... | Yes |
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}}^{ * } \... | 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 a... | No |
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 \( s \) a... | 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... | Yes |
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 \( {GF}\left( 2\right) \) . | Proof. If \( C \) is an alternating cycle, then form a word \( {\omega }^{\prime } \) by listing the vertex at the end of each \ | No |
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 \( {\operatorname{rk}}_{2}\left( {A\left( Y\right) }\right) = \) \( n + 1 - s \) . | 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 \( {C}_{s} \), an... | Yes |
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 us... | 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 \ | No |
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 \( x \) . | 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 \( \... | Yes |
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 a... | This is a special case of the Heine-Borel theorem. See the discussion before Lemma 1 in Section 1.4, page 20. | No |
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. | 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, th... | Yes |
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. | 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... | Yes |
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. | 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, ... | Yes |
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. | 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... | Yes |
Theorem 4. The inverse image of a closed set by a continuous map is closed. | 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... | Yes |
Theorem 5. If a series in a Banach space is absolutely convergent, then all rearrangements of the series converge to a common value. | 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 ... | Yes |
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. | 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}... | Yes |
Lemma 2. A closed subset of a compact set is compact. | 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. | Yes |
Corollary 1. Every finite-dimensional normed linear space is complete. | 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{... | Yes |
Corollary 2. Every finite-dimensional subspace in a normed linear space is closed. | 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 ... | Yes |
Theorem 2. If the unit ball in a normed linear space is compact, then the space has finite dimension. | 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 subs... | Yes |
If \( X = {\mathbb{R}}^{n} \) and \( Y = {\mathbb{R}}^{m} \), then each linear map of \( X \) into \( Y \) is of the form \( f\left( x\right) = y \) , | \[ y\left( i\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}x\left( j\right) \;\left( {1 \leq i \leq m}\right) \] where the \( {a}_{ij} \) are certain real numbers that form an \( m \times n \) matrix. | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 1. A linear transformation acting between normed linear spaces is continuous if and only if it is continuous at zero. | 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 \; \Rig... | Yes |
Theorem 2. A linear transformation acting between normed linear spaces is continuous if and only if it is bounded. | 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, ... | Yes |
Theorem 2. A linear transformation acting between normed linear spaces is continuous if and only if it is bounded. | 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,... | Yes |
Corollary 2. Every linear transformation from a finite-dimensional normed space to another normed space is continuous. | 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}... | Yes |
Corollary 3. All norms on a finite-dimensional vector space are equivalent, as defined in Problem 3, page 23. | 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. H... | Yes |
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. | 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... | Yes |
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\[{\left( I - A\right) }^{-1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }{A}^{k}\] | 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 ... | Yes |
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. | 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. | No |
Theorem 2. Every nontrivial vector space has a Hamel base. | 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 coll... | Yes |
Theorem 2. Every nontrivial vector space has a Hamel base. | 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 coll... | Yes |
Theorem 2. Every nontrivial vector space has a Hamel base. | 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 coll... | Yes |
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 abov... | Proof. Use the Hahn-Banach Theorem with \( p\left( x\right) = M\parallel x\parallel \) . | Yes |
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 \... | 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 \... | Yes |
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 \( \phi \left( w\right) = \parallel w\parallel \) | Proof. In Corollary 2, take \( Y \) to be the 0 -subspace. | No |
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) = \mat... | \[ \parallel \phi \parallel = \mathop{\sup }\limits_{{\parallel x{\parallel }_{\infty } \leq 1}}\left| {\mathop{\sum }\limits_{{i = 1}}^{n}u\left( i\right) x\left( i\right) }\right| = \mathop{\sum }\limits_{{i = 1}}^{n}\left| {u\left( i\right) }\right| = \parallel u{\parallel }_{1} \] | Yes |
Corollary 4. For each \( x \) in a normed linear space \( X \), we have\n\n\[ \parallel x\parallel = \max \left\{ {\left| {\phi \left( x\right) }\right| : \phi \in {X}^{ * },\parallel \phi \parallel = 1}\right\} \] | 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 =... | Yes |
Theorem 3. A subset in a normed space is fundamental if and only if its annihilator is \( \{ 0\} \) . | 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 \( \phi \in {Y}^{ \bot } \) . Hen... | Yes |
Theorem 3. A subset in a normed space is fundamental if and only if its annihilator is \( \\{ 0\\} \) . | 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... | Yes |
Theorem 4. If \( X \) is a normed linear space (not necessarily complete)\nthen its conjugate space \( {X}^{ * } \) is complete. | Proof. This follows from Theorem 4 in Section 1.5, page 27, by letting \( Y = \mathbb{R} \) in that theorem. | Yes |
Theorem 1. Baire's Theorem. In a complete metric space, the intersection of a countable family of open dense sets is dense. | Proof. (A set is \ | No |
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 \( \{... | 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 ... | Yes |
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. | 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\rbr... | Yes |
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... | 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... | Yes |
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... | 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... | Yes |
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 }\li... | 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\r... | Yes |
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 \( {x}^{\prime } = y \) . | 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 }_{... | Yes |
Corollary 1. If an algebraic isomorphism of one Banach space onto another is continuous, then its inverse is continuous. | 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}^... | Yes |
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. | 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 ... | Yes |
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