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Let \( D = \left( {{d}_{1},{d}_{2},\ldots ,{d}_{n}}\right) \) be a sequence of positive integers where \( n \geq 2 \). (a) Suppose \( D \) is a list of the degrees of vertices of some \( n \) -vertex tree \( T \), that is, \( {d}_{i} \) is the degree of the \( i \) th vertex of \( T \). Explain why \[ \mathop{\sum }\li... | (11.9) | No |
Prove Corollary 11.10.12: If all edges in a finite weighted graph have distinct weights, then the graph has a unique MST. | Hint: Suppose \( M \) and \( N \) were different MST’s of the same graph. Let \( e \) be the smallest edge in one and not the other, say \( e \in M - N \), and observe that \( N + e \) must have a cycle. | No |
Theorem 12.3.1 (Euler's Formula). If a connected graph has a planar embedding, then\n\n\[ v - e + f = 2 \]\n\nwhere \( v \) is the number of vertices, \( e \) is the number of edges, and \( f \) is the number of faces. | Proof. The proof is by structural induction on the definition of planar embeddings. Let \( P\left( \mathcal{E}\right) \) be the proposition that \( v - e + f = 2 \) for an embedding, \( \mathcal{E} \) .\n\nBase case ( \( \mathcal{E} \) is the one-vertex planar embedding): By definition, \( v = 1, e = 0 \) , and \( f = ... | Yes |
Theorem 12.4.3. Suppose a connected planar graph has \( v \geq 3 \) vertices and \( e \) edges. Then\n\n\[ e \leq {3v} - 6\text{.} \] | Proof. By definition, a connected graph is planar iff it has a planar embedding. So suppose a connected graph with \( v \) vertices and \( e \) edges has a planar embedding with \( f \) faces. By Lemma 12.4.1, every edge has exactly two occurrences in the face boundaries. So the sum of the lengths of the face boundarie... | Yes |
Corollary 12.5.1. \( {K}_{5} \) is not planar. | Proof. \( {K}_{5} \) is connected and has 5 vertices and 10 edges. But since \( {10} > 3 \cdot 5 - 6 \) , \( {K}_{5} \) does not satisfy the inequality (12.3) that holds in all planar graphs. | Yes |
Lemma 12.5.2. In a planar embedding of a connected bipartite graph with at least 3 vertices, each face has length at least 4. | Proof. By Lemma 12.4.2, every face of a planar embedding of the graph has length at least 3. But by Lemma 11.7.2 and Theorem 11.9.3.3, a bipartite graph can't have odd length closed walks. Since the faces of a planar embedding are closed walks, there can't be any faces of length 3 in a bipartite embedding. So every fac... | Yes |
Theorem 12.5.3. Suppose a connected bipartite graph with \( v \geq 3 \) vertices and e edges is planar. Then\n\n\[ e \leq {2v} - 4\text{.} \] | Proof. Lemma 12.5.2 implies that all the faces of an embedding of the graph have length at least 4. Now arguing as in the proof of Theorem 12.4.3, we find that the sum of the lengths of the face boundaries is exactly \( {2e} \) and at least \( {4f} \) . Hence,\n\n\[ {4f} \leq {2e} \]\n\nfor any embedding of a planar bi... | Yes |
Corollary 12.5.4. \( {K}_{3,3} \) is not planar. | Proof. \( {K}_{3,3} \) is connected, bipartite and has 6 vertices and 9 edges. But since \( 9 > \) \( 2 \cdot 6 - 4,{K}_{3,3} \) does not satisfy the inequality (12.3) that holds in all bipartite planar graphs. | Yes |
Lemma 12.6.2. Merging two adjacent vertices of a planar graph leaves another planar graph. | Merging two adjacent vertices, \( {n}_{1} \) and \( {n}_{2} \) of a graph means deleting the two vertices and then replacing them by a new \ | No |
Lemma 12.6.3. Every planar graph has a vertex of degree at most five. | Proof. Assuming to the contrary that every vertex of some planar graph had degree at least 6, then the sum of the vertex degrees is at least \( {6v} \) . But the sum of the vertex degrees equals \( {2e} \) by the Handshake Lemma 11.2.1, so we have \( e \geq {3v} \) contradicting the fact that \( e \leq {3v} - 6 < {3v} ... | Yes |
Theorem 12.6.4. Every planar graph is five-colorable. | Proof. The proof will be by strong induction on the number, \( v \), of vertices, with induction hypothesis:\n\nEvery planar graph with \( v \) vertices is five-colorable.\n\nBase cases \( \left( {v \leq 5}\right) \) : immediate.\n\nInductive case: Suppose \( G \) is a planar graph with \( v + 1 \) vertices. We will de... | Yes |
What are the discrete faces of the following two graphs? | Write each cycle as a sequence of letters without spaces, starting with the alphabetically earliest letter in the clockwise direction, for example \ | No |
Theorem 13.1.1. If \( \left| x\right| < 1 \), then\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{\infty }{x}^{i} = \frac{1}{1 - x} \] | Proof.\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{\infty }{x}^{i} : \mathrel{\text{:=}} \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{i = 0}}^{n}{x}^{i} \]\n\n\[ = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1 - {x}^{n + 1}}{1 - x} \]\n(by equation 13.2)\n\n\[ = \frac{1}{1 - x}\text{.} \]\... | Yes |
Theorem 13.1.2. If \( \\left| x\\right| < 1 \\), then\n\n\[ \n\\mathop{\\sum }\\limits_{{i = 1}}^{\\infty }i{x}^{i} = \\frac{x}{{\\left( 1 - x\\right) }^{2}}\n\]\n\n(13.13) | As a consequence, suppose that there is an annuity that pays \( {im} \) dollars at the end of each year \( i \), forever. For example, if \( m = \\$ {50},{000} \), then the payouts are \( \\$ {50},{000} \) and then \( \\$ {100},{000} \) and then \( \\$ {150},{000} \) and so on. It is hard to believe that the value of t... | Yes |
Theorem 13.3.2. Let \( f : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) be a weakly increasing function. Define\n\n\[ S : \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}f\left( i\right) \]\n\nand\n\n\[ I : \mathrel{\text{:=}} {\int }_{1}^{n}f\left( x\right) {dx}. \]\n\nThen\n\n\[ I + f\left( 1\right) \l... | Proof. Suppose \( f : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) is weakly increasing. The value of the sum \( S \) in (13.15) is the sum of the areas of \( n \) unit-width rectangles of heights \( f\left( 1\right), f\left( 2\right) ,\ldots, f\left( n\right) \) . This area of these rectangles is shown shaded ... | No |
Theorem 13.1.1\n\n\[ \n= \frac{\left( {1 - {xy}}\right) - x\left( {1 - y}\right) }{\left( {1 - x}\right) \left( {1 - y}\right) \left( {1 - {xy}}\right) } \n\] | \[ \n= \frac{1 - x}{\left( {1 - x}\right) \left( {1 - y}\right) \left( {1 - {xy}}\right) } \n\]\n\n\[ \n= \frac{1}{\left( {1 - y}\right) \left( {1 - {xy}}\right) }\text{.} \n\] | Yes |
Lemma 13.7.2. \( {x}^{a} = o\left( {x}^{b}\right) \) for all nonnegative constants \( a < b \) . | Using the familiar fact that \( \log x < x \) for all \( x > 1 \), we can prove | No |
Lemma 13.7.3. \( \log x = o\left( {x}^{\epsilon }\right) \) for all \( \epsilon > 0 \) . | Proof. Choose \( \epsilon > \delta > 0 \) and let \( x = {z}^{\delta } \) in the inequality \( \log x < x \) . This implies\n\n\[ \log z < {z}^{\delta }/\delta = o\left( {z}^{\epsilon }\right) \;\text{ by Lemma 13.7.2. } \] | No |
Lemma 13.7.7. If \( f = o\left( g\right) \) or \( f \sim g \), then \( f = O\left( g\right) \) . | Proof. \( \lim f/g = 0 \) or \( \lim f/g = 1 \) implies \( \lim f/g < \infty \), so by Lemma 13.7.6, \( \lim \sup f/g < \infty \) . | Yes |
Lemma 13.7.8. If \( f = o\left( g\right) \), then it is not true that \( g = O\left( f\right) \) . | Proof. \[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{g\left( x\right) }{f\left( x\right) } = \frac{1}{\mathop{\lim }\limits_{{x \rightarrow \infty }}f\left( x\right) /g\left( x\right) } = \frac{1}{0} = \infty , \] so by Lemma 13.7.6, \( g \neq O\left( f\right) \) . | Yes |
Proposition 13.7.11. \( {x}^{2} + {100x} + {10} = O\left( {x}^{2}\right) \) . | Proof. \( \left( {{x}^{2} + {100x} + {10}}\right) /{x}^{2} = 1 + {100}/x + {10}/{x}^{2} \) and so its limit as \( x \) approaches infinity is \( 1 + 0 + 0 = 1 \) . So in fact, \( {x}^{2} + {100x} + {10} \sim {x}^{2} \), and therefore \( {x}^{2} + {100x} + \) \( {10} = O\left( {x}^{2}\right) \) . Indeed, it’s conversely... | Yes |
Proposition 13.7.12. \( {a}_{k}{x}^{k} + {a}_{k - 1}{x}^{k - 1} + \cdots + {a}_{1}x + {a}_{0} = O\left( {x}^{k}\right) \) . | We'll omit the routine proof. | No |
Let \( f : {\mathbb{R}}^{ + } \rightarrow {\mathbb{R}}^{ + } \) be a weakly decreasing function. Define\n\n\[ S : \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}f\left( i\right) \]\n\nand\n\n\[ I : \mathrel{\text{:=}} {\int }_{1}^{n}f\left( x\right) {dx}. \]\n\nProve that\n\n\[ I + f\left( n\right) \leq S \leq ... | (Proof by very clear picture is OK.) | No |
False Claim.\n\n\[ \n{2}^{n} = O\left( 1\right) \n\] | Explain why the claim is false. Then identify and explain the mistake in the following bogus proof.\n\nBogus proof. The proof is by induction on \( n \) where the induction hypothesis, \( P\left( n\right) \) , is the assertion (13.31).\n\nbase case: \( P\left( 0\right) \) holds trivially.\n\ninductive step: We may assu... | Yes |
Lemma 14.1.1. The number of ways to select \( n \) donuts when \( k \) flavors are available is the same as the number of binary sequences with exactly \( n \) zeroes and \( k - 1 \) ones. | This example demonstrates the power of the bijection rule. We managed to prove that two very different sets are actually the same size-even though we don't know exactly how big either one is. But as soon as we figure out the size of one set, we'll immediately know the size of the other. | No |
Lemma 14.10.1 (Pascal's Triangle Identity).\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) = \left( \begin{array}{l} n - 1 \\ k - 1 \end{array}\right) + \left( \begin{matrix} n - 1 \\ k \end{matrix}\right) \] | We proved Pascal's Triangle Identity without any algebra! Instead, we relied purely on counting techniques.\n\n## 14.10.2 Giving a Combinatorial Proof\n\nA combinatorial proof is an argument that establishes an algebraic fact by relying on counting principles. Many such proofs follow the same basic outline:\n\n1. Defin... | No |
Theorem 14.10.2.\n\[ \n\\mathop{\\sum }\\limits_{{r = 0}}^{n}\\left( \\begin{array}{l} n \\\\ r \\end{array}\\right) \\left( \\begin{matrix} {2n} \\\\ n - r \\end{matrix}\\right) = \\left( \\begin{matrix} {3n} \\\\ n \\end{matrix}\\right) \n\] | Proof. We give a combinatorial proof. Let \( S \) be all \( n \) -card hands that can be dealt from a deck containing \( n \) different red cards and \( {2n} \) different black cards. First, note that every \( {3n} \) -element set has\n\n\[ \n\\left| S\\right| = \\left( \\begin{matrix} {3n} \\\\ n \\end{matrix}\\right)... | Yes |
There is a simple relationship between the degree sequence of an \( n \)-vertex numbered tree and the occurrence sequence of its code. Describe this relationship and explain why it holds. Conclude that counting \( n \)-vertex numbered trees with a given degree sequence is the same as counting the number of length \( n ... | Hint: How many times does a vertex of degree, \( d \), occur in the code? | No |
Problem 14.39. (a) Show that the Magician could not pull off the trick with a deck larger than 124 cards. | Hint: Compare the number of 5-card hands in an \( n \) -card deck with the number of 4-card sequences. | No |
Give a combinatorial proof for this identity:\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{n}\left( \begin{matrix} k + i \\ k \end{matrix}\right) = \left( \begin{matrix} k + n + 1 \\ k + 1 \end{matrix}\right) \] | Hint: Let \( {S}_{i} \) be the set of binary sequences with exactly \( n \) zeroes, \( k + 1 \) ones, and a total of exactly \( i \) occurrences of zeroes appearing before the rightmost occurrence of a one. | No |
Each day, an MIT student selects a breakfast from among \( b \) possibilities, lunch from among \( l \) possibilities, and dinner from among \( d \) possibilities. In each case one of the possibilities is Doritos. However, a legimate daily menu may include Doritos for at most one meal. Give a combinatorial (not algebra... | \[ \n{bld} - \\left\\lbrack {\\left( {b - 1}\\right) + \\left( {l - 1}\\right) + \\left( {d - 1}\\right) + 1}\\right\\rbrack \n\]\n\n\[ \n= \\left( {l - 1}\\right) \\left( {d - 1}\\right) + \\left( {b - 1}\\right) \\left( {d - 1}\\right) + \\left( {b - 1}\\right) \\left( {l - 1}\\right) + \\left( {b - 1}\\right) \\left... | Yes |
Theorem 15.2.1 (Maclaurin's Theorem).\n\n\[ f\left( x\right) = f\left( 0\right) + {f}^{\prime }\left( 0\right) x + \frac{{f}^{\prime \prime }\left( 0\right) }{2!}{x}^{2} + \frac{{f}^{\prime \prime \prime }\left( 0\right) }{3!}{x}^{3} + \cdots + \frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n} + \cdots . \] | This theorem says that the \( n \) th coefficient of \( 1/{\left( 1 - x\right) }^{k} \) is equal to its \( n \) th derivative evaluated at 0 and divided by \( n \) !. Computing the \( n \) th derivative turns out not to be very difficult\n\n\[ \frac{{d}^{n}}{{d}^{n}x}\frac{1}{{\left( 1 - x\right) }^{k}} = k\left( {k + ... | Yes |
What is the coefficient of \( {x}^{n} \) in the generating function series for \( S\left( x\right) \) ? | \[ S\left( x\right) : \mathrel{\text{:=}} \frac{{x}^{2} + x}{{\left( 1 - x\right) }^{3}}. \] | No |
We will use generating functions to determine how many ways there are to use pennies, nickels, dimes, quarters, and half-dollars to give \( n \) cents change. | (a) Write the generating function \( P\left( x\right) \) for for the number of ways to use only pennies to make \( n \) cents.\n\n(b) Write the generating function \( N\left( x\right) \) for the number of ways to use only nickels to make \( n \) cents.\n\n(c) Write the generating function for the number of ways to use ... | No |
Let \( {P}_{n} \) denote the number of different collections of \( n \) pets that can accompany her, where we regard chihuahuas and labradors leashed in different orders as different collections. Verify that \( P\left( x\right) = \frac{4{x}^{6}}{{\left( 1 - x\right) }^{2}\left( {1 - {2x}}\right) }.\) | \[ P\left( x\right) = \frac{4{x}^{6}}{{\left( 1 - x\right) }^{2}\left( {1 - {2x}}\right) }.\] | No |
(a) Let \( B\left( x\right) \) be the generating function for the number of ways to bring \( n \) burgers, \( F\left( x\right) \) for the number of ways to bring \( n \) pairs of flip flops, \( T\left( x\right) \) for towels, and \( A\left( x\right) \) for Afghans. Write simple formulas for each of these. | (b) Let \( {g}_{n} \) be the the number of different ways for T-Pain to bring \( n \) items (burgers, pairs of flip flops, towels, and/or afghans) on his boat trip. Let \( G\left( x\right) \) be the generating function \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{g}_{n}{x}^{n} \) . Verify that\n\n\[ G\left( x\right) = ... | No |
(a) Find a recursive definition for \( {c}_{n} \) in terms of \( {c}_{0},{c}_{1},\ldots {c}_{n - 1} \) for \( n \geq 1 \) . (Hint: assuming \( s \) and \( t \) are in RecMatch, what are the possible pairs of integers describing the numbers of left brackets in each such that [] has exactly \( n \) left brackets in total... | Solving for \( C\left( x\right) \) in (15.24) shows that one of the two possible generating functions corresponding to the Catalan numbers is\n\n\[ C\left( x\right) = \frac{1 - \sqrt{1 - {4x}}}{2x} \] | No |
Verify that in the ring of formal power series, \[ r \otimes \left( {{g}_{0},{g}_{1},{g}_{2},\ldots }\right) = \left( {r{g}_{0}, r{g}_{1}, r{g}_{2},\ldots }\right) . \] | In particular, \[ - \left( {{g}_{0},{g}_{1},{g}_{2},\ldots }\right) = - 1 \otimes \left( {{g}_{0},{g}_{1},{g}_{2},\ldots }\right) . \] | No |
Prove the following probabilistic inequality, referred to as the Union Bound.\n\nLet \( {A}_{1},{A}_{2},\ldots ,{A}_{n},\ldots \) be events. Then\n\n\[ \Pr \left\lbrack {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right\rbrack \leq \mathop{\sum }\limits_{{n \in \mathbb{N}}}\Pr \left\lbrack {A}_{n}\right\rbrac... | Hint: Replace the \( {A}_{n} \) ’s by pairwise disjoint events and use the Sum Rule. | No |
Theorem 17.4.1 (Bayes' Rule).\n\n\[ \Pr \left\lbrack {B \mid A}\right\rbrack = \frac{\Pr \left\lbrack {A \mid B}\right\rbrack \cdot \Pr \left\lbrack B\right\rbrack }{\Pr \left\lbrack A\right\rbrack } \] | Proof. We have\n\n\[ \Pr \left\lbrack {B \mid A}\right\rbrack \cdot \Pr \left\lbrack A\right\rbrack = \Pr \left\lbrack {A \cap B}\right\rbrack = \Pr \left\lbrack {A \mid B}\right\rbrack \cdot \Pr \left\lbrack B\right\rbrack \]\n\nby definition of conditional probability. Dividing by \( \Pr \left\lbrack A\right\rbrack \... | Yes |
The Conditional Probability Product Rule for \( n \) Events is\n\nRule.\n\n\[ \Pr \left\lbrack {{E}_{1} \cap {E}_{2} \cap \ldots \cap {E}_{n}}\right\rbrack = \Pr \left\lbrack {E}_{1}\right\rbrack \cdot \Pr \left\lbrack {{E}_{2} \mid {E}_{1}}\right\rbrack \cdot \Pr \left\lbrack {{E}_{3} \mid {E}_{1} \cap {E}_{2}}\right\... | (b) Prove it by induction. | No |
Lemma 18.2.1. Two events are independent iff their indicator variables are independent. | The simple proof is left to Problem 18.1. | No |
Lemma 18.2.2. Let \( R \) and \( S \) be independent random variables, and \( f \) and \( g \) be functions such that \( \operatorname{domain}\left( f\right) = \operatorname{codomain}\left( R\right) \) and \( \operatorname{domain}\left( g\right) = \operatorname{codomain}\left( S\right) \) . Then \( f\left( R\right) \) ... | The proof is another simple exercise left to Problem 18.30. | No |
Lemma 18.4.2. If \( {I}_{A} \) is the indicator random variable for event \( A \), then\n\n\[ \operatorname{Ex}\left\lbrack {I}_{A}\right\rbrack = \Pr \left\lbrack A\right\rbrack \] | Proof.\n\n\[ \operatorname{Ex}\left\lbrack {I}_{A}\right\rbrack = 1 \cdot \Pr \left\lbrack {{I}_{A} = 1}\right\rbrack + 0 \cdot \Pr \left\lbrack {{I}_{A} = 0}\right\rbrack = \Pr \left\lbrack {{I}_{A} = 1}\right\rbrack \]\n\n\[ = \Pr \left\lbrack A\right\rbrack .\;\left( {\text{ def of }{I}_{A}}\right) \] | Yes |
Theorem 18.4.3. For any random variable \( R \) ,\n\n\[ \operatorname{Ex}\left\lbrack R\right\rbrack = \mathop{\sum }\limits_{{x \in \operatorname{range}\left( R\right) }}x \cdot \Pr \left\lbrack {R = x}\right\rbrack \]\n\n(18.3) | The proof of Theorem 18.4.3, like many of the elementary proofs about expectation in this chapter, follows by regrouping of terms in equation (18.2):\n\nProof. Suppose \( R \) is defined on a sample space \( \mathcal{S} \) . Then,\n\n\[ \operatorname{Ex}\left\lbrack R\right\rbrack : \mathrel{\text{:=}} \mathop{\sum }\l... | Yes |
Theorem 18.4.5 (Law of Total Expectation). Let \( R \) be a random variable on a sample space \( \mathcal{S} \), and suppose that \( {A}_{1},{A}_{2},\ldots \), is a partition of \( \mathcal{S} \) . Then\n\n\[ \operatorname{Ex}\left\lbrack R\right\rbrack = \mathop{\sum }\limits_{i}\operatorname{Ex}\left\lbrack {R \mid {... | Proof.\n\n\[ \operatorname{Ex}\left\lbrack R\right\rbrack = \mathop{\sum }\limits_{{r \in \operatorname{range}\left( R\right) }}r \cdot \Pr \left\lbrack {R = r}\right\rbrack \] \n\n(by 18.3)\n\n\[ = \mathop{\sum }\limits_{r}r \cdot \mathop{\sum }\limits_{i}\Pr \left\lbrack {R = r \mid {A}_{i}}\right\rbrack \Pr \left\lb... | Yes |
Theorem 18.5.1. For any random variables \( {R}_{1} \) and \( {R}_{2} \) ,\n\n\[ \operatorname{Ex}\left\lbrack {{R}_{1} + {R}_{2}}\right\rbrack = \operatorname{Ex}\left\lbrack {R}_{1}\right\rbrack + \operatorname{Ex}\left\lbrack {R}_{2}\right\rbrack \] | Proof. Let \( T : \mathrel{\text{:=}} {R}_{1} + {R}_{2} \) . The proof follows straightforwardly by rearranging terms in equation (18.2) in the definition of expectation:\n\n\[ \operatorname{Ex}\left\lbrack T\right\rbrack : \mathrel{\text{:=}} \mathop{\sum }\limits_{{\omega \in \mathcal{S}}}T\left( \omega \right) \cdot... | Yes |
Theorem 18.5.4. Given any collection of events \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \), the expected number of events that will occur is\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\Pr \left\lbrack {A}_{i}\right\rbrack \] | Proof. Define \( {R}_{i} \) to be the indicator random variable for \( {A}_{i} \), where \( {R}_{i}\left( \omega \right) = 1 \) if \( w \in {A}_{i} \) and \( {R}_{i}\left( \omega \right) = 0 \) if \( w \notin {A}_{i} \) . Let \( R = {R}_{1} + {R}_{2} + \cdots + {R}_{n} \) . Then\n\n\[ \operatorname{Ex}\left\lbrack R\ri... | Yes |
Theorem 18.5.5 (Linearity of Expectation). Let \( {R}_{0},{R}_{1},\ldots \), be random variables such that\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{\infty }\operatorname{Ex}\left\lbrack \left| {R}_{i}\right| \right\rbrack \]\n\nconverges. Then\n\n\[ \operatorname{Ex}\left\lbrack {\mathop{\sum }\limits_{{i = 0}}^{\infty }... | Proof. Let \( T : \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 0}}^{\infty }{R}_{i} \) .\n\nWe leave it to the reader to verify that, under the given convergence hypothesis, all the sums in the following derivation are absolutely convergent, which justifies rearranging them as follows:\n\n\[ \mathop{\sum }\limits_{{... | No |
Theorem 18.5.6. For any two independent random variables \( {R}_{1},{R}_{2} \) , \n\n\[ \n\operatorname{Ex}\left\lbrack {{R}_{1} \cdot {R}_{2}}\right\rbrack = \operatorname{Ex}\left\lbrack {R}_{1}\right\r\brack \cdot \operatorname{Ex}\left\lbrack {R}_{2}\right\r\brack \n\] | The proof follows by rearrangement of terms in the sum that defines \( \operatorname{Ex}\left\lbrack {{R}_{1} \cdot {R}_{2}}\right\rbrack \) . Details appear in Problem 18.25. | No |
Are you convinced by this argument? Write out a careful proof of (18.17). | The probability that \( R = S \) is the same as the probability that \( R \) takes whatever value \( S \) happens to have, therefore\n\n\[ \Pr \left\lbrack {R = S}\right\rbrack = \frac{1}{\left| V\right| } \]\n\n(18.17)\n\nHint: The event \( \left\lbrack {R = S}\right\rbrack \) is a disjoint union of events\n\n\[ \left... | No |
Problem 18.26.\n\nApplying linearity of expectation to the binomial distribution \( {f}_{n, p} \) immediately yielded the identity 18.13:\n\n\[ \n\operatorname{Ex}\left\lbrack {f}_{n, p}\right\rbrack : \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 0}}^{n}k\left( \begin{array}{l} n \\ k \end{array}\right) {p}^{k}{\lef... | (b) Now conclude equation (*). | No |
Let \( R \) and \( S \) be independent random variables, and \( f \) and \( g \) be any functions such that \( \operatorname{domain}\left( f\right) = \operatorname{codomain}\left( R\right) \) and \( \operatorname{domain}\left( g\right) = \operatorname{codomain}\left( S\right) \) . Prove that \( f\left( R\right) \) and ... | Hint: The event \( \left\lbrack {f\left( R\right) = a}\right\rbrack \) is the disjoint union of all the events \( \left\lbrack {R = r}\right\rbrack \) for \( r \) such that \( f\left( r\right) = a \) . | No |
Theorem 19.1.1 (Markov's Theorem). If \( R \) is a nonnegative random variable, then for all \( x > 0 \)\n\n\[ \Pr \left\lbrack {R \geq x}\right\rbrack \leq \frac{\operatorname{Ex}\left\lbrack R\right\rbrack }{x} \] | Proof. Let \( y \) vary over the range of \( R \) . Then for any \( x > 0 \)\n\n\[ \operatorname{Ex}\left\lbrack R\right\rbrack : \mathrel{\text{:=}} \mathop{\sum }\limits_{y}y\Pr \left\lbrack {R = y}\right\rbrack \]\n\n\[ \geq \mathop{\sum }\limits_{{y \geq x}}y\Pr \left\lbrack {R = y}\right\rbrack \geq \mathop{\sum }... | Yes |
Corollary 19.1.2. If \( R \) is a nonnegative random variable, then for all \( c \geq 1 \n\n\[ \n\Pr \left\lbrack {R \geq c \cdot \operatorname{Ex}\left\lbrack R\right\rbrack }\right\rbrack \leq \frac{1}{c} \n\] \n\n(19.3) | This Corollary follows immediately from Markov's Theorem(19.1.1) by letting \( x \) be \( c \cdot \operatorname{Ex}\left\lbrack R\right\rbrack \) . | Yes |
Theorem 19.2.3 (Chebyshev). Let \( R \) be a random variable and \( x \in {\mathbb{R}}^{ + } \) . Then\n\n\[ \Pr \left\lbrack {\left| {R - \operatorname{Ex}\left\lbrack R\right\rbrack }\right| \geq x}\right\rbrack \leq \frac{\operatorname{Var}\left\lbrack R\right\rbrack }{{x}^{2}}. \] | The expression \( \operatorname{Ex}\left\lbrack {\left( R - \operatorname{Ex}\left\lbrack R\right\rbrack \right) }^{2}\right\rbrack \) for variance is a bit cryptic; the best approach is to work through it from the inside out. The innermost expression, \( R - \operatorname{Ex}\left\lbrack R\right\rbrack \), is precisel... | No |
The standard deviation of the payoff in Game B is: | \[ {\sigma }_{B} = \sqrt{\operatorname{Var}\left\lbrack B\right\rbrack } = \sqrt{2,{004},{002}} \approx {1416}. \] | Yes |
Corollary 19.2.6. Let \( R \) be a random variable, and let \( c \) be a positive real number. | \[ \Pr \left\lbrack {\left| {R - \operatorname{Ex}\left\lbrack R\right\rbrack }\right| \geq c{\sigma }_{R}}\right\rbrack \leq \frac{1}{{c}^{2}}. \] | Yes |
Lemma 19.3.1.\n\n\\[ \n\\operatorname{Var}\\left\\lbrack R\\right\\rbrack = \\operatorname{Ex}\\left\\lbrack {R}^{2}\\right\\rbrack - {\\operatorname{Ex}}^{2}\\left\\lbrack R\\right\\rbrack \n\\]\n\nfor any random variable, \\( R \\) . | Proof. Let \\( \\mu = \\operatorname{Ex}\\left\\lbrack R\\right\\rbrack \\) . Then\n\n\\[ \n\\operatorname{Var}\\left\\lbrack R\\right\\rbrack = \\operatorname{Ex}\\left\\lbrack {\\left( R - \\operatorname{Ex}\\left\\lbrack R\\right\\rbrack \\right) }^{2}\\right\\rbrack \n\\]\n(Def 19.2.2 of variance)\n\n\\[ \n= \\oper... | Yes |
Corollary 19.3.2. If \( B \) is a Bernoulli variable where \( p : \mathrel{\text{:=}} \Pr \left\lbrack {B = 1}\right\rbrack \), then\n\n\[ \operatorname{Var}\left\lbrack B\right\rbrack = p - {p}^{2} = p\left( {1 - p}\right) . \] | Proof. By Lemma 18.4.2, \( \operatorname{Ex}\left\lbrack B\right\rbrack = p \) . But \( B \) only takes values 0 and 1, so \( {B}^{2} = B \) and equation (19.6) follows immediately from Lemma 19.3.1. | Yes |
Lemma 19.3.3. If failures occur with probability \( p \) independently at each step, and \( C \) is the number of steps until the first failure \( {}^{2} \), then | \[ \operatorname{Var}\left\lbrack C\right\rbrack = \frac{1 - p}{{p}^{2}} \] | No |
Theorem 19.3.4. [Square Multiple Rule for Variance] Let \( R \) be a random variable and a a constant. Then\n\n\[ \operatorname{Var}\left\lbrack {aR}\right\rbrack = {a}^{2}\operatorname{Var}\left\lbrack R\right\rbrack \] | Proof. Beginning with the definition of variance and repeatedly applying linearity of expectation, we have:\n\n\[ \operatorname{Var}\left\lbrack {aR}\right\rbrack : \mathrel{\text{:=}} \operatorname{Ex}\left\lbrack {\left( aR - \operatorname{Ex}\left\lbrack aR\right\rbrack \right) }^{2}\right\rbrack \]\n\n\[ = \mathrm{... | Yes |
Theorem 19.3.7. If \( R \) and \( S \) are independent random variables, then\n\n\[ \operatorname{Var}\left\lbrack {R + S}\right\rbrack = \operatorname{Var}\left\lbrack R\right\rbrack + \operatorname{Var}\left\lbrack S\right\rbrack \] | Proof. We may assume that \( \operatorname{Ex}\left\lbrack R\right\rbrack = 0 \), since we could always replace \( R \) by \( R - \operatorname{Ex}\left\lbrack R\right\rbrack \) in equation (19.11); likewise for \( S \) . This substitution preserves the independence of the variables, and by Theorem 19.3.5, does not cha... | Yes |
Theorem 19.3.8. [Pairwise Independent Additivity of Variance] If \( {R}_{1},{R}_{2},\ldots ,{R}_{n} \) are pairwise independent random variables, then\n\n\[ \operatorname{Var}\left\lbrack {{R}_{1} + {R}_{2} + \cdots + {R}_{n}}\right\rbrack = \operatorname{Var}\left\lbrack {R}_{1}\right\rbrack + \operatorname{Var}\left\... | Now we have a simple way of computing the variance of a variable, \( J \), that has an \( \left( {n, p}\right) \) -binomial distribution. We know that \( J = \mathop{\sum }\limits_{{k = 1}}^{n}{I}_{k} \) where the \( {I}_{k} \) are mutually independent indicator variables with \( \Pr \left\lbrack {{I}_{k} = 1}\right\rb... | No |
Lemma 19.3.9 (Variance of the Binomial Distribution). If \( J \) has the \( \left( {n, p}\right) \) -binomial distribution, then | \n\[\n\operatorname{Var}\left\lbrack J\right\rbrack = n\operatorname{Var}\left\lbrack {I}_{k}\right\rbrack = {np}\left( {1 - p}\right) .\n\]\n\n(19.15) | Yes |
Theorem 19.4.1 (Pairwise Independent Sampling). Let \( {G}_{1},\ldots ,{G}_{n} \) be pairwise independent variables with the same mean, \( \mu \), and deviation, \( \sigma \) . Define\n\n\[ \n{S}_{n} : \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}{G}_{i} \n\]\n\nThen\n\[ \n\Pr \left\lbrack {\left| {\frac{{S}_... | Proof. We observe first that the expectation of \( {S}_{n}/n \) is \( \mu \) :\n\n\[ \n\operatorname{Ex}\left\lbrack \frac{{S}_{n}}{n}\right\rbrack = \operatorname{Ex}\left\lbrack \frac{\mathop{\sum }\limits_{{i = 1}}^{n}{G}_{i}}{n}\right\rbrack \n\]\n\n(def of \( {S}_{n} \) )\n\n\( = \frac{\mathop{\sum }\limits_{{i = ... | Yes |
Lemma 19.6.3.\n\n\\[ \operatorname{Ex}\left\lbrack {c}^{{T}_{i}}\right\rbrack \leq {e}^{\left( {c - 1}\right) \operatorname{Ex}\left\lbrack {T}_{i}\right\rbrack } \\] | Proof. All summations below range over values \\( v \\) taken by the random variable \\( {T}_{i} \\) , which are all required to be in the interval \\( \\left\lbrack {0,1}\\right\rbrack \\) .\n\n\\[ \operatorname{Ex}\left\lbrack {c}^{{T}_{i}}\right\rbrack = \sum {c}^{v}\Pr \\left\lbrack {{T}_{i} = v}\\right\rbrack \\]\... | Yes |
Theorem 19.6.4 (Murphy’s Law). Let \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) be mutually independent events. Let \( {T}_{i} \) be the indicator random variable for \( {A}_{i} \) and define\n\n\[ T : \mathrel{\text{:=}} {T}_{1} + {T}_{2} + \cdots + {T}_{n} \]\n\nto be the number of events that occur. Then\n\n\[ \Pr \left\lb... | Proof.\n\n\[ \Pr \left\lbrack {T = 0}\right\rbrack = \Pr \left\lbrack {{\bar{A}}_{1} \cap {\bar{A}}_{2} \cap \ldots \cap {\bar{A}}_{n}}\right\rbrack \]\n\[ \left( {T = 0\text{ iff no }{A}_{i}\text{ occurs }}\right) \]\n\[ = \mathop{\prod }\limits_{{i = 1}}^{n}\Pr \left\lbrack {\bar{A}}_{i}\right\rbrack \]\n\[ \text{(in... | Yes |
The vast majority of people have an above average number of fingers. Which of the following statements explain why this is true? Explain your reasoning. | 1. Most people have a super secret extra bonus finger of which they are unaware.\n\n2. A pedantic minority don't count their thumbs as fingers, while the majority of people do.\n\n3. Polydactyly is rarer than amputation.\n\n4. When you add up the total number of fingers among the world's population and then divide by t... | No |
Prove that \( \Pr \left\lbrack {\text{no}{A}_{n}\text{occurs}}\right\rbrack = 0 \) . | Hint: \( {B}_{k} \) the event that no \( {A}_{n} \) with \( n \leq k \) occurs. So the event that no \( {A}_{n} \) occurs is\n\n\[ B : \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{k \in \mathbb{N}}}{B}_{k} \]\n\nApply Murphy’s Law, Theorem 19.6.4, to \( {B}_{k} \) . | No |
Corollary 20.1.2. In the Gambler’s Ruin game with initial capital, \( n \), target, \( T \) , and probability \( p < 1/2 \) of winning each individual bet,\n\n\[ \Pr \left\lbrack \text{ the gambler wins }\right\rbrack < {\left( \frac{1}{r}\right) }^{T - n} \]\n\nwhere \( r : \mathrel{\text{:=}} q/p > 1 \) . | So the gambler gains his intended profit before going broke with probability at most \( 1/r \) raised to the intended profit power. Notice that this upper bound does not depend on the gambler's starting capital, but only on his intended profit. This has the amazing consequence we announced above: no matter how much mon... | Yes |
Theorem 20.1.3. In the Gambler’s Ruin game with initial capital \( n \), target \( T \), and probability \( p \) of winning each bet,\n\n\[ \operatorname{Ex}\left\lbrack \text{ number of bets }\right\rbrack = \begin{cases} n\left( {T - n}\right) & \text{ for }p = \frac{1}{2}, \\ \frac{{w}_{n} \cdot T - n}{p - q} & \tex... | In the unbiased case, (20.11) can be rephrased simply as\n\n\[ \operatorname{Ex}\left\lbrack \text{ number of fair bets }\right\rbrack = \text{ initial capital } \cdot \text{ intended profit. } \] | Yes |
Lemma 20.1.4. If the gambler starts with one or more dollars and plays a fair unbounded game, then he will go broke with probability 1. | Proof. If the gambler has initial capital \( n \) and goes broke in a game without reaching a target \( T \), then he would also go broke if he were playing and ignored the target. So the probability that he will lose if he keeps playing without stopping at any target \( T \) must be at least as large as the probabilit... | Yes |
Lemma 20.1.5. If the gambler starts with one or more dollars and plays a fair unbounded game, then his expected number of plays is infinite. | ## A proof appears in Problem 20.2. | No |
Theorem 21.3.1. If \( f\left( n\right) \) and \( g\left( n\right) \) are both solutions to a homogeneous linear recurrence, then \( h\left( n\right) = {sf}\left( n\right) + {tg}\left( n\right) \) is also a solution for all \( s, t \in \mathbb{R} \) . | Proof.\n\n\[ h\left( n\right) = {sf}\left( n\right) + \operatorname{tg}\left( n\right) \]\n\n\[ = s\left( {{a}_{1}f\left( {n - 1}\right) + \cdots + {a}_{d}f\left( {n - d}\right) }\right) + t\left( {{a}_{1}g\left( {n - 1}\right) + \cdots + {a}_{d}g\left( {n - d}\right) }\right) \]\n\n\[ = {a}_{1}\left( {{sf}\left( {n - ... | Yes |
Theorem 21.4.2 (Master Theorem). Let \( T \) be a recurrence of the form\n\n\[ T\left( n\right) = {aT}\left( \frac{n}{b}\right) + g\left( n\right) . \]\n\nCase 1: If \( g\left( n\right) = O\left( {n}^{{\log }_{b}\left( a\right) - \epsilon }\right) \) for some constant \( \epsilon > 0 \), then\n\n\[ T\left( n\right) = \... | The Master Theorem can be proved by induction on \( n \) or, more easily, as a corollary of Theorem 21.4.1. We will not include the details here. | No |
Problem 5.10. Define the set of subformulas of a formula \( \varphi \) of a first-order language \( \mathcal{L} \) . | For example, if \( \varphi \) is\n\n\[ \left( {\left( {\left( {\neg \forall {v}_{1}\left( {\neg = {v}_{1}{c}_{7}}\right) }\right) \rightarrow {P}_{3}^{2}{v}_{5}{v}_{8}}\right) \rightarrow \forall {v}_{8}\left( { = {v}_{8}{f}_{5}^{3}{c}_{0}{v}_{1}{v}_{5} \rightarrow {P}_{2}^{1}{v}_{8}}\right) }\right) \]\n\nin the langu... | Yes |
Example 6.1. Consider the structure \( \mathfrak{R} = \left( {\mathbb{R},0,1,+, \cdot }\right) \) for \( {\mathcal{L}}_{F} \) . Each of the following functions \( V \rightarrow \mathbb{R} \) is an assignment for \( \mathfrak{R} \) : (1) \( p\left( {v}_{n}\right) = \pi \) for each \( n \) , (2) \( r\left( {v}_{n}\right)... | In fact, every function \( V \rightarrow \mathbb{R} \) is an assignment for \( \mathfrak{R} \) . | Yes |
Proposition 8.6. If \( \mathfrak{M} \) is a structure, then \( \operatorname{Th}\left( \mathfrak{M}\right) \) is a maximally consistent set of sentences. | EXAMPLE 8.1. \( \mathfrak{M} = \left( {\{ 5\} }\right) \) is a structure for \( {\mathcal{L}}_{ = } \), so \( \operatorname{Th}\left( \mathfrak{M}\right) \) is a maximally consistent set of sentences. Since it turns out that \( \operatorname{Th}\left( \mathfrak{M}\right) = \) Th \( \left( {\{ \forall x\forall {yx} = y\... | No |
Problem 13.11. Informally, define a computable function which must be different from every primitive recursive function. | Unbounded minimalization. The last of our three method of building computable functions from computable functions is unbounded minimalization. The functions which can be defined from the initial functions using unbounded minimalization, as well as composition and primitive recursion, turn out to be precisely the Turing... | Yes |
Problem 16.1. Pick a short sequence of short formulas of \( {\mathcal{L}}_{N} \) and find the code of the sequence. | A particular integer \( n \) may simultaneously be the Gödel code of a symbol, a sequence of symbols, and a sequence of sequences of symbols of \( {\mathcal{L}}_{N} \) . We shall rely on context to avoid confusion, but, with some more work, one could set things up so that no integer was the code of more than one kind o... | No |
Problem 17.6. Show that the initial functions are representable in \( \operatorname{Th}\left( \mathcal{A}\right) \) : | (1) The zero function \( \mathrm{O}\left( n\right) = 0 \) .\n\n(2) The successor function \( \mathrm{S}\left( n\right) = n + 1 \) .\n\n(3) For every positive \( k \) and \( i \leq k \), the projection function \( {\pi }_{i}^{k} \) . | Yes |
Is the converse to Proposition 18.8 true? | THEOREM 18.10 (Tarski’s Undefinability Theorem). \( \ulcorner \mathrm{{Th}}\left( \mathfrak{N}\right) \urcorner \) is not definable in \( \mathfrak{N} \). | No |
Example 1.2. Let \( \mathbb{N} \) be the set of natural numbers \( \mathbb{N} = \{ 0,1,2,3\ldots \} \) . For \( n, k \in \mathbb{N} \) we define \( n{ \leq }_{D}k \) iff \( n \) divides \( k \) .\n\nIs this a partial order? | Check for P1: Every \( n \) divides \( n \), so each \( n{ \leq }_{D}n \) .\n\nCheck for P2: If \( n \) divides \( k \) then \( n \leq k \) (where \( \leq \) is the usual order). If \( k \) divides \( n \), then \( k \leq n \) . We know that if \( n \leq k \) and \( k \leq n \) then \( n = k \) . Hence if \( n \) divid... | Yes |
Example 1.3. Let \( X \) be any set and, for \( x, y \in X \), define \( x{ \leq }_{E}y \) iff \( x = y \) for all \( x, y \in X \). Is \( { \leq }_{E} \) a partial order? | Check for P1: Since each \( x = x \), each \( x{ \leq }_{E}x \). Check for P2: If \( x{ \leq }_{E}y{ \leq }_{E}x \) then \( x = y \) and (redundantly) \( y = x \). Check for P3: If \( x{ \leq }_{E}y{ \leq }_{E}z \) then \( x = y \) and \( y = z \), so \( x = z \). Hence \( x{ \leq }_{E}z \). | Yes |
Example 1.4. Let \( X \) be any collection of sets, and for all \( x, y \in X \) define \( x{ \leq }_{S}y \) iff \( x \subseteq y{.}^{2} \) | The proof that example 4 is a partial order is left to the reader. | No |
Example 1.5. Consider the set of real numbers \( \mathbb{R} \). For \( x, y \in \mathbb{R} \) we define \( x \preccurlyeq y \) iff \( \exists z{z}^{2} = y - x \). | In fact, in example \( {1.5}, x \preccurlyeq y \) iff \( x \leq y \) in the usual sense. We will come back to this when we discuss models of first order theories. | No |
Is \( { \leq }_{e} \) a partial order? | Check for P1: Immediate from the definition.\n\nCheck for P2: If \( \sigma { \leq }_{e}\tau \) then \( l\left( \sigma \right) \leq l\left( \tau \right) \) . So if \( \sigma { \leq }_{e}\tau { \leq }_{e}\sigma \) then \( l\left( \sigma \right) = l\left( \tau \right) \) . Hence, since there is no \( \rho \neq \varnothing... | Yes |
Example 1.7. Let \( S \) be a set, and let \( X \) be the set of all finite sequences (including the empty sequence) whose elements are in \( S \). For \( \sigma ,\tau \in X,\sigma { \leq }_{e}\tau \) iff there is some \( \rho \in X \) with \( \tau = \sigma \cap \rho \). | That \( { \leq }_{e} \) is also a partial order is left to the reader. | No |
Example 1.9. Let \( X = {\mathbb{N}}^{\mathbb{N}} \), i.e., the set of all functions from \( \mathbb{N} \) to \( \mathbb{N} \) . Define \( f{ \leq }_{ae}g \) iff \( \{ n \) : \( f\left( n\right) > g\left( n\right) \} \) is finite. \( {}^{4} \) | Let’s check P2: Suppose \( f\left( 3\right) \neq g\left( 3\right), f\left( n\right) = g\left( n\right) \) for all \( n \neq 3 \) . Then \( f{ \leq }_{ae}g{ \leq }_{ae}f \) but \( f \neq g \) . So \( \mathrm{P}2 \) fails and \( { \leq }_{ae} \) is not a partial order. | Yes |
Theorem 1.13. \( X \) is linear iff the lexicographic order on \( \operatorname{FIN}\left( X\right) \) is linear. | Proof. Suppose \( X \) is not linear. Let \( x, y \) be incomparable elements of \( X \) . Then \( \left( x\right) ,\left( y\right) \) are incomparable elements of \( {FIN}\left( X\right) \) . Now suppose \( X \) is linear under the order \( \leq \) . Let \( \sigma ,\tau \in \operatorname{FIN}\left( X\right) \) . If \(... | Yes |
If \( X \) is a pre-order under \( \leq \), we define \( x{ \equiv }_{ \leq }y \) iff \( x \leq y \leq x \) . | We show that \( { \equiv }_{ \leq } \) is an equivalence relation:\n\nE1: \( x \leq x \leq x \)\n\nE2: \( x{ \equiv }_{ \leq }y \) iff \( x \leq y \leq x \) iff \( y \leq x \leq y \) iff \( y{ \equiv }_{ \leq }x \)\n\nE3: If \( x \leq y \leq x \) and \( y \leq z \leq y \) then, since \( \leq \) is transitive, \( x \leq... | Yes |
Suppose \( \preccurlyeq \) is a pre-order on \( X \) For \( \left\lbrack x\right\rbrack ,\left\lbrack y\right\rbrack \in X/{ \equiv }_{ \preccurlyeq } \) we write \( \left\lbrack x\right\rbrack { \leq }_{ \preccurlyeq }\left\lbrack y\right\rbrack \) iff \( x \preccurlyeq y \) . | We show that example 1.21 is well-defined, i.e., that it does not depend on the choice of representative for the equivalence class: Suppose \( {x}^{\prime }{ \equiv }_{ \preccurlyeq }x,{y}^{\prime }{ \equiv }_{ \preccurlyeq }y \) . If \( x \preccurlyeq y \) then \( {x}^{\prime } \preccurlyeq x \preccurlyeq y \preccurly... | Yes |
Proposition 1.25. Assume \( \mathfrak{P} \) is a partition of \( X \) and \( \equiv \) is an equivalence relation on \( X \) . (a) \( { \equiv }_{\mathfrak{P}} \) is an equivalence relation on \( X \) . | Proof. We only prove (a) and (d), and leave the rest to the reader. For (a): E1: if \( x \in P \in \mathfrak{P} \) then \( x \in P \) . E2: If \( x, y \in P \in \mathfrak{P} \) then \( y, x \in P \) . E3: If \( x, y \in P \in \mathfrak{P} \) and \( y, z \in {P}^{\prime } \in \mathfrak{P} \), then \( P = {P}^{\prime } \... | No |
Example 1.28. \( X = \left\{ {n + \frac{m}{m + 1} : n, m \in \mathbb{N}}\right\} \) . | Let's give a formal proof that example 1.28 is well-ordered:\n\nSuppose \( Y \subseteq X \) . Because \( \mathbb{N} \) is well-ordered, there is a least \( n \) so that \( \exists {mn} + \frac{m}{m + 1} \in Y \) . For this \( n \), because \( \mathbb{N} \) is well-ordered, there is a least \( m \) so that \( n + \frac{... | No |
Let \( X \) be a partially ordered set, and, for \( n \in \mathbb{N} \), let \( {L}_{n}\left( X\right) \) be the sequences of length \( \leq n \) with elements in \( X \), under the lexicographic order. | We prove that if \( X \) is well-ordered, so is \( {L}_{n}\left( X\right) \) : Let \( Y \subseteq {L}_{n}\left( X\right) \) . Let \( {x}_{1} \) be the least element in \( X \) so that some \( \left( {{x}_{1}{y}_{2}\ldots {y}_{k}}\right) \in Y \) . If \( \left( {x}_{1}\right) \in Y \), we’ve found our minimum. Otherwise... | Yes |
Proposition 1.30. Every subset of a well-ordered set is well-ordered. | Proof. Suppose \( Y \subseteq X, X \) is well-ordered. If \( Z \subseteq Y \) then \( Z \subseteq X \), so \( Z \) has a minimum. Hence every subset of \( Y \) has a minimum. | Yes |
Theorem 1.31. A linear order \( X \) is well-ordered iff it has no infinite descending chain. | Proof. An infinite descending chain is a set \( C = \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) where each \( {x}_{n} > {x}_{n + 1} \) . So suppose \( X \) has such a chain \( C \) . Then \( C \) has no minimum and \( X \) is not well-ordered. For the other direction, suppose \( X \) is not well-ordered. Let \( Y \... | Yes |
Proposition 1.34. Let \( X \) be well-ordered, \( x \in X \) . Either \( x \) is a maximum, or \( x \) has a successor. | Proof. If \( x \) is not a maximum, then \( S = \{ y \in X : y > x\} \neq \varnothing \) . The minimum of \( S \) is the successor of \( x \) . | Yes |
Theorem 1.36. Let \( X \) be a partial order, \( x \in X \) . If \( n < m \) and \( {S}^{m}\left( x\right) \) exists, then \( {S}^{n}\left( x\right) < {S}^{m}\left( x\right) \) . | Proof. Fix \( x \) . Fix \( n \) so \( {S}^{n}\left( x\right) \) exists. By definition 1.32, \( {S}^{n}\left( x\right) < {S}^{n + 1}\left( x\right) < {S}^{n + 2}\left( x\right) \) ... for all \( k \) with \( {S}^{n + k}\left( x\right) \) defined. | Yes |
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