Functions and Cardinality of Sets
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A Proof of Cantor's Theorem
Cantor’s Theorem Joe Roussos 1 Preliminary ideas Two sets have the same number of elements (are equinumerous, or have the same cardinality) iff there is a bijection between the two sets. Mappings: A mapping, or function, is a rule that associates elements of one set with elements of another set. We write this f : X ! Y , f is called the function/mapping, the set X is called the domain, and Y is called the codomain. We specify what the rule is by writing f(x) = y or f : x 7! y. e.g. X = f1; 2; 3g;Y = f2; 4; 6g, the map f(x) = 2x associates each element x 2 X with the element in Y that is double it. A bijection is a mapping that is injective and surjective.1 • Injective (one-to-one): A function is injective if it takes each element of the do- main onto at most one element of the codomain. It never maps more than one element in the domain onto the same element in the codomain. Formally, if f is a function between set X and set Y , then f is injective iff 8a; b 2 X; f(a) = f(b) ! a = b • Surjective (onto): A function is surjective if it maps something onto every element of the codomain. It can map more than one thing onto the same element in the codomain, but it needs to hit everything in the codomain. Formally, if f is a function between set X and set Y , then f is surjective iff 8y 2 Y; 9x 2 X; f(x) = y Figure 1: Injective map. -
The Matroid Theorem We First Review Our Definitions: a Subset System Is A
CMPSCI611: The Matroid Theorem Lecture 5 We first review our definitions: A subset system is a set E together with a set of subsets of E, called I, such that I is closed under inclusion. This means that if X ⊆ Y and Y ∈ I, then X ∈ I. The optimization problem for a subset system (E, I) has as input a positive weight for each element of E. Its output is a set X ∈ I such that X has at least as much total weight as any other set in I. A subset system is a matroid if it satisfies the exchange property: If i and i0 are sets in I and i has fewer elements than i0, then there exists an element e ∈ i0 \ i such that i ∪ {e} ∈ I. 1 The Generic Greedy Algorithm Given any finite subset system (E, I), we find a set in I as follows: • Set X to ∅. • Sort the elements of E by weight, heaviest first. • For each element of E in this order, add it to X iff the result is in I. • Return X. Today we prove: Theorem: For any subset system (E, I), the greedy al- gorithm solves the optimization problem for (E, I) if and only if (E, I) is a matroid. 2 Theorem: For any subset system (E, I), the greedy al- gorithm solves the optimization problem for (E, I) if and only if (E, I) is a matroid. Proof: We will show first that if (E, I) is a matroid, then the greedy algorithm is correct. Assume that (E, I) satisfies the exchange property. -
Linear Transformation (Sections 1.8, 1.9) General View: Given an Input, the Transformation Produces an Output
Linear Transformation (Sections 1.8, 1.9) General view: Given an input, the transformation produces an output. In this sense, a function is also a transformation. 1 4 3 1 3 Example. Let A = and x = 1 . Describe matrix-vector multiplication Ax 2 0 5 1 1 1 in the language of transformation. 1 4 3 1 31 5 Ax b 2 0 5 11 8 1 Vector x is transformed into vector b by left matrix multiplication Definition and terminologies. Transformation (or function or mapping) T from Rn to Rm is a rule that assigns to each vector x in Rn a vector T(x) in Rm. • Notation: T: Rn → Rm • Rn is the domain of T • Rm is the codomain of T • T(x) is the image of vector x • The set of all images T(x) is the range of T • When T(x) = Ax, A is a m×n size matrix. Range of T = Span{ column vectors of A} (HW1.8.7) See class notes for other examples. Linear Transformation --- Existence and Uniqueness Questions (Section 1.9) Definition 1: T: Rn → Rm is onto if each b in Rm is the image of at least one x in Rn. • i.e. codomain Rm = range of T • When solve T(x) = b for x (or Ax=b, A is the standard matrix), there exists at least one solution (Existence question). Definition 2: T: Rn → Rm is one-to-one if each b in Rm is the image of at most one x in Rn. • i.e. When solve T(x) = b for x (or Ax=b, A is the standard matrix), there exists either a unique solution or none at all (Uniqueness question). -
I0 and Rank-Into-Rank Axioms
I0 and rank-into-rank axioms Vincenzo Dimonte∗ July 11, 2017 Abstract Just a survey on I0. Keywords: Large cardinals; Axiom I0; rank-into-rank axioms; elemen- tary embeddings; relative constructibility; embedding lifting; singular car- dinals combinatorics. 2010 Mathematics Subject Classifications: 03E55 (03E05 03E35 03E45) 1 Informal Introduction to the Introduction Ok, that’s it. People who know me also know that it is years that I am ranting about a book about I0, how it is important to write it and publish it, etc. I realized that this particular moment is still right for me to write it, for two reasons: time is scarce, and probably there are still not enough results (or anyway not enough different lines of research). This has the potential of being a very nasty vicious circle: there are not enough results about I0 to write a book, but if nobody writes a book the diffusion of I0 is too limited, and so the number of results does not increase much. To avoid this deadlock, I decided to divulge a first draft of such a hypothetical book, hoping to start a “conversation” with that. It is literally a first draft, so it is full of errors and in a liquid state. For example, I still haven’t decided whether to call it I0 or I0, both notations are used in literature. I feel like it still lacks a vision of the future, a map on where the research could and should going about I0. Many proofs are old but never published, and therefore reconstructed by me, so maybe they are wrong. -
Functions and Inverses
Functions and Inverses CS 2800: Discrete Structures, Spring 2015 Sid Chaudhuri Recap: Relations and Functions ● A relation between sets A !the domain) and B !the codomain" is a set of ordered pairs (a, b) such that a ∈ A, b ∈ B !i.e. it is a subset o# A × B" Cartesian product – The relation maps each a to the corresponding b ● Neither all possible a%s, nor all possible b%s, need be covered – Can be one-one, one&'an(, man(&one, man(&man( Alice CS 2800 Bob A Carol CS 2110 B David CS 3110 Recap: Relations and Functions ● ) function is a relation that maps each element of A to a single element of B – Can be one-one or man(&one – )ll elements o# A must be covered, though not necessaril( all elements o# B – Subset o# B covered b( the #unction is its range/image Alice Balch Bob A Carol Jameson B David Mews Recap: Relations and Functions ● Instead of writing the #unction f as a set of pairs, e usually speci#y its domain and codomain as: f : A → B * and the mapping via a rule such as: f (Heads) = 0.5, f (Tails) = 0.5 or f : x ↦ x2 +he function f maps x to x2 Recap: Relations and Functions ● Instead of writing the #unction f as a set of pairs, e usually speci#y its domain and codomain as: f : A → B * and the mapping via a rule such as: f (Heads) = 0.5, f (Tails) = 0.5 2 or f : x ↦ x f(x) ● Note: the function is f, not f(x), – f(x) is the value assigned b( f the #unction f to input x x Recap: Injectivity ● ) function is injective (one-to-one) if every element in the domain has a unique i'age in the codomain – +hat is, f(x) = f(y) implies x = y Albany NY New York A MA Sacramento B CA Boston .. -
Primitive Recursive Functions Are Recursively Enumerable
AN ENUMERATION OF THE PRIMITIVE RECURSIVE FUNCTIONS WITHOUT REPETITION SHIH-CHAO LIU (Received April 15,1900) In a theorem and its corollary [1] Friedberg gave an enumeration of all the recursively enumerable sets without repetition and an enumeration of all the partial recursive functions without repetition. This note is to prove a similar theorem for the primitive recursive functions. The proof is only a classical one. We shall show that the theorem is intuitionistically unprovable in the sense of Kleene [2]. For similar reason the theorem by Friedberg is also intuitionistical- ly unprovable, which is not stated in his paper. THEOREM. There is a general recursive function ψ(n, a) such that the sequence ψ(0, a), ψ(l, α), is an enumeration of all the primitive recursive functions of one variable without repetition. PROOF. Let φ(n9 a) be an enumerating function of all the primitive recursive functions of one variable, (See [3].) We define a general recursive function v(a) as follows. v(0) = 0, v(n + 1) = μy, where μy is the least y such that for each j < n + 1, φ(y, a) =[= φ(v(j), a) for some a < n + 1. It is noted that the value v(n + 1) can be found by a constructive method, for obviously there exists some number y such that the primitive recursive function <p(y> a) takes a value greater than all the numbers φ(v(0), 0), φ(y(Ϋ), 0), , φ(v(n\ 0) f or a = 0 Put ψ(n, a) = φ{v{n), a). -
The Cardinality of a Finite Set
1 Math 300 Introduction to Mathematical Reasoning Fall 2018 Handout 12: More About Finite Sets Please read this handout after Section 9.1 in the textbook. The Cardinality of a Finite Set Our textbook defines a set A to be finite if either A is empty or A ≈ Nk for some natural number k, where Nk = {1,...,k} (see page 455). It then goes on to say that A has cardinality k if A ≈ Nk, and the empty set has cardinality 0. These are standard definitions. But there is one important point that the book left out: Before we can say that the cardinality of a finite set is a well-defined number, we have to ensure that it is not possible for the same set A to be equivalent to Nn and Nm for two different natural numbers m and n. This may seem obvious, but it turns out to be a little trickier to prove than you might expect. The purpose of this handout is to prove that fact. The crux of the proof is the following lemma about subsets of the natural numbers. Lemma 1. Suppose m and n are natural numbers. If there exists an injective function from Nm to Nn, then m ≤ n. Proof. For each natural number n, let P (n) be the following statement: For every m ∈ N, if there is an injective function from Nm to Nn, then m ≤ n. We will prove by induction on n that P (n) is true for every natural number n. We begin with the base case, n = 1. -
Elementary Functions Part 1, Functions Lecture 1.6D, Function Inverses: One-To-One and Onto Functions
Elementary Functions Part 1, Functions Lecture 1.6d, Function Inverses: One-to-one and onto functions Dr. Ken W. Smith Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 26 / 33 Function Inverses When does a function f have an inverse? It turns out that there are two critical properties necessary for a function f to be invertible. The function needs to be \one-to-one" and \onto". Smith (SHSU) Elementary Functions 2013 27 / 33 Function Inverses When does a function f have an inverse? It turns out that there are two critical properties necessary for a function f to be invertible. The function needs to be \one-to-one" and \onto". Smith (SHSU) Elementary Functions 2013 27 / 33 Function Inverses When does a function f have an inverse? It turns out that there are two critical properties necessary for a function f to be invertible. The function needs to be \one-to-one" and \onto". Smith (SHSU) Elementary Functions 2013 27 / 33 Function Inverses When does a function f have an inverse? It turns out that there are two critical properties necessary for a function f to be invertible. The function needs to be \one-to-one" and \onto". Smith (SHSU) Elementary Functions 2013 27 / 33 One-to-one functions* A function like f(x) = x2 maps two different inputs, x = −5 and x = 5, to the same output, y = 25. But with a one-to-one function no pair of inputs give the same output. Here is a function we saw earlier. This function is not one-to-one since both the inputs x = 2 and x = 3 give the output y = C: Smith (SHSU) Elementary Functions 2013 28 / 33 One-to-one functions* A function like f(x) = x2 maps two different inputs, x = −5 and x = 5, to the same output, y = 25. -
Polya Enumeration Theorem
Polya Enumeration Theorem Sasha Patotski Cornell University [email protected] December 11, 2015 Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 1 / 10 Cosets A left coset of H in G is gH where g 2 G (H is on the right). A right coset of H in G is Hg where g 2 G (H is on the left). Theorem If two left cosets of H in G intersect, then they coincide, and similarly for right cosets. Thus, G is a disjoint union of left cosets of H and also a disjoint union of right cosets of H. Corollary(Lagrange's theorem) If G is a finite group and H is a subgroup of G, then the order of H divides the order of G. In particular, the order of every element of G divides the order of G. Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 2 / 10 Applications of Lagrange's Theorem Theorem n! For any integers n ≥ 0 and 0 ≤ m ≤ n, the number m!(n−m)! is an integer. Theorem (ab)! (ab)! For any positive integers a; b the ratios (a!)b and (a!)bb! are integers. Theorem For an integer m > 1 let '(m) be the number of invertible numbers modulo m. For m ≥ 3 the number '(m) is even. Sasha Patotski (Cornell University) Polya Enumeration Theorem December 11, 2015 3 / 10 Polya's Enumeration Theorem Theorem Suppose that a finite group G acts on a finite set X . Then the number of colorings of X in n colors inequivalent under the action of G is 1 X N(n) = nc(g) jGj g2G where c(g) is the number of cycles of g as a permutation of X . -
Cardinality of Sets
Cardinality of Sets MAT231 Transition to Higher Mathematics Fall 2014 MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 1 / 15 Outline 1 Sets with Equal Cardinality 2 Countable and Uncountable Sets MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 2 / 15 Sets with Equal Cardinality Definition Two sets A and B have the same cardinality, written jAj = jBj, if there exists a bijective function f : A ! B. If no such bijective function exists, then the sets have unequal cardinalities, that is, jAj 6= jBj. Another way to say this is that jAj = jBj if there is a one-to-one correspondence between the elements of A and the elements of B. For example, to show that the set A = f1; 2; 3; 4g and the set B = {♠; ~; }; |g have the same cardinality it is sufficient to construct a bijective function between them. 1 2 3 4 ♠ ~ } | MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 3 / 15 Sets with Equal Cardinality Consider the following: This definition does not involve the number of elements in the sets. It works equally well for finite and infinite sets. Any bijection between the sets is sufficient. MAT231 (Transition to Higher Math) Cardinality of Sets Fall 2014 4 / 15 The set Z contains all the numbers in N as well as numbers not in N. So maybe Z is larger than N... On the other hand, both sets are infinite, so maybe Z is the same size as N... This is just the sort of ambiguity we want to avoid, so we appeal to the definition of \same cardinality." The answer to our question boils down to \Can we find a bijection between N and Z?" Does jNj = jZj? True or false: Z is larger than N. -
Enumeration of Finite Automata 1 Z(A) = 1
INFOI~MATION AND CONTROL 10, 499-508 (1967) Enumeration of Finite Automata 1 FRANK HARARY AND ED PALMER Department of Mathematics, University of Michigan, Ann Arbor, Michigan Harary ( 1960, 1964), in a survey of 27 unsolved problems in graphical enumeration, asked for the number of different finite automata. Re- cently, Harrison (1965) solved this problem, but without considering automata with initial and final states. With the aid of the Power Group Enumeration Theorem (Harary and Palmer, 1965, 1966) the entire problem can be handled routinely. The method involves a confrontation of several different operations on permutation groups. To set the stage, we enumerate ordered pairs of functions with respect to the product of two power groups. Finite automata are then concisely defined as certain ordered pah's of functions. We review the enumeration of automata in the natural setting of the power group, and then extend this result to enumerate automata with initial and terminal states. I. ENUMERATION THEOREM For completeness we require a number of definitions, which are now given. Let A be a permutation group of order m = ]A I and degree d acting on the set X = Ix1, x~, -.. , xa}. The cycle index of A, denoted Z(A), is defined as follows. Let jk(a) be the number of cycles of length k in the disjoint cycle decomposition of any permutation a in A. Let al, a2, ... , aa be variables. Then the cycle index, which is a poly- nomial in the variables a~, is given by d Z(A) = 1_ ~ H ~,~(°~ . (1) ~$ a EA k=l We sometimes write Z(A; al, as, .. -
The Axiom of Choice and Its Implications
THE AXIOM OF CHOICE AND ITS IMPLICATIONS KEVIN BARNUM Abstract. In this paper we will look at the Axiom of Choice and some of the various implications it has. These implications include a number of equivalent statements, and also some less accepted ideas. The proofs discussed will give us an idea of why the Axiom of Choice is so powerful, but also so controversial. Contents 1. Introduction 1 2. The Axiom of Choice and Its Equivalents 1 2.1. The Axiom of Choice and its Well-known Equivalents 1 2.2. Some Other Less Well-known Equivalents of the Axiom of Choice 3 3. Applications of the Axiom of Choice 5 3.1. Equivalence Between The Axiom of Choice and the Claim that Every Vector Space has a Basis 5 3.2. Some More Applications of the Axiom of Choice 6 4. Controversial Results 10 Acknowledgments 11 References 11 1. Introduction The Axiom of Choice states that for any family of nonempty disjoint sets, there exists a set that consists of exactly one element from each element of the family. It seems strange at first that such an innocuous sounding idea can be so powerful and controversial, but it certainly is both. To understand why, we will start by looking at some statements that are equivalent to the axiom of choice. Many of these equivalences are very useful, and we devote much time to one, namely, that every vector space has a basis. We go on from there to see a few more applications of the Axiom of Choice and its equivalents, and finish by looking at some of the reasons why the Axiom of Choice is so controversial.