I0 and Rank-Into-Rank Axioms
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Even Ordinals and the Kunen Inconsistency∗
Even ordinals and the Kunen inconsistency∗ Gabriel Goldberg Evans Hall University Drive Berkeley, CA 94720 July 23, 2021 Abstract This paper contributes to the theory of large cardinals beyond the Kunen inconsistency, or choiceless large cardinal axioms, in the context where the Axiom of Choice is not assumed. The first part of the paper investigates a periodicity phenomenon: assuming choiceless large cardinal axioms, the properties of the cumulative hierarchy turn out to alternate between even and odd ranks. The second part of the paper explores the structure of ultrafilters under choiceless large cardinal axioms, exploiting the fact that these axioms imply a weak form of the author's Ultrapower Axiom [1]. The third and final part of the paper examines the consistency strength of choiceless large cardinals, including a proof that assuming DC, the existence of an elementary embedding j : Vλ+3 ! Vλ+3 implies the consistency of ZFC + I0. embedding j : Vλ+3 ! Vλ+3 implies that every subset of Vλ+1 has a sharp. We show that the existence of an elementary embedding from Vλ+2 to Vλ+2 is equiconsistent with the existence of an elementary embedding from L(Vλ+2) to L(Vλ+2) with critical point below λ. We show that assuming DC, the existence of an elementary embedding j : Vλ+3 ! Vλ+3 implies the consistency of ZFC + I0. By a recent result of Schlutzenberg [2], an elementary embedding from Vλ+2 to Vλ+2 does not suffice. 1 Introduction Assuming the Axiom of Choice, the large cardinal hierarchy comes to an abrupt halt in the vicinity of an !-huge cardinal. -
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. -
Arxiv:2101.07455V2 [Math.LO] 13 Feb 2021 Ilas Ics Ti Hssection
HOW STRONG IS A REINHARDT SET OVER EXTENSIONS OF CZF? HANUL JEON Abstract. We investigate the lower bound of the consistency strength of CZF with Full Separation Sep and a Reinhardt set, a constructive analogue of Reinhardt cardinals. We show that CZF + Sep with a Reinhardt set interprets ZF− with a cofinal elementary embedding j : V ≺ V . We also see that CZF + Sep with a − Reinhardt set interprets ZF with a model of ZF + WA0, the Wholeness axiom for bounded formulas. 1. Introduction Large cardinals are one of the important topics of set theory: the linear hierarchy of large cardinals provides a scale to fathom the consistency strength of a given set-theoretic statement. Reinhardt introduced a quite strong notion of large cardinal, now known as Reinhardt cardinal. Unfortunately, Kunen [15] showed that Reinhardt cardinals do not exist in ZFC. However, Kunen’s proof relied on the Axiom of Choice, so it remains the hope that Reinhardt cardinals are consistent if we dispose of the Axiom of Choice. Recently, there are attempts to study Reinhardt cardinals over ZF and find intrinsic evidence for choiceless large cardinals, cardinals that are incompatible with the Axiom of Choice. We may ask choiceless large cardinals are actually consistent with subtheories of ZFC, and there are some positive answers for this question. For example, Schultzenberg [21] showed that ZF with an elementary embedding j : Vλ+2 ≺ Vλ+2 is consistent if ZFC with I0 is. Matthews [17] proved that Reinhardt cardinal is − compatible with ZFC if we assume the consistency of ZFC with I1. We may also ask the lower bound of the consistency strength of large cardinals over subtheories of ZFC, which is quite non-trivial if we remove axioms other than choice. -
Monomorphism - Wikipedia, the Free Encyclopedia
Monomorphism - Wikipedia, the free encyclopedia http://en.wikipedia.org/wiki/Monomorphism Monomorphism From Wikipedia, the free encyclopedia In the context of abstract algebra or universal algebra, a monomorphism is an injective homomorphism. A monomorphism from X to Y is often denoted with the notation . In the more general setting of category theory, a monomorphism (also called a monic morphism or a mono) is a left-cancellative morphism, that is, an arrow f : X → Y such that, for all morphisms g1, g2 : Z → X, Monomorphisms are a categorical generalization of injective functions (also called "one-to-one functions"); in some categories the notions coincide, but monomorphisms are more general, as in the examples below. The categorical dual of a monomorphism is an epimorphism, i.e. a monomorphism in a category C is an epimorphism in the dual category Cop. Every section is a monomorphism, and every retraction is an epimorphism. Contents 1 Relation to invertibility 2 Examples 3 Properties 4 Related concepts 5 Terminology 6 See also 7 References Relation to invertibility Left invertible morphisms are necessarily monic: if l is a left inverse for f (meaning l is a morphism and ), then f is monic, as A left invertible morphism is called a split mono. However, a monomorphism need not be left-invertible. For example, in the category Group of all groups and group morphisms among them, if H is a subgroup of G then the inclusion f : H → G is always a monomorphism; but f has a left inverse in the category if and only if H has a normal complement in G. -
Solutions to Homework Set 5
MAT320/640, Fall 2020 Renato Ghini Bettiol Solutions to Homework Set 5 1. Decide if each of the statements below is true or false. If it is true, give a complete proof; if it is false, give an explicit counter-example. (Recall that a function f : X ! Y is called surjective, or onto, if every point of Y belongs to its image, that is, if f(X) = Y .) (a) There exists a continuous function f : R n f0g ! R. (b) There exists a continuous surjective function f : R n f0g ! R. (c) There exists a continuous function f : [0; 1] ! R. (d) There exists a continuous surjective function f : [0; 1] ! R. (e) The function f : R ! R, f(x) = x2, is uniformly continuous. (f) The function f : [0; 1] ! R, f(x) = x2, uniformly continuous. Solution: (a) TRUE: For example, any constant function f : R n f0g ! R is continuous; e.g., f(x) = 0 for all x 2 R n f0g. (b) TRUE: For example, consider the function f : R n f0g ! R given by ( x; if x < 0; f(x) = x − 1; if x > 0: Since the domain of f is R n f0g, it is continuous on its domain. Moreover, f is clearly surjective (draw its graph). (c) TRUE: For example, any constant function f : [0; 1] ! R is continuous; e.g., f(x) = 0 for all x 2 [0; 1]. (d) FALSE: Suppose, by contradiction, that such a function f : [0; 1] ! R exists. Since [0; 1] is compact and f : [0; 1] ! R is continuous, its image f([0; 1]) is compact. -
Logic and Proof Release 3.18.4
Logic and Proof Release 3.18.4 Jeremy Avigad, Robert Y. Lewis, and Floris van Doorn Sep 10, 2021 CONTENTS 1 Introduction 1 1.1 Mathematical Proof ............................................ 1 1.2 Symbolic Logic .............................................. 2 1.3 Interactive Theorem Proving ....................................... 4 1.4 The Semantic Point of View ....................................... 5 1.5 Goals Summarized ............................................ 6 1.6 About this Textbook ........................................... 6 2 Propositional Logic 7 2.1 A Puzzle ................................................. 7 2.2 A Solution ................................................ 7 2.3 Rules of Inference ............................................ 8 2.4 The Language of Propositional Logic ................................... 15 2.5 Exercises ................................................. 16 3 Natural Deduction for Propositional Logic 17 3.1 Derivations in Natural Deduction ..................................... 17 3.2 Examples ................................................. 19 3.3 Forward and Backward Reasoning .................................... 20 3.4 Reasoning by Cases ............................................ 22 3.5 Some Logical Identities .......................................... 23 3.6 Exercises ................................................. 24 4 Propositional Logic in Lean 25 4.1 Expressions for Propositions and Proofs ................................. 25 4.2 More commands ............................................ -
A New Characterization of Injective and Surjective Functions and Group Homomorphisms
A new characterization of injective and surjective functions and group homomorphisms ANDREI TIBERIU PANTEA* Abstract. A model of a function f between two non-empty sets is defined to be a factorization f ¼ i, where ¼ is a surjective function and i is an injective Æ ± function. In this note we shall prove that a function f is injective (respectively surjective) if and only if it has a final (respectively initial) model. A similar result, for groups, is also proven. Keywords: factorization of morphisms, injective/surjective maps, initial/final objects MSC: Primary 03E20; Secondary 20A99 1 Introduction and preliminary remarks In this paper we consider the sets taken into account to be non-empty and groups denoted differently to be disjoint. For basic concepts on sets and functions see [1]. It is a well known fact that any function f : A B can be written as the composition of a surjective function ! followed by an injective function. Indeed, f f 0 id , where f 0 : A Im(f ) is the restriction Æ ± B ! of f to Im(f ) is one such factorization. Moreover, it is an elementary exercise to prove that this factorization is unique up to an isomorphism: i.e. if f i ¼, i : A X , ¼: X B is 1 ¡ Æ ±¢ ! 1 ! another factorization for f , then we easily see that X i ¡ Im(f ) and that ¼ i ¡ f 0. The Æ Æ ± purpose of this note is to study the dual problem. Let f : A B be a function. We shall define ! a model of the function f to be a triple (X ,i,¼) made up of a set X , an injective function i : A X and a surjective function ¼: X B such that f ¼ i, that is, the following diagram ! ! Æ ± is commutative. -
Proofs and Mathematical Reasoning
Proofs and Mathematical Reasoning University of Birmingham Author: Supervisors: Agata Stefanowicz Joe Kyle Michael Grove September 2014 c University of Birmingham 2014 Contents 1 Introduction 6 2 Mathematical language and symbols 6 2.1 Mathematics is a language . .6 2.2 Greek alphabet . .6 2.3 Symbols . .6 2.4 Words in mathematics . .7 3 What is a proof? 9 3.1 Writer versus reader . .9 3.2 Methods of proofs . .9 3.3 Implications and if and only if statements . 10 4 Direct proof 11 4.1 Description of method . 11 4.2 Hard parts? . 11 4.3 Examples . 11 4.4 Fallacious \proofs" . 15 4.5 Counterexamples . 16 5 Proof by cases 17 5.1 Method . 17 5.2 Hard parts? . 17 5.3 Examples of proof by cases . 17 6 Mathematical Induction 19 6.1 Method . 19 6.2 Versions of induction. 19 6.3 Hard parts? . 20 6.4 Examples of mathematical induction . 20 7 Contradiction 26 7.1 Method . 26 7.2 Hard parts? . 26 7.3 Examples of proof by contradiction . 26 8 Contrapositive 29 8.1 Method . 29 8.2 Hard parts? . 29 8.3 Examples . 29 9 Tips 31 9.1 What common mistakes do students make when trying to present the proofs? . 31 9.2 What are the reasons for mistakes? . 32 9.3 Advice to students for writing good proofs . 32 9.4 Friendly reminder . 32 c University of Birmingham 2014 10 Sets 34 10.1 Basics . 34 10.2 Subsets and power sets . 34 10.3 Cardinality and equality . -
Foundations of Abstract Mathematics Paul L. Bailey
Foundations of Abstract Mathematics Paul L. Bailey Department of Mathematics, Southern Arkansas University E-mail address: [email protected] Date: January 21, 2009 i Contents Preface vii 1. Purpose vii 2. Description of the Blocks vii 2.1. Commentary vii 2.2. Developmental Blocks vii 2.3. Orientation Blocks viii 3. Description of the Volumes viii 3.1. Foundations of Abstract Mathematics viii 3.2. Objects of Abstract Mathematics viii 3.3. Categories of Abstract Mathematics viii 4. Description of this Volume viii 4.1. Goals viii 4.2. Similar Treatments ix Chapter 1. Sets 1 1. Orientation 1 2. Axioms of Set Theory 2 3. Subsets 3 4. Set Operations 4 5. Cartesian Product 7 6. Relations 9 7. Numbers 10 8. Problems and Exercises 11 Chapter 2. Functions 13 1. Mappings 13 2. Functions 14 3. Images and Preimages 15 4. Injective and Surjective Functions 15 5. Restrictions and Constrictions 16 6. Composition of Functions 17 7. Identities and Inverses 17 8. Retractions and Sections 18 9. Monos and Epis 19 10. Inclusions and Projections 20 11. Problems and Exercises 21 Chapter 3. Collections 23 1. Collections 23 2. Power Sets 23 iii iv CONTENTS 3. Mapping Sets 24 4. Function Sets 24 5. Characteristic Functions 25 6. Products 26 7. Coproducts 27 8. Families 28 9. Partitions 29 10. Problems and Exercises 30 Chapter 4. Relations 31 1. Relations 31 2. Properties of Relations 31 3. Order Relations 32 4. Equivalence Relations 34 5. Equivalence Classes 35 6. Equivalence Relations and Partitions 35 7. Equivalence Relations and Functions 37 8. -
Midterm 2 Review the Exam Will Cover Chapters 4 (Functions
Midterm 2 Review The exam will cover Chapters 4 (Functions, injections, surjections, bijections, cardinality of sets etc.) and 13 (sequences, convergence, bounded, monotone etc.). 1 As I stated in class, the first problem on the test will be to state the definition of a sequence (an)n=1 converging to a limit L 2 R. You should also be able to state the other definitions in these chapters, and the exam will also ask for some of these definitions. Here is a list of definitions you should be able to state: • injective function • surjective function • bijective function • cardinality of a finite set • countable set • composition of two functions • convergent sequence • bounded sequence • monotone sequence • supremum and infimum of a set/sequence You should also be able to produce examples of each of these objects, and also examples which have various combinations of these properties (e.g. a function which is injective but not surjective, or a sequence which is bounded but not monotone), as well as when there exist no examples (e.g. an unbounded convergent sequence, which cannot exist because every convergent sequence is bounded). For example, if X = f1; 2; 3g and Y = f4; 5; 6; 7g, then an example of a function f : X ! Y which is injective and not surjective would be to define f as follows: f(1) = 5, f(2) = 4, f(3) = 7. n For example, a sequence which is bounded but not monotone is an = (−1) . You should be able to prove that a given sequence converges, and good practice for this would be (−1)n 1 1 1 p1 showing that sequences like n , n2 , n3 , n , n all converge. -
Consider the “Function” F(X) = 1 X − 4 . in Fact, This Is Not a Function Until I Have Told You It's Domain and Codomain
Consider the “function” 1 f(x) = . x − 4 In fact, this is not a function until I have told you it’s domain and codomain. More accurately we have to talk about the function 1 f : R \{4} → R, f(x) = . x − 4 It’s important to get in the habit of specifying the domain and codomain. In fact the definition of function f : A → B specifies it as a subset of the cartesian produce A × B. From this point of view, our function is the set {(x, 1/(x − 4) : x ∈ R \{4}} ⊂ R \{4} × R 1 Note: my definition of function is different from that of the book. Book does not require that the function be defined everywhere in A. Range The range of a function f : A → B is the subset of B consisting of values of the function. A function is surjective (or onto) if the range equals the codomain. The function f : R \{4} → R given by f(x) = 1/(x − 4) is not surjective because it misses the value 0. That is, the equation g(x) = 0 has no solution in R \{4}. 2 However, if we define a new function g : R \{4} → R \{0}, using the same formula g(x) = 1/(x−4), then g is a surjective function. Note that although the formula is the same, this is a different function from f because we have changed the codomain. To prove g is surjective, we must show that if y ∈ R \{0}, then the equation g(x) = y has a solution x ∈ R\{4}. -
Large Cardinals Beyond Choice
Large Cardinals beyond Choice Joan Bagaria XXVI incontro dell’Associazione Italiana di Logica e sue Applicazioni 25-28 September 2017 Padova This is a report on joint work with Peter Koellner and W. Hugh Woodin Jensen’s Dichotomy Ronald Jensen proved in the early 70’s that either V is very close to L, or very far from it. More precisely: Theorem (R. Jensen) Exactly one of the following hold: 1. Every singular cardinal γ is singular in L, and (γ+)L = γ+. 2. Every uncountable cardinal is inaccessible in L. Woodin’s Dichotomy (Weak form) A similar dichotomy for the model HOD was proved by Woodin a few years ago. Theorem (W. H. Woodin) Suppose that there is an extendible cardinal κ. Then exactly one of the following hold: 1. Every singular cardinal γ > κ is singular in HOD, and (γ+)HOD = γ+. 2. Every regular cardinal γ > κ is measurable in HOD. Two possible futures The existence of large cardinals (e.g.,0 ]) implies the second side of Jensen’s dichotomy, whereas no traditional large cardinals imply that the second side of the HOD dichotomy must hold (because every traditional large cardinal is compatible with V = HOD). Assuming the existence on an extendible cardinal we have two futures: First future: The first alternative of the HOD dichotomy holds and so HOD is close to V . Second future: The second alternative of the HOD dichotomy holds and HOD is far from V . Which future will it be? Two possible outcomes: I The Inner Model Program (as outlined by Woodin) is successful, and then the first future must hold.