This document is a thesis submitted by Di Wu to the University Honors Program at the University of Minnesota-Twin Cities in partial fulfillment of the requirements for the degree of Bachelor of Arts, summa cum laude in Mathematics. The thesis explores whether there exists a set in the real numbers that is not almost open but still measurable. It begins by defining concepts such as nowhere dense sets, meager sets, almost open sets, and negligible sets. It then discusses examples like the Cantor set that are nowhere dense and meager. The goal is to use these concepts and examples to construct a set that is not almost open but still measurable.
The document discusses different types of sequences and series that the author has learned, including:
1. Arithmetic sequences, where each term is obtained by adding a constant to the preceding term. The author provides examples of finding the common difference and the nth term.
2. Geometric sequences, where each term is obtained by multiplying the preceding term by a constant ratio. Examples are given for finding the common ratio and the nth term.
3. Harmonic sequences, where the reciprocals of the terms form an arithmetic sequence.
4. Fibonacci sequences, where each term is the sum of the two preceding terms, following the pattern of rabbit populations.
The author reflects on appreciating patterns in nature
The document is a guide for teachers on teaching limits and continuity in Years 11-12 mathematics. It begins by motivating the concepts of limits and continuity through examples of functions. It then covers key topics like the limit of a sequence, limiting sums, the limit of a function at infinity, and the limit at a point. It also briefly discusses continuity and the continuity of piecewise functions. The document provides examples and exercises throughout and concludes by discussing links to more advanced concepts and the history of limits.
The document contains information about SAT exams, scoring, and target scores for students at ATHS. It provides 10 tips for scoring higher on the SAT and states that the minimum number of correct answers needed in each section to reach the target score is 10 questions. It also includes a sample SAT worksheet with 20 multiple choice questions testing various math and reasoning skills.
Class 10 Cbse Maths 2010 Sample Paper Model 3 Sunaina Rawat
The document provides information on the design of a mathematics question paper for Class X. It specifies:
1) The weightage and distribution of marks for different content units and forms of questions. Number systems, algebra and geometry make up the bulk of the content with the highest marks.
2) The paper will contain very short answer questions worth 1 mark each, short answer questions worth 2-3 marks each, and long answer questions worth 6 marks.
3) Some questions will provide internal choices while maintaining the overall scheme.
4) Questions will be evenly distributed between easy, average, and difficult levels in terms of marks.
5) Sample papers and blueprints are included based on this design to
This document provides an overview of methods for solving higher order ordinary differential equations (ODEs), including variation of parameters and Laplace transforms. It presents the general steps for each method and works through examples of applying each method to solve higher order nonhomogeneous ODEs. The document also discusses applications of higher order ODEs to model physical systems like a bungee jumper attached to a cord. It works through several problems modeling the bungee jumper's motion to determine properties like the maximum jump height.
This document outlines the contents and references for the unit on linear differential equations of second and higher order. The unit covers topics such as complementary functions, particular integrals, Cauchy's linear equations, Legendre's linear equations, and the method of variation of parameters. It provides 18 slides covering these topics, including examples of solving differential equations using each method. The unit is part of a course on engineering mathematics for first year students.
Vipul was studying late at night when the lights went out. He lit two candles of equal initial length - a thick candle that was supposed to last 6 hours and a thinner candle that was to last 2 hours less. When Vipul finally went to sleep, the thick candle was twice as long as the thin candle. To determine how long Vipul studied, it is assumed he studied for X hours. Equations are then set up relating the amount each candle burned in X hours to their initial and remaining lengths. Solving these equations reveals Vipul studied for 3 hours.
Okay, let's break this down step-by-step:
* Original salary = Rs. X
* He got a 5% raise on his original salary
* 5% of X is 0.05X
* So his salary after the raise was X + 0.05X = 1.05X
* The next year he got a 2.5% cut on the salary after the raise
* 2.5% of 1.05X is 0.025 * 1.05X = 0.02625X
* So his salary after the cut was 1.05X - 0.02625X = 1.02375X
* We're given this salary is Rs. 22702.
The document discusses different types of sequences and series that the author has learned, including:
1. Arithmetic sequences, where each term is obtained by adding a constant to the preceding term. The author provides examples of finding the common difference and the nth term.
2. Geometric sequences, where each term is obtained by multiplying the preceding term by a constant ratio. Examples are given for finding the common ratio and the nth term.
3. Harmonic sequences, where the reciprocals of the terms form an arithmetic sequence.
4. Fibonacci sequences, where each term is the sum of the two preceding terms, following the pattern of rabbit populations.
The author reflects on appreciating patterns in nature
The document is a guide for teachers on teaching limits and continuity in Years 11-12 mathematics. It begins by motivating the concepts of limits and continuity through examples of functions. It then covers key topics like the limit of a sequence, limiting sums, the limit of a function at infinity, and the limit at a point. It also briefly discusses continuity and the continuity of piecewise functions. The document provides examples and exercises throughout and concludes by discussing links to more advanced concepts and the history of limits.
The document contains information about SAT exams, scoring, and target scores for students at ATHS. It provides 10 tips for scoring higher on the SAT and states that the minimum number of correct answers needed in each section to reach the target score is 10 questions. It also includes a sample SAT worksheet with 20 multiple choice questions testing various math and reasoning skills.
Class 10 Cbse Maths 2010 Sample Paper Model 3 Sunaina Rawat
The document provides information on the design of a mathematics question paper for Class X. It specifies:
1) The weightage and distribution of marks for different content units and forms of questions. Number systems, algebra and geometry make up the bulk of the content with the highest marks.
2) The paper will contain very short answer questions worth 1 mark each, short answer questions worth 2-3 marks each, and long answer questions worth 6 marks.
3) Some questions will provide internal choices while maintaining the overall scheme.
4) Questions will be evenly distributed between easy, average, and difficult levels in terms of marks.
5) Sample papers and blueprints are included based on this design to
This document provides an overview of methods for solving higher order ordinary differential equations (ODEs), including variation of parameters and Laplace transforms. It presents the general steps for each method and works through examples of applying each method to solve higher order nonhomogeneous ODEs. The document also discusses applications of higher order ODEs to model physical systems like a bungee jumper attached to a cord. It works through several problems modeling the bungee jumper's motion to determine properties like the maximum jump height.
This document outlines the contents and references for the unit on linear differential equations of second and higher order. The unit covers topics such as complementary functions, particular integrals, Cauchy's linear equations, Legendre's linear equations, and the method of variation of parameters. It provides 18 slides covering these topics, including examples of solving differential equations using each method. The unit is part of a course on engineering mathematics for first year students.
Vipul was studying late at night when the lights went out. He lit two candles of equal initial length - a thick candle that was supposed to last 6 hours and a thinner candle that was to last 2 hours less. When Vipul finally went to sleep, the thick candle was twice as long as the thin candle. To determine how long Vipul studied, it is assumed he studied for X hours. Equations are then set up relating the amount each candle burned in X hours to their initial and remaining lengths. Solving these equations reveals Vipul studied for 3 hours.
Okay, let's break this down step-by-step:
* Original salary = Rs. X
* He got a 5% raise on his original salary
* 5% of X is 0.05X
* So his salary after the raise was X + 0.05X = 1.05X
* The next year he got a 2.5% cut on the salary after the raise
* 2.5% of 1.05X is 0.025 * 1.05X = 0.02625X
* So his salary after the cut was 1.05X - 0.02625X = 1.02375X
* We're given this salary is Rs. 22702.
The document describes a math problem involving dividing bullets equally among three friends and determining the original number divided based on information about how many bullets remained after each friend shot 4 bullets. The assistant provides the step-by-step working and determines that the original number of bullets divided was 18.
Bahan ajar materi spltv kelas x semester 1MartiwiFarisa
Pengembangan bahan ajar dibuat dengan tujuan menambah referensi belajar siswa SMA kelas X tentang materi Sistem Persamaan Linear Tiga Variabel (SPLTV). Di dalam modul ini terdapat 4 metode penyelesaian SPLTV beserta langkah-langkahnya. Semoga bermanfaat..
This article provides the existence and uniqueness of a common fixed point for a pair of self-mappings, positive integers powers of a pair, and a sequence of self-mappings over a closed subset of a Hilbert space satisfying various contraction conditions involving rational expressions.
13 3 arithmetic and geometric series and their sumshisema01
This document discusses arithmetic and geometric series. It defines a series as the sum of the terms of a sequence, and defines notation for representing the sum of n terms. It provides the formulas for calculating the sum of finite arithmetic and geometric series, and includes examples of applying the formulas to find sums. It also includes practice problems for readers to solve.
This document discusses using shortcuts and properties of right triangles to solve for unknown side lengths. It explains that right isosceles triangles have two congruent legs and a 45-45-90 angle relationship. The shortcut for these triangles is that the hypotenuse is equal to one leg times the square root of 2. It also discusses 30-60-90 triangles having a short leg opposite the 30 degree angle, a hypotenuse twice as long as the short leg, and a long leg equal to the short leg times the square root of 3. Several examples are provided to demonstrate using these properties and shortcuts to find missing side lengths of right triangles.
An infinite sequence is a function whose domain is the set of natural numbers, while a finite sequence has a domain of natural numbers up to some limit. A sequence can be described by its general term, which gives a rule for calculating each term based on its position in the sequence. The sum of the terms of a sequence is called a series, which is finite if it includes a finite number of terms and infinite if it includes all terms.
The document provides instructions for solving a system of equations based on a graph. It explains how to identify the y-intercepts and slopes of each line from the graph. It then shows how to write the equations in slope-intercept form and determine if the lines are parallel, intersecting, or overlapping based on having the same or different y-intercepts and slopes.
This document outlines the contents and references for a mathematics textbook. The contents include topics like ordinary differential equations, linear differential equations, mean value theorems, functions of several variables, vector calculus, and series and sequences. It provides details of 3 textbooks and 3 references that can be used. It also includes 17 lecture slides on mean value theorems, Taylor's theorem, functions of several variables, maxima and minima, and Lagrange's method of undetermined multipliers.
Study Material Numerical Solution of Odinary Differential EquationsMeenakshisundaram N
1. The document provides information about a numerical methods course for physics majors at Vivekananda College in Tiruvedakam West, including the reference textbook and details about Unit V on numerical solutions of ordinary differential equations.
2. It introduces the concept of using Taylor series approximations to find numerical solutions to differential equations, providing the general Taylor series expansion formula and explaining how to derive the terms needed to solve specific differential equations.
3. It gives examples of using the Taylor series method to solve sample ordinary differential equations, finding approximate values of y at increasing values of x to several decimal places.
First order linear differential equationNofal Umair
1. A differential equation relates an unknown function and its derivatives, and can be ordinary (involving one variable) or partial (involving partial derivatives).
2. Linear differential equations have dependent variables and derivatives that are of degree one, and coefficients that do not depend on the dependent variable.
3. Common methods for solving first-order linear differential equations include separation of variables, homogeneous equations, and exact equations.
This document is a student's work on sequences and series. It introduces arithmetic and geometric sequences, explaining how to find subsequent terms and sums. Examples are provided to demonstrate finding terms of sequences, common differences or ratios, and summing finite and infinite sequences. Key concepts covered include the nth term formulas, using formulas to solve problems, and modeling real-world scenarios like population growth with sequences.
The document discusses solving systems of linear equations with two or three variables. There are three possible cases for the solution: 1) a unique solution, 2) infinitely many solutions (a dependent system), or 3) no solution. The document demonstrates solving systems using substitution and elimination methods, and provides examples of each case. Graphically, case 1 corresponds to intersecting lines or planes, case 2 to coinciding lines or intersecting planes, and case 3 to parallel lines or non-intersecting planes.
The document discusses various methods for solving quadratic equations, including factoring, square root method, completing the square, and the quadratic formula. It also covers solving other types of equations that are quadratic in form, such as radical equations, through transformations. The objectives are to solve quadratic, radical, and other equations that are quadratic in form and to find sums and products of roots, the quadratic equation given roots, and solve application problems involving these equation types.
This document is from IFET College of Engineering and presents information on solving second order linear differential equations with constant coefficients. It defines such an equation as one where the highest order derivative is of order 2 and all coefficients are constants. The general solution is described as the sum of the complementary function and particular integral. Various cases are discussed for the complementary function depending on whether the roots are real/complex and distinct or repeated. Methods like variation of parameters and Cauchy's and Legendre's equations are also mentioned for solving related problems.
The document shows the step-by-step work to solve the linear equation 3x^2 + 4 = 13. It begins by multiplying both sides by 2 to eliminate the fraction. It then distributes and combines like terms before subtracting and dividing to isolate x. The final solution is x = 6.
This document provides an overview of key concepts in real numbers and geometry. It defines sets and set operations like union, intersection, and difference. It describes the properties of real numbers, including natural numbers, integers, fractions, algebraic numbers, and transcendental numbers. The document also covers inequalities, absolute value, the number line, distance and midpoint formulas, and representations of conic sections like circles, parabolas, ellipses, and hyperbolas. Examples are provided to illustrate solving inequalities with absolute value, finding distance and midpoint, and the standard forms of conic sections. In conclusion, it references two textbooks on calculus and geometry.
1. There are two characteristics that dictate an exponential behavior: the base x must be a decimal with an absolute value less than 1, and it must be a negative number.
2. The document provides examples and explanations for why certain exponential expressions with a negative decimal base will always be positive or negative.
3. It evaluates various inequalities involving exponential expressions with a negative decimal base to determine relationships between the expressions.
1. The angles labeled (2a)° and (5a + 5)° are supplementary and add up to 180°. Solving for a gives a = 25. Similarly, the angles labeled (4b +10)° and (2b – 10)° are supplementary and add up to 180°. Solving for b gives b = 30. Therefore, a + b = 25 + 30 = 55.
2. Two parallel lines intersected by a transversal form eight angles. The acute angles are equal and the obtuse angles are equal. The acute angles are supplementary to the obtuse angles. Solving the equation relating the angles gives the answer.
3. The relationship between angles formed when
We have established a new general method to build doubly even magic squares. New types of magic squares are built. The method is aesthetic and easy to understand and has remarkable topological properties.
1. Assume that an algorithm to solve a problem takes f(n) microse.docxSONU61709
1. Assume that an algorithm to solve a problem takes f(n) microseconds for some function f of the input size n. For each time t labeled across the top, determine the exact largest value of n which can be solved in time f(n) where f(n) ≤ t. Use a calculator! You will find it helpful to convert the t values to microseconds, and you may find it helpful to insert a row for n. Note that “lg n” is the log2 n. Note that the only row you can’t write out the values for fully is the “lg n” row—only there may you write 2x for the appropriate value of x. Use the Windows built-in scientific calculator (under Accessories menu) as necessary. A couple values are filled in to get you started. Important: “exact values” means precisely that. Check your answers with values above and below!
Time t =
f(n) =
1 second
1 hour
1 day
1 month
=30 days
n2
1,609,968
lg n
n3
2n
n lg n
2,755,147,513
2. Use loop counting to give a O( ) characterization of each of the following loops basing each upon the size of its input:
a. Algorithm Loop1(n):
s ← 0
for i ← 1 to n do
s ← s + i
b. Algorithm Loop2(p):
p ← 1
for i ← 1 to 2n do
p ← p * i
c. Algorithm Loop3(n):
p ← 1
for j ← 1 to n2 do
p ← p * i
d. Algorithm Loop4(n):
s ← 0
for j ← 1 to 2n do
for k ← 1 to j do
s ← s + j
e. Algorithm Loop5(n):
k ← 0
for r ← 1 to n2 do
for s ← 1 to r do
k ← k + r
3. Order the following functions from smallest to largest by their big-O notation—you can use the letters in your answer rather than copying each formula. Be clear which is smallest and which is largest, and which functions are asymptotically equivalent. For example, if g, h, and m are all O(n lg n), you would write g = h = m = O(n lg n).
a. 562 log3 108
b. n3
c. 2n lg n
d. lg nn
e. n3 lg n
f. (n3 lg n3)/2
g. nn
h. 56n
i. log5 (n!)
j. ncos n
k. n / lg n
l. lg* n
m.
4. a. Which of these equations is true, and why?
b. Which of these is smaller for very large n?
Trisecting the Circle: A Case for Euclidean Geometry
Author(s): Alfred S. Posamentier
Source: The Mathematics Teacher, Vol. 99, No. 6 (FEBRUARY 2006), pp. 414-418
Published by: National Council of Teachers of Mathematics
Stable URL: http://www.jstor.org/stable/27972006
Accessed: 09-02-2018 18:19 UTC
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide
range of content in a trusted digital archive. We use information technology and tools to increase productivity and
facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at
http://about.jstor.org/terms
National Council of Teachers of Mathematics is collaborating with JSTOR to digitize,
preserve and extend access to The Mathematics Teacher
This content downloaded ...
This document provides an overview of the topics that will be covered in a finite mathematics course, including residue arithmetic, elements of finite groups/rings/fields, number theory concepts like the Euclidean algorithm and Chinese Remainder Theorem, and basics of finite vector spaces and fields. The style of the course will be leisurely and discursive, focusing on mathematical thinking and discovery. While the mathematics is classical, it will be new to students. The goal is to emphasize elegance and aesthetics over utility alone.
The document describes a math problem involving dividing bullets equally among three friends and determining the original number divided based on information about how many bullets remained after each friend shot 4 bullets. The assistant provides the step-by-step working and determines that the original number of bullets divided was 18.
Bahan ajar materi spltv kelas x semester 1MartiwiFarisa
Pengembangan bahan ajar dibuat dengan tujuan menambah referensi belajar siswa SMA kelas X tentang materi Sistem Persamaan Linear Tiga Variabel (SPLTV). Di dalam modul ini terdapat 4 metode penyelesaian SPLTV beserta langkah-langkahnya. Semoga bermanfaat..
This article provides the existence and uniqueness of a common fixed point for a pair of self-mappings, positive integers powers of a pair, and a sequence of self-mappings over a closed subset of a Hilbert space satisfying various contraction conditions involving rational expressions.
13 3 arithmetic and geometric series and their sumshisema01
This document discusses arithmetic and geometric series. It defines a series as the sum of the terms of a sequence, and defines notation for representing the sum of n terms. It provides the formulas for calculating the sum of finite arithmetic and geometric series, and includes examples of applying the formulas to find sums. It also includes practice problems for readers to solve.
This document discusses using shortcuts and properties of right triangles to solve for unknown side lengths. It explains that right isosceles triangles have two congruent legs and a 45-45-90 angle relationship. The shortcut for these triangles is that the hypotenuse is equal to one leg times the square root of 2. It also discusses 30-60-90 triangles having a short leg opposite the 30 degree angle, a hypotenuse twice as long as the short leg, and a long leg equal to the short leg times the square root of 3. Several examples are provided to demonstrate using these properties and shortcuts to find missing side lengths of right triangles.
An infinite sequence is a function whose domain is the set of natural numbers, while a finite sequence has a domain of natural numbers up to some limit. A sequence can be described by its general term, which gives a rule for calculating each term based on its position in the sequence. The sum of the terms of a sequence is called a series, which is finite if it includes a finite number of terms and infinite if it includes all terms.
The document provides instructions for solving a system of equations based on a graph. It explains how to identify the y-intercepts and slopes of each line from the graph. It then shows how to write the equations in slope-intercept form and determine if the lines are parallel, intersecting, or overlapping based on having the same or different y-intercepts and slopes.
This document outlines the contents and references for a mathematics textbook. The contents include topics like ordinary differential equations, linear differential equations, mean value theorems, functions of several variables, vector calculus, and series and sequences. It provides details of 3 textbooks and 3 references that can be used. It also includes 17 lecture slides on mean value theorems, Taylor's theorem, functions of several variables, maxima and minima, and Lagrange's method of undetermined multipliers.
Study Material Numerical Solution of Odinary Differential EquationsMeenakshisundaram N
1. The document provides information about a numerical methods course for physics majors at Vivekananda College in Tiruvedakam West, including the reference textbook and details about Unit V on numerical solutions of ordinary differential equations.
2. It introduces the concept of using Taylor series approximations to find numerical solutions to differential equations, providing the general Taylor series expansion formula and explaining how to derive the terms needed to solve specific differential equations.
3. It gives examples of using the Taylor series method to solve sample ordinary differential equations, finding approximate values of y at increasing values of x to several decimal places.
First order linear differential equationNofal Umair
1. A differential equation relates an unknown function and its derivatives, and can be ordinary (involving one variable) or partial (involving partial derivatives).
2. Linear differential equations have dependent variables and derivatives that are of degree one, and coefficients that do not depend on the dependent variable.
3. Common methods for solving first-order linear differential equations include separation of variables, homogeneous equations, and exact equations.
This document is a student's work on sequences and series. It introduces arithmetic and geometric sequences, explaining how to find subsequent terms and sums. Examples are provided to demonstrate finding terms of sequences, common differences or ratios, and summing finite and infinite sequences. Key concepts covered include the nth term formulas, using formulas to solve problems, and modeling real-world scenarios like population growth with sequences.
The document discusses solving systems of linear equations with two or three variables. There are three possible cases for the solution: 1) a unique solution, 2) infinitely many solutions (a dependent system), or 3) no solution. The document demonstrates solving systems using substitution and elimination methods, and provides examples of each case. Graphically, case 1 corresponds to intersecting lines or planes, case 2 to coinciding lines or intersecting planes, and case 3 to parallel lines or non-intersecting planes.
The document discusses various methods for solving quadratic equations, including factoring, square root method, completing the square, and the quadratic formula. It also covers solving other types of equations that are quadratic in form, such as radical equations, through transformations. The objectives are to solve quadratic, radical, and other equations that are quadratic in form and to find sums and products of roots, the quadratic equation given roots, and solve application problems involving these equation types.
This document is from IFET College of Engineering and presents information on solving second order linear differential equations with constant coefficients. It defines such an equation as one where the highest order derivative is of order 2 and all coefficients are constants. The general solution is described as the sum of the complementary function and particular integral. Various cases are discussed for the complementary function depending on whether the roots are real/complex and distinct or repeated. Methods like variation of parameters and Cauchy's and Legendre's equations are also mentioned for solving related problems.
The document shows the step-by-step work to solve the linear equation 3x^2 + 4 = 13. It begins by multiplying both sides by 2 to eliminate the fraction. It then distributes and combines like terms before subtracting and dividing to isolate x. The final solution is x = 6.
This document provides an overview of key concepts in real numbers and geometry. It defines sets and set operations like union, intersection, and difference. It describes the properties of real numbers, including natural numbers, integers, fractions, algebraic numbers, and transcendental numbers. The document also covers inequalities, absolute value, the number line, distance and midpoint formulas, and representations of conic sections like circles, parabolas, ellipses, and hyperbolas. Examples are provided to illustrate solving inequalities with absolute value, finding distance and midpoint, and the standard forms of conic sections. In conclusion, it references two textbooks on calculus and geometry.
1. There are two characteristics that dictate an exponential behavior: the base x must be a decimal with an absolute value less than 1, and it must be a negative number.
2. The document provides examples and explanations for why certain exponential expressions with a negative decimal base will always be positive or negative.
3. It evaluates various inequalities involving exponential expressions with a negative decimal base to determine relationships between the expressions.
1. The angles labeled (2a)° and (5a + 5)° are supplementary and add up to 180°. Solving for a gives a = 25. Similarly, the angles labeled (4b +10)° and (2b – 10)° are supplementary and add up to 180°. Solving for b gives b = 30. Therefore, a + b = 25 + 30 = 55.
2. Two parallel lines intersected by a transversal form eight angles. The acute angles are equal and the obtuse angles are equal. The acute angles are supplementary to the obtuse angles. Solving the equation relating the angles gives the answer.
3. The relationship between angles formed when
We have established a new general method to build doubly even magic squares. New types of magic squares are built. The method is aesthetic and easy to understand and has remarkable topological properties.
1. Assume that an algorithm to solve a problem takes f(n) microse.docxSONU61709
1. Assume that an algorithm to solve a problem takes f(n) microseconds for some function f of the input size n. For each time t labeled across the top, determine the exact largest value of n which can be solved in time f(n) where f(n) ≤ t. Use a calculator! You will find it helpful to convert the t values to microseconds, and you may find it helpful to insert a row for n. Note that “lg n” is the log2 n. Note that the only row you can’t write out the values for fully is the “lg n” row—only there may you write 2x for the appropriate value of x. Use the Windows built-in scientific calculator (under Accessories menu) as necessary. A couple values are filled in to get you started. Important: “exact values” means precisely that. Check your answers with values above and below!
Time t =
f(n) =
1 second
1 hour
1 day
1 month
=30 days
n2
1,609,968
lg n
n3
2n
n lg n
2,755,147,513
2. Use loop counting to give a O( ) characterization of each of the following loops basing each upon the size of its input:
a. Algorithm Loop1(n):
s ← 0
for i ← 1 to n do
s ← s + i
b. Algorithm Loop2(p):
p ← 1
for i ← 1 to 2n do
p ← p * i
c. Algorithm Loop3(n):
p ← 1
for j ← 1 to n2 do
p ← p * i
d. Algorithm Loop4(n):
s ← 0
for j ← 1 to 2n do
for k ← 1 to j do
s ← s + j
e. Algorithm Loop5(n):
k ← 0
for r ← 1 to n2 do
for s ← 1 to r do
k ← k + r
3. Order the following functions from smallest to largest by their big-O notation—you can use the letters in your answer rather than copying each formula. Be clear which is smallest and which is largest, and which functions are asymptotically equivalent. For example, if g, h, and m are all O(n lg n), you would write g = h = m = O(n lg n).
a. 562 log3 108
b. n3
c. 2n lg n
d. lg nn
e. n3 lg n
f. (n3 lg n3)/2
g. nn
h. 56n
i. log5 (n!)
j. ncos n
k. n / lg n
l. lg* n
m.
4. a. Which of these equations is true, and why?
b. Which of these is smaller for very large n?
Trisecting the Circle: A Case for Euclidean Geometry
Author(s): Alfred S. Posamentier
Source: The Mathematics Teacher, Vol. 99, No. 6 (FEBRUARY 2006), pp. 414-418
Published by: National Council of Teachers of Mathematics
Stable URL: http://www.jstor.org/stable/27972006
Accessed: 09-02-2018 18:19 UTC
JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide
range of content in a trusted digital archive. We use information technology and tools to increase productivity and
facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]
Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at
http://about.jstor.org/terms
National Council of Teachers of Mathematics is collaborating with JSTOR to digitize,
preserve and extend access to The Mathematics Teacher
This content downloaded ...
This document provides an overview of the topics that will be covered in a finite mathematics course, including residue arithmetic, elements of finite groups/rings/fields, number theory concepts like the Euclidean algorithm and Chinese Remainder Theorem, and basics of finite vector spaces and fields. The style of the course will be leisurely and discursive, focusing on mathematical thinking and discovery. While the mathematics is classical, it will be new to students. The goal is to emphasize elegance and aesthetics over utility alone.
THE MIDLINE THEOREM-.pptx GRADE 9 MATHEMATICS THIRD QUARTERRicksCeleste
1. The document discusses classroom rules which include listening when others are speaking, raising your hand to speak or get up, and being respectful.
2. It then provides an example of using the midline theorem to solve several geometry problems involving triangles. The midline theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
3. Finally, it discusses how the midline theorem can be applied in real life contexts like architecture and engineering for structures like rooftops, bridges, and buildings.
Solutions Manual for An Introduction To Abstract Algebra With Notes To The Fu...Aladdinew
This document provides solutions to exercises from Chapter 1 of a textbook on abstract algebra. The exercises cover topics from sections 1.1 and 1.2 such as proofs by induction, properties of integers (commutativity, associativity, etc.), divisibility, and finding the greatest common divisor. The solutions demonstrate techniques like proof by contradiction and distributing operations. The document is intended for students to check their work and for instructors to help explain the concepts.
This document provides an overview of sequences and series. It defines what a sequence is, discusses concepts like convergence and divergence of sequences, and introduces common types of sequences like an = n and an = 1/n. It also presents rules for determining limits of sequences, such as the sum and product rules. Examples are provided to demonstrate applying these rules and the squeeze theorem for limits.
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1. IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILL
MEASURABLE IN R
Di Wu
Submitted under the supervision of Professor Scot Adams to the University
Honors Program at the University of Minnesota-Twin Cities in partial fulfillment
of the requirements for the degree of Bachelor of Arts, summa cum laude in
Mathematics.
Department of Mathematics
University of Minnesota-Twin Cities
The United States
May, 9th , 2016
2. IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILL
MEASURABLE IN R?
Di Wu
A THESIS
in
Mathematics
Presented to the Faculties of the University of Minnesota-Twin Cities
in Partial Fulfillment of the Requirements for the Honors Degree of
Bachelor of Arts, summa cum laude in Mathematics.
Spring, 2016
Supervisor of Dissertation
Scot Adams, Professor of Mathematics
Thesis Readers:
Karel Prikry, Professor of Mathematics
Wayne Richter, Professor of Mathematics
3. Acknowledgments
First and foremost, I have to thank my thesis supervisors, Professor Scot Adams.
Without his assistance and dedicated involvement in every step throughout the
process, this paper would have never been accomplished. I would like to thank you
very much for your support and understanding over the past year.
I would also like to show gratitude to my committee, including Professor Karel
Prikry and Professor Wayne Richter. Without the help of Professor Karel Prikry,
we would not be able to figure out such an easier way to construct a not almost
open and measurable set.
The first time that I met Professor Scot Adams is through his lectures for GRE
math subject at the University of Minnesota. His teaching style and enthusiasm
for the topic made a strong impression on me and I have always carried positive
memories of his lectures with me. After several meetings, I decided to discuss the
possible Honors Thesis topics with Professor Scot Adams. He thought the possible
topics very carefully and raised many precious points in our discussion.
Getting through my thesis required more than academic support, and I have
ii
4. many, many people to thank for listening to and, at times, having to tolerate me
over the past three years. I cannot begin to express my gratitude and appreciation
for their friendship. Jun Li, Hanyu Feng, Eva Lian, Ronald Siegel, Carme Calderer,
Rina Ashkenazi and Andy Whitman have been unwavering in their personal and
professional support during the time I spent at the University. For many memorable
evenings out and in, I must thank everyone above as well as Sue Steinberg, and Matt
Hanson. I would also like to thank Eunice Lee and Christina Gee who opened both
their home and heart to me when I first arrived in the city.
Most importantly, none of this could have happened without my family. My
dad, who offered his encouragement during my hardest time. With his care, I
finally went through a really tough period of my life. To my parents it would be an
understatement to say that, as a family, we have experienced some ups and downs
in the past three years. Every time I was ready to quit, you did not let me and I
am forever grateful. This thesis stands as a testament to your unconditional love
and encouragement.
iii
5. ABSTRACT
IS THERE A SET THAT IS NOT ALMOST OPEN BUT STILL
MEASURABLE IN R
Di Wu
Professor Scot Adams, Supervisor
This paper demonstrates a simple way to construct a not almost open and
nonmeasurable set by taking the union of a Bernstein set and a meager and conull
set.
iv
6. Contents
1 Introduction 1
2 Not Almost Open and Measurable 2
2.1 Nowhere Dense and Almost Open . . . . . . . . . . . . . . . . . . . 2
2.2 Split and Continuum Cardinality . . . . . . . . . . . . . . . . . . . 10
2.3 A Not Almost Open but Measurable Set . . . . . . . . . . . . . . . 16
v
7. Chapter 1
Introduction
The motivations for this research comes from an analogy between measure theory
and the Baire category theory. In measure theory, to describe something“small”,
we have the concept of measure 0. While in the Baire category theory, to describe
“small” sets, we have the concept of meager sets. In addition, the idea of measurable
set is analogical to the concept of almost open in Baire category theory. However,
most of sets we deal with are often both measurable and almost open. Moreover,
when we try to construct a not almost open set, it is very easy to make this set non-
measurable simultaneously. So an interesting topic comes up: Could we construct
a not almost open but measure set?
1
8. Chapter 2
Not Almost Open and Measurable
2.1 Nowhere Dense and Almost Open
Definition 2.1. Let S ⊆ R. Then S is nowhere dense if (S)o
=∅.
Notes:
1) Loosely speaking, nowhere dense set is a set whose elements are not tightly
clustered anywhere in the defined topological space.
2) This definition can also be interpreted as saying that S is nowhere dense in the
defined topological space T if S contains no nonempty open set of T.
3) Any subset of a nowhere dense set is nowhere dense, obviously from the definition
and the elementary properties of closure and interior.
3) Some examples:
i) ∅ is nowhere dense since (∅)o
=∅.
2
9. ii) The boundary of an open (closed) set is nowhere dense ( Lemma 2.1).
iii) The Cantor set is nowhere dense ( Lemma 2.2).
Lemma 2.1. Let U ⊆ R be open or closed. Then ∂U is closed and nowhere dense.
Proof. On one hand, ∂U= U (Uo
), where U is closed and Uo
is open. Since we
know a closed set a open set= a closed set, then ∂U is closed. On the other hand,
(∂U)o
=(∂U)o
=(U (Uo
))o
= (U)o
∩ ((Uo
)c
)o
=(U)o
∩ ((Uo))c
=((U)o
) (Uo).
1). If U is open, then Uo
= U. Thus (∂U)o
=((U)o
) (Uo)=((U)o
) U=∅.
2). If U is closed, then U=U . Thus (∂U)o
=((U)o
) (Uo)=(Uo
) (Uo)=∅.
So ∂U is nowhere dense. Thus ∂U is closed and nowhere dense.
Lemma 2.2. The (standard or ternary) Cantor set C is nowhere dense and has
Lebesgue measure 0 (Steen,1995).
Proof. Construction of Cantor ternary set is created by deleting the open middle
third from each of a set of line segment repeatedly. I start with C0 = [0, 1].
Step 1. Delete the open middle third (1
3
, 2
3
) with Lebesgue measure (20
∗ 1
31 = 1
3
)
from [0, 1]. The remainder is the union of two line segments [0, 1
3
] ∪ [2
3
, 1]. Call this
set C1.
Step 2. Delete the open middle third of each of these remaining segments, i.e.
(1
9
, 2
9
) ∪ (7
9
, 8
9
) with Lebesgue measure (21
∗ 1
32 = 2
9
). The remainder is the union of
four line segments [0, 1
9
] ∪ [2
9
, 1
3
] ∪ [2
3
, 7
9
] ∪ [8
9
, 1]. Call this setC2.
...
3
10. Step n. Continue to delete the open middle third of each of these remaining seg-
ments. The removed length at this step is (2n−1
∗ 1
3n = 2n−1
3n ) and the remainder is
Cn = Cn−1
3
∪ (2
3
+ Cn−1
3
).
...
Continue this infinite process, the Cantor ternary set is defines as:
C := lim∞
n=1 Cn=∩∞
n=1Cn =[0, 1] ∪∞
n=1 ∪2n−1−1
m=0 (3m+1
3n , 3m+2
3n ).
The Cantor set is the set of points in [0, 1], which are not removed. By the construc-
tion, the Cantor set cannot contain any interval with positive length. Otherwise,
suppose it contains some > 0 length interval. The reason is ∀ integer n >= 0, Cn
contains the interval of length 1/3n
. Since C = ∩∞
n=1Cn and 1/3n
→ 0, as n → ∞,
C contains no interval of positive length.
Definition 2.2. Let S ⊆ R. Then S is meager means:
∃S1, S2, ... ⊆ S, s.t. S =S1 ∪ S2 ∪ ... and s.t. ∀j ∈ N, Sj is nowhere dense.
Fact: Any subset of meager set is meager.
Proof. Let S be a meager set. Then ∃S1, S2, ... ⊆ S, s.t. S =S1 ∪ S2 ∪ ... and s.t.
4
11. ∀j ∈ N, Sj is nowhere dense. For any subset A of S, and Sj ∩ A is nowhere dense
since Sj ∩ A is the subset of nowhere dense set Sj. Therefore, any subset of meager
set is meager.
Definition 2.3. Let S ⊆ R. Then S is called an almost open set if ∃ an open set U
⊆ R, ∃ meager sets Z, Z ⊆ R s.t. S=(U Z) ∪ Z .
Notes:
1) I will denote A =∗
B for convenience iff ∃ meager sets Z, Z , s.t.(AZ)∪Z = B.
Figure 2.1: A =∗
B
2) Fact: A =∗
B iff A B and B A are meager.
Proof. On one hand, if A =∗
B, choose meager sets Z, Z s.t. (A Z) ∪ Z = B.
A B =A [(A Z) ∪ Z ] ⊆ A (A Z) ⊆ Z.
B A=[(A Z) ∪ Z ] A ⊆ (A ∪ Z ) A ⊆ Z .
Since A B and B A are subsets of meager sets, A B and B A are meager.
On the other hand, if AB and B A are meager, since [A(AB)]∪(B A) = B,
we know A =∗
B.
5
12. 3) Based on Fact 2), obviously we have A =∗
B iff A B is meager.
4) Fact: If A =∗
B, then (R A)=*(R B).
Proof. Because A =∗
B, A B and B A are meager.
Then (R A) (R B)=Ac
Bc
=Ac
∩ (Bc
)c
=B ∩ Ac
= B A.
Similarly, (R B) (R A)= A B. So (R A)=*(R B).
3) This definition can be easily applied to define almost closed by changing the open
set U to some closed set C from Definition 1.3.
Definition 2.4. Let S ⊆ R, S is almost closed if ∃ clsoed set C ⊆ R s.t. S =∗
C.
Lemma 2.3. Let S ⊆ R be almost open. Then R S is almost open.
Proof. 1) First I will prove that R S is almost closed.
Since S ⊆ R be almost open, ∃ an open set U ⊆ R, ∃ meager sets Z, Z ⊆ R, s.t.
S=(U Z) ∪ Z , written as S=*U. By the previous fact, R S=*R U, where R U
is a close set.
2) Then I will prove that almost closed also implies almost open.
Recall the lemma 1.1, ∂U is closed and nowhere dense.
Thus R S=*((R U) ∂U) ∪ ∂U=*(R U) ∪ ∂U=* R U. Since R U is an open
set, R S is almost open.
Lemma 2.4. Let S1, S2, ... ⊆ R all be almost open. Then S1 ∪S2 ∪... is almost open.
Proof. Since S1, S2, ... ⊆ R are all almost open, ∃ open sets U1, U2, ..., s.t. S1=*U1, S2=*U2....
Then for i = 1, 2, ..., Si Ui and Ui Si are meager. What we want to prove is
6
13. (∪∞
i=1Si) (∪∞
i=1Ui) and (∪∞
i=1Ui) (∪∞
i=1Si) are meager.
(∪∞
i=1Si) (∪∞
i=1Ui) = (∪∞
i=1Si) ∩ (∪∞
i=1Ui)c
= (∪∞
i=1Si) ∩ (∩∞
i=1Uc
i ) ⊆ ∪∞
i=1(Si ∩ Uc
i ).
By the definition of meager set, it is obvious that a countable union of meager sets
is still meager. Also by the symmetry of (∪∞
i=1Si)(∪∞
i=1Ui) and (∪∞
i=1Ui)(∪∞
i=1Si),
we get what we want. Therefore S1 ∪ S2 ∪ ... is almost open.
Definition 2.5. Let S ⊆ R. Then S is negligible means : ∀ > 0, ∃ intervals
I1, I2, ... ⊆ R, s.t. S ⊆
∞
j=1
Ij and
∞
j=1
[(sup Ij) − (inf Ij)] < .
Fact: Let S1, S2, ... ⊆ R all be negligible. Then S1 ∪ S2 ∪ ... is negligible.
Proof. Let > 0 be given, we seek a covering of S1 ∩ S2 ∩ S3... by countable many
intervals, whose total lengths < . Since for any given i ∈ N, Si is negligible, for each
Si, choose intervals Ii1, Ii2... s.t. Si ⊆
∞
j=1
Iij and s.t.
∞
j=1
[(sup Iij) − (inf Iij)] < 2i .
Then
∞
i=1
Si ⊆
∞
i=1
∞
j=1
Iij, and
∞
i=1
∞
j=1
[(sup Iij) − (inf Iij)] <
∞
i=1
2i = . So
∞
i=1
Si is
negligible.
Intuitively, both meager and negligible are used to describe “small” sets. However,
essentially they are very different from each other. In addition, one can find a mea-
ger set whose complement is negligible. Two examples are given below. Before we
get into these examples, I would like to introduce a collection of important sets,
called “Fat Cantor set”, which can be viewed as an extension of the Cantor set.
The Fat Cantor Set (Smith-Volterra-Cantor set, wiki). The construction of the
Fat Cantor set is very similar to the construction of the Cantor set. The Fat Cantor
7
14. set is constructed by removing certain intervals from the unite interval [0,1]. For
example, we are able to make a Fat Cantor set of measure 1
2
through the following
process.
Step 1. Remove the middle 1
4
from the interval [0,1] and the remainder is
[0, 3
8
] ∪ [5
8
, 1], called C1.
Step 2. Remove subintervals of width 1
16
from the middle of each segment intervals,
and the remainder is [0, 5
32
] ∪ [ 7
32
, 3
8
] ∪ [5
8
, 25
32
] ∪ [27
32
, 1], called C2.
...
Step n. Continuing removing subintervals of width 1
22n from the middle of each of
the 2n−1
remaining segments, called Cn.
...
Continuing this infinite process, the Fat Cantor set is defined as: C = ∩∞
i=1Ci.
Each subsequent iteration removes proportionally less from the remainder while
the Cantor set removes proportion as a constant from each remainder. The total
measure removed is equal to
∞
n=0
2n
22n+2 = 1
2
, so C has measure 1
2
, which is positive.
Also C is “ topologically small” in that no open interval is contained in C .
In general, for any given p ∈ (0, 1), one can construct a Fat Cantor set with measure
p in a more flexible way. For example, one can construct a Fat Cantor set with mea-
sure 3
4
by using a decreasing sequence that converges to 1 like 5
4
, 6
5
, 7
6
, 8
7
, . . .. For the
construction, we also require that every term of sequence is < 4
3
. This then gives a
8
15. decreasing sequence (3
4
)(5
4
), (3
4
)(6
5
), (3
4
)(7
6
), . . . that converge to 3
4
. A Fat Cantor set
of measure 3
4
can be constructed in the following process:
Step 1. We want the first remainder to have measure (3
4
)(5
4
) = 15
16
. So we can
remove an open interval of 1 − 15
16
= 1
16
from the middle of interval [0,1]and call the
remainder C1 . Then C1 has measure 15
16
, and is a union of two disjoint compact
intervals.
Step 2. We want the second remainder to have measure (3
4
)(6
5
) = 9
10
. So we want
to remove a total measure of 15
16
− 9
10
= 3
80
from C1 . This means we should remove
intervals of length 3
80
/21
= 3
160
from each of the two intervals in C1 . Call the re-
minder C2
...
Step n. Continuing, we remove subintervals of length (3
4
) 1
(n+2)(n+3)
/2n−1
from the
middle of each of the 2n−1
remaining intervals in Cn−1. to create Cn.
...
Continuing this infinite process, the Fat Cantor set is defined as: C = ∩∞
i=1Ci .
Then this Fat Cantor set C has measure 3
4
.
Example 1. ∃P ⊆ [0, 1] s.t. P is meager, but [0, 1] P is negligible.
Solution: Construct a sequence of Fat Cantor sets P1, P2, . . . with measure 1
2
, 2
3
, 3
4
. . ..
Pn. Let P =
∞
n=1
Pn. Then P is meager but has measure 1. Thus P is meager but
[0, 1] P is negligible.
9
16. Example 2. ∃Q ⊆ R s.t. Q is meager, but R Q is negligible.
Solution: Let P be as in Example 1. Then let Q =
∞
k=−∞
(P + k), where P + k =
{x + k|x ∈ P} ⊆ [k, k + 1].
2.2 Split and Continuum Cardinality
Definition 2.6. Let A, C be sets, Then C splits A means: A∩C = ∅ and C A = ∅.
Lemma 2.5. Let A1, A2, A3, ... ⊆ R. Assume A1, A2, . . . are all infinite. Then ∃ an
infinitely countable set C ⊆ R s.t., ∀j ∈ N, C splits Aj.
Proof. Step 1. Choose v1, w1 ∈ A1, s.t. v1 = w1.
Step 2. Choose v2, w2 ∈ A2 {v1, w1}, s.t. v2 = w2.
...
Step n. Choose vn, wn ∈ An {v1, w1, v2, w2, ...vn−1, wn−1}, s.t. vn = wn.
...
Continue this process countably many times .
Let B := {v1, v2, ...} and C := {w1, w2, ...}.
Then B ∩ C = ∅ and, for any given j ∈ N, we have , Aj ∩ B = ∅ and Aj ∩ C = ∅.
∀j ∈ N, Aj ∩ B ⊆ Aj C, then Aj C = ∅. Thus C splits Aj.
The following lemma is just another way to express the previous Lemma 2.5, but it
is helpful for understanding more generalized theorems later.
10
17. Lemma 2.6. Let J := N, R := R, and let A : J → 2R
. Assume ∀j ∈ J, ∃ an injection
J → Aj. Then ∃h : {1, 2} × J → R, s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. Step 1. Choose v1, w1 ∈ A1, s.t. v1 = w1.
Step 2. Choose v2, w2 ∈ A2 {v1, w1}, s.t. v2 = w2.
...
Step n. Choose vn, wn ∈ An {v1, w1, v2, w2, ...vn−1, wn−1}, s.t. vn = wn.
...
Define h : {1, 2} × J → R by hkj =
vj if k = 1
wj if k = 2
Corollary 2.2.1. Let J := N, R := R, and A ⊆ 2R
. Assume ∃ a surjection J A
and ∀A ∈ A, ∃ an injection J → A. Then ∃ an infinitely countable set C ⊆ R s.t.
∀A ∈ A, C splits A, and ∃ a bijection J → C.
Proof. Choose a surjection A : J A. ∀j ∈ J, ∃ an injection J → Aj s.t. Aj is
infinite. For every j ∈ J, construct h1j, h2j in the previous way as in how we chose
vj, wj in Lemma 1.6.
This yields h : {1, 2} × J → R s.t. ∀j ∈ J, h1j, h2j ∈ Aj.
Define C := {h2j|j ∈ J}. Then j → h2j is a bijection J → C.
∀j ∈ J, h2j ∈ Aj ∩ C so Aj ∩ C = ∅.
∀j ∈ J, h1j ∈ Aj C so Aj C = ∅.
Thus ∀j ∈ J, C splits Aj. Therefore C splits every set in A.
11
18. Define A0 := {(a, b) ⊆ R|a, b ∈ Q, a < b}. Note that A0 is a countable set.
Fact: ∃ an infinitely countable set C ⊂ R s.t. ∀A ∈ A0 , C splits A. Proof: Just
apply the previous Corrollary 2.2.1 replacing A with A0.
Define B0 := { non empty open subsets of R}.
Now we want to consider the question whether ∃ an infinite countable set C ⊆ R,
s.t. ∀B ∈ B0, C splits B or not? The answer is yes. The main reason is that every
non-empty open set in R contains an interval with rational end points.
Remark 2.2.2. ∀B ∈ B0, ∃A ∈ A0 s.t. A ⊆ B.
Remark 2.2.3. Let A, B, C be sets. Assume C splits A and A ⊆ B. Then C splits
B.
Proof. ∅ = A ∩ C ⊆ B ∩ C and ∅ = A C ⊆ B C.
Remark 2.2.4. Denote J := N, I := {I ⊆ J| ∃j0 ∈ J s.t. I = {j ∈ J| j < j0}}∪{J}.
Then I = {∅, {1}, {1, 2}, ...} ∩ {J}.
Now Let us prove Lemma 2.6 in a different but more fancy way, which can lead
us to notice that J can be any infinite set and R can be any set. Before we go to
the details of the proof, let us review Zorn’s Lemma and Well-ordering Theorem.
Zorn’s Lemma : Suppose a partially ordered set P has the property that every
chain has an upper bound in P. Then set P contains at least one maximal element.
Well-ordering Theorem : Every set can be well-ordered.
12
19. Definition 2.7. Let S be a set with cardinal well-ordering <. We say < is a cardinal
well-ordering if ∀s ∈ S, {t ∈ S|t < s} is not bijective with S.
Remark 2.2.5. Every set has a cardinal well-ordering.
Lemma 2.7. Let J := N, R := R, and let A : J → 2R
. Assume ∀j ∈ J, ∃ an injection
J → Aj. Then ∃h : {1, 2} × J → R, s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. J := N, I := {I ⊆ J| ∃j0 ∈ J s.t. I = {j ∈ J| j < j0}} ∪ {J} ⊆ 2J
.
∀I ∈ I, FI := {f : {1, 2} × I → R s.t. ∀i ∈ I, f1i, f2i ∈ Ai}
What we want now: FJ = ∅.
F := ∪I∈IFI.
∀f, g ∈ F, f g means dom[f] ⊆ dom[g] and g|(dom[f]) = f. By Zorn’s Lemma,
we can choose h ∈ F s.t. h is -maximal.
Choose I ∈ I s.t. h ∈ FI. To prove the lemma, we want to show : I = J.
Proof by contradiction: Suppose I = J, then I J.
Pick j1 =min (J I).
Define I1 := {j ∈ J |j ≤ j1}, i.e, I1 = I ∪ {j1}.
Choose v, w ∈ Aj1 (im[h]) s.t. v = w.
Define f := {1, 2} × I1 → R by fki =
hki if i ∈ I
v if i = j1 and k = 1
w if i = j1 and k = 2
Then f ∈ FI1 ⊆ F and h f (contradiction).
Theorem 2.2.6. Let J be an infinite set , R be a set, and let A : J → 2R
. Assume
13
20. ∀j ∈ J, ∃ an injection J → Aj. Then ∃h : {1, 2} × J → R s.t. ∀j ∈ J, h1j, h2j ∈ A.
Proof. Let < be a cardinal well ordering on J. Let I := {I ⊆ J| ∃j0 ∈ J s.t.
I = {j ∈ J| j < j0}} ∪ {J} ⊆ 2J
. (The following part of the proof is the same as
Lemma 2.7.)
∀I ∈ I, FI := {f : {1, 2} × I → R s.t. ∀i ∈ I, f1i, f2i ∈ Ai}
What we want now: FJ = ∅.
F := ∪I∈IFI.
∀f, g ∈ F, f g means dom[f] ⊆ dom[g] and g|(dom[f]) = f. By Zorn’s Lemma,
we can choose h ∈ F s.t. h is -maximal.
Choose I ∈ I s.t. h ∈ FI. To prove the lemma, we want to show : I = J.
Proof by contradiction: Suppose I = J, then I J.
Pick j1 =min (J I).
Define I1 := {j ∈ J |j ≤ j1}, i.e, I1 = I ∪ {j1}.
Notice that card (im[h])=2(card(I)) < card(J) ≤ card (Aj).
Because < is a cardinal well-ordering, im[h] Aj.
Choose v, w ∈ Aj (im[h]) s.t. v = w.
Define f := {1, 2} × I1 → R by fki =
hki if i ∈ I
v if i = j1 and k = 1
w if i = j1 and k = 2
Then f ∈ FI1 ⊆ F and h f (contradiction).
Corollary 2.2.7. Let J be an infinite set, R be a set, and A ⊆ 2R
. Assume ∃
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21. a surjection J A and ∀A ∈ A, ∃ an injection J → A. Then ∃ an infinitely
countable set C ⊆ R s.t. ∀A ∈ A, C splits A, and ∃ a bijection J → C.
Proof. (The following proof is very similar to Corollary 2.2.1.) Choose a surjection
A : J A, ∀j ∈ J, ∃ an injection J → Aj.
By Theorem 2.2.5, choose h : {1, 2} × J → R s.t. ∀j ∈ J, h1j, h2j ∈ Aj.
Define C := {h2j|j ∈ J}. Then j → h2j is a bijection J → C.
∀j ∈ J, h2j ∈ Aj ∩ C so Aj ∩ C = ∅.
∀j ∈ J, h1j ∈ Aj C so Aj C = ∅.
Thus ∀j ∈ J, C splits Aj. Therefore C simultaneously splits A.
Now let us move to discuss Continuum Cardinality.
Definition 2.8. Let A be a set. Then A has continuum cardinality means ∃ a
bijection: 2N
→ A.
Remark 2.2.8. R has continuum cardinality.
Define A1 := { closed subsets of R with continuum cardinality}.
Fact:
1) ∃ a surjection: 2N
{open subsets of R}.
Proof. Let B := {(a, b)| a, b ∈ Q, a < b}. Claim ∃ a surjection B : N B.
Define f : 2N
→{open subsets of R} by f(S) = ∪i∈SBi. Since B is a basis for the
topology on R, f is onto.
15
22. 2) ∃ a bijection :{open subsets of R} → { closed subsets of R}. Take the complement
mapping which maps every open set to its complement.
3) ∃ a surjection: { closed subsets of R} A1. This is obvious since A1 ⊂ { closed
subsets of R}.
Thus by 1)- 3), we know ∃ a surjection :2N
A1.
Remark 2.2.9. ∀A ∈ A1, ∃ a bijection :2N
→ A.
Fact: ∃C ⊆ R s.t. ∀A ∈ A1, C splits A ( by Corollary 2.2.7 with J := 2N
and
R := R ).
2.3 A Not Almost Open but Measurable Set
Define: B1 := {S ⊆ R|S is almost open and S is nonmeager}.
Remark 2.3.1. ∀B ∈ B1, ∃A ∈ A1 s.t.A ⊆ B.
Proof. This follows from Proposition 8.23 (i) ⇐⇒ (ii) in Kechris.
16
23. Theorem 2.3.2. ∃C ⊆ R s.t. ∀B ∈ B1 , C splits B.
Proof. By the previous Fact, choose ∃C ⊆ R s.t. ∀A ∈ A1, C splits A. Let B ∈ B1
for given. By Remark 2.3.1, choose A ∈ A1 s.t.A ⊆ B. Since C splits A and A ⊆ B,
it follows by Remark 2.2.3 that C splits B.
Theorem 2.3.3. ∃C ⊆ R s.t. C is not almost open.
Proof. We follows Example 8.24 from Kechris.
Choose C ⊆ R s.t. ∀B ∈ B1, C splits B.
Proof by contradiction: Assume C is almost open.
∀B ∈ B1, C splits B. So C = B.
This implies C /∈ B1. So C is meager.
Since C is almost open, R C is also almost open.
Also R C is comeager and thus nonmeager. So R C ∈ B1.
Then C splits R C.
C splits R C implies C ∩ (R C) = ∅, but C ∩ (R C) = ∅ (contradiction).
Now we are able to construct a not almost open but measurable set.
Take C is not almost open (Theorem 2.3.3) and Q ⊆ R s.t. Q is meager, R Q is
negligible (Fat Cantor set Example 2). Let C1: =C ∪ Q.
Claim: C1 is not almost open but measurable.
Proof. Suppose C1=C ∪ Q is almost open. Since Q is meager, this implies C is
almost open (contradiction). So C1 is not almost open. Take the complement of
17
24. C1, Cc
1 = Cc
∩ Qc
⊆ Qc
. Since Qc
has measure 0, Cc
1 also has measure 0, so Cc
1 is
measurable. Then C1 is measurable.
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25. Bibliography
[1] Kechris, A. S. “Polish Spaces.” Classical Descriptive Set Theory. New York:
Springer-Verlag, 1995. 47-48. Print.
[2] Smith-Volterra-Cantor set. (n.d.). In Wikipedia. Retrieved May 5, 2016, from
https : //en.wikipedia.org/wiki/Smith − V olterra − Cantor − set
[3] Steen, Lynn Arthur; Seebach, J. Arthur Jr.(1995)[1978]. Counterexamples in
Topology, Berlin, New York: Springer-Verlag, ISBN 978-0-486-68735-3.
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