### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Class Note for MATH 281 at GW

### View Full Document

## 18

## 0

## Popular in Course

## Popular in Department

This 7 page Class Notes was uploaded by an elite notetaker on Saturday February 7, 2015. The Class Notes belongs to a course at George Washington University taught by a professor in Fall. Since its upload, it has received 18 views.

## Reviews for Class Note for MATH 281 at GW

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/07/15

Math 281 General Topology Notes De nition 1 Relation Let A and B be sets a relation from A a B is a subset R C A X B A relation on A is a relation from A to itself De nition 2 Composition of Relations Given R C A X B S C B X C the composition S o R C A X C de ned as SoR ac3bEB with ab ER and bc ES SoR m U aRb means ab E R De nition 3 Inverse Relation IfR A X B then R 1 B X A R 1 bala7 b E R De nition 4 Identity Relation I C A X A A E A RoIR 038 always Notice that R o R 1 a I in general Example 1 R is the empty relation If R C A X B is a function write R A a B then R 1 a function if and only if R is bijective De nition 5 Equivalence Relation R is an equivalence relation if its reflexive transitive and symmetric De nition 6 Antisymmetric if aRb and bRa then a b De nition 7 Partial Order R is a partial order if it is reflective transitive and antisymmetric De nition 8 Total Order A Partial Order is a total order or a linear order or an order if Va y E A either dRy or ny Munkres gives axioms for lt7 rather than S De nition 9 Quasi Order A relation 3 on A is a quasi order or preorder if it is reflective and transitive Example 2 Let A some collection of sets For ST E A de ne S S T to mean 3 an injective function f S a T This is a quasi order where as containment is a partial order Theorem 1 Schroder Cantor Bernstein HA 3 B and B S A then 3 a bijective from A to B De nition 10 Given a set A a partition 7r ofA is a collection of nonempty pairwise disjoint subsets ofA having union equal to A Note A partition 7r of A is itself a set of sets of A The elements of 7r are called blocks Partition of A gt Equivalence Relation Given a partition 7r 6 HA set of all partitions of 7139 De ne a NW y to mean Ly belong to the same block7 Vw7y E A Then NW is an equivalence relation De nition 11 Equivalence Class Let N be an equivalence relation on A then y E A y N w is called the equivalence class of w Given an equivalence relation N on A7 we obtain a partition A N of A with blocks De nition 12 Quotient Set The partition A N is called the quotient set ofA by N So we have a bijection HA lt gtEquivA7 7r gt gtN7r and A Ngt gtN They have inverse correspondences 7r gt gtN7rgt gt A NW 71 De nition 13 Image ofa Function and Kernel of a Function Given afunction between f A a B we have image imf w E A Q B and the kernel kerf f 1b b E imf Corresponding equivalence relation7 a N y if and only if fw fy Given a group homomorphism7 f G a H7 Kerf f 1e Q G Kerf K is a normal subgroup of G Relationship7 kerf is the partition of G into cosets of K Kerf Suppose f A a B is an abitrary function7 let p be the following map p A a kerf p is called the canonical surjection j imf a B a gt gt w j is called the inclusion mapinjection f kerf a imf a gt gt fV E f is a bijection 7 The canonical factorization of f is f j o f o p The commutative diagram is A L B pl l ker f L im f September 8 2006 De nition 14 Order Preserving If P7Q are partially ordered sets with relations ltp ltQ then a function f P a Q is order preserving if ltp y gt f ltQ y De nition 15 Isomomorphism Let P and Q be partially ordered sets Then a function f P a Q is an isomorphism iff is a bijection and f and f 1 are order preserving De ne a relation on Q by a y ltgt a S yandy S a It is easy to check that this is an equivalence relation and if we de ne a relation 5 on the set of equivalence classes Q by j y ltgt a S y7 then 5 is a well de ned partial order Example 3 Let S all set in some universal set U De ne A S B to mean 3 an injective function f A a B Equivalence classes A are called cardinal numbers or cardinalities We can then restate Schroder Cantor Bernstein as A B ltgt 3 a bijection A lt gt B S is a quasi order 5 is a partial order linear order on the set of all cardinal numbers If P and Q are isomorphic ordered sets7 then all other theoretic properties that hold for P also hold for Q Example 4 Z S Z Z has a minimum element Z doesn t ZS Q ForallaltcEQ73bEQwithaltbltc NottrueinZ What about 1R7 R 9 Yes pZR R gt gtez What about 711701 Yes 7170 01 1 2 gt gt What about 71717R Yes 7171 R x x l gt 717 2 So a7 b R7Va7b 6 IR If PC are posets7 the direct product ordering on P X Q is given by 11751 S 12752 ltgt lt11 S lt12 and b1 S b2 3 17a 17b De nition 16 Dictionary Order If PC are linearly ordered the dictionary order lexicographic order on P X Q is given by a17b1lta27b2ltgt either a1 lt a2 or al a2 and b1 lt b2 De nition 17 Maximal Element Suppose P is a poset and A Q P a E A is maximal ifx S an E A De nition 18 Maximum Element Suppose P is a poset and A Q P a E A is a maximum in A if it is the unique maximum element De nition 19 Bounded Above Suppose P is a poset and A Q P Then A is bounded above if 3p 6 P such that x S p7Vx E A Minimal7 minimum7 and bounded below de ned similarly De nition 20 Least Upper Bound y is a least upper bound for A ify is an upper bound for A and y is the minimal element of the set of all upper bound for A Notation7 least upper bound y lubA supA Greatest lower bound7 in mum7 glbA infA de ned similarly Example 5 A 071 C IR has lubA 1 A x E le lt has no lub in Q n IR lubA Theorem 2 Completeness oflR Every subset oflR which is bounded above has a least upper bound Note The least upper bound property implies the greatest lower bound property 4 September 13 2006 Properties of IR 0 Risa eld o lfxltygtx2ltyzV2 o ifxltygtx2ltyzVzgt0 o completeness every nonempty subset of IR which is bounded above has a least upper bound or supremum onlt2ER73ywithxltyltz De nition 21 Natural Numbers Z lN 1237 The natural numbers has the following properties 0 Principal of Mathematical Induction Suppose A Q lN ifl E A and n E A gt n 1 E A then A lN o Well Ordering Every nonempty subset ole has a least element Note IR7 lt is not well ordered Theorem 3 I ltgt De nition 22 Cartesian Product Suppose A is a collection of sets An indexing function for this family is a surjectiue function f I a A where I is some set called an indexing set We write Ai for fiVi E I Generalizing notation A17A27 An this is the case I 17 27 Note f being surjective means every A E A is equal to Ai for some i E I f is not necessarily injective7 this mean we can have Ai A7 for i a j De ne LJAi7 WAi7 ifI 17 27 H7727 then Uid Ai U221 Ai A1UA2 U UAn write Aihd ieI ieI or AZ 7 for the indexed family De nition 23 I tuple Given an indexed family Aiken an I tuple is afunction x I a UAi such that E A 7Vi E 2 61 I write xi for xi and x id for I tuple x In the case ofI 17237 H7727 then we have x id x17 xn7x E A2171 1 n Cartesian product H Ai consists of all I tuples x ieb xi 6 Ai 2 61 ForI 17237 n7HA HAZ A1 gtlt A2 gtlt gtlt An xL7 6 Ai for i S i S ieI i1 we have Al X A2 X 1727 2 E E ielN i21 i1 HA Av12 ThenHA A X A X A w um i1 n AgtltAgtlt gtltA z Hit it A De nition 24 Finite Set A set A is nite ltgt 3 a bijection A lt gt 127 for some n E N ltgt there is no bijection A lt gt B where B Q A ltgt every injection A a A is bijective ltgt every smjection is bijective De nition 25 A set A is in nite if not nite Example 6 N is in nite f N a 2N n gt gt 2n is bijective De nition 26 Cardinality The cardinality lAl of a set A to be the equivalence class consisting of all sets B such that 3 a bijection B lt gt A Write A B if 3 a bijection A lt gt B ie if lAl Write lAl n if A 177n for some n De nition 27 lAl 3 Bl lAl 3 Bl if 3 an injection from A a Beqnivalently if 3 a snrjection from B a A Theorem 4 S is a partial order on the set of all cardinal numbers Proof re exive lAl S lAl by the identity A a A transitiVity Composition of injections is an injection antisymmetry lAl 3 Bl and lBl S lAl gt lAl lBl is Schoder Cantor Bernstein Uses axiom of choice D The ordering S on cardinals corresponds to the usual ordering for nite cardinals 17 27 We write No for llNl rst in nite cardinal We write N1 for Question ls there a cardinal number 7 With N0 lt 7N1 Continuum Hypothesis There is no such cardinal Theorem 5 Cohen Continuum Hypothesis is neither true of false It is independent of the axioms of set theory De nition 28 Countably In nite A set A is countably in nite if A N0 HN means A is in nite and we can write A 0417127 De nition 29 Countable A set A is countable if it is nite or countably in nite ie if A S No So A is countable ltgt 3 an injection A a N ltgt 3 a surjection N a A Theorem 6 o ifA is countable and B Q A then B is countable o ifA and B are countable then A X B is countable Question If A17 A27 are countable7 is H AZ No 2 1 Proof 1 Given B gt A N D Proof 2 Given A 04170427 Given B b17b27 So7 11751 11752 11753 12751 12752 12753 A X B 13751 13752 13753 So A X B aL7 b17 a27 b17 aL7 b27 a37 b17 a27 b27 aL7 b37 aL7 b47 de nes a bijection A X B a N D

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I made $350 in just two days after posting my first study guide."

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.