Review: reltions A inry reltion on set A is suset R Ñ A ˆ A, where elements p, q re written s. Exmple: A Z nd R t pmod nqu. A inry reltion on set A is... (R) reflexive if for ll P A; (S) symmetric if implies ; (T) trnsitive if nd c implies c, i.e. p ^ cq ñ c An equivlence reltion on set A is inry reltion tht is reflexive, symmetric, nd trnsitive. Review: set theoretic definition of the numers. Nturl numers: Let 0 H. Given n, definethesuccessor to n s Spnq n Ytnu. (By successor to n we siclly men n ` 1.) Let Z 0 e the set of ll sets generted y 0 nd S. Integers: Define Z y formlly letting Z 0 t n n P Z 0 u, where 0 0; nd Z Z 0 Y Z 0.ExtendS : Z Ñ Z y defining Sp q for ny P N t0u s Sp q, where Spq. Some opertions: Define ` : N ˆ N Ñ N y, for ll, P N, y ` 0 nd ` Spq Sp ` q. Define : N ˆ N Ñ N y, for ll, P N, n 0 0 nd Spq p q`.
Review: Some properties of ` nd (we present without proof): 1. Addition nd multipliction stisfy commuttivity, ssocitivity, nd distriutivity. 2. We still hve ` 0 0 ` (dditive identity) nd 1 1 (multiplictive identity) forll P Z. 3. We lso hve `p q 0 (prove). (dditive inverses) We cll ny numer system tht hs n ddition nd multipliction tht stisfy ll these properties (commuttive) ring. Order: For, P Z, wesy if SpSp Spq qq. Properties of order (we present without proof): (i) For ll, P N, wehve or. (ii) If nd, then. (iii) If nd c, then c. (iv) If then ` c ` c. (v) If then c c. Rtionl numers Let Q Z ˆpZ t0uq, nd define n equivlence reltion on Q y p, q px, x q for ll x P Z t0u. Under this equivlence reltion, write rp, qs. Then rtionl numers re! ) Q ˇ, P Z, 0. (Note tht we get lzy, nd write 1.) Define ` : Q ˆ Q Ñ Q nd : Q ˆ Q Ñ Q y ` c d d ` c nd d c d c d.
Let Q Z ˆpZ t0uq nd define n equivlence reltion on Q y p, q px, x q for ll x P Z t0u. Under this equivlence reltion, write rp, qs (writing 1 ). Then rtionl numers re! ) Q ˇ, P Z, 0. Define ` : Q ˆ Q Ñ Q nd : Q ˆ Q Ñ Q y ` c d d ` c nd d c d c d. Agin... 1. Addition nd multipliction still stisfy commuttivity, ssocitivity, nd distriutivity. 2. We still hve x ` 0 x (dditive identity) ndx 1 x (multiplictive identity) forllxpq. 3. We lso hve tht x `p xq 0. (dditive inverses) So Q is lso (commuttive) ring. In ddition, for ll { P Q with 0, 1 This mkes Q field (gin, modern lger). (multiplictive inverses). Order on Q Define (you cn show á ). We define on Q y the following: for,, c, d P N, wehve 1. c d 2. c d ;nd whenever d c; 3. c d whenever c d. Then, gin, (i) For ll, P N, wehve or. (ii) If nd, then. (iii) If nd c, then c. (iv) If then ` c ` c. (v) If nd 0 c, thenc c. This mkes Q into n ordered field.
Let X e n ordered set of numers (think N, Z, Q, nd, eventully, R), nd let S e nonempty suset of X. () If S contins lrgest element x (x P S nd for ll y P S, y x), then we cll x mxpsq the mximum if S. Creful writing note: Compre to, The mximum of S is n element x P S stisfying y x for ll y P S. Wht s wrong? () If S contins smllest element x (x P S nd for ll y P S, y x), then we cll x minpsq the minimum of S. Ex: If S t 2, 1{2, 100{3u, thenminpsq 2, mxpsq 100{3. Ex: If S Ñ Z 0 is finite, then mxpsq s nd minpsq s. sps sps Ex: If S is finite, then mxpsq nd minpsq exist. Ex: For S Z 0, minpsq 1 nd mxpsq does not exist. Ex: For S Q 0, minpsq nd mxpsq do not exist. Note: Min/mx don t depend on the set X! Let X e n ordered set of numers (think N, Z, Q, nd, eventully, R), nd let S e nonempty suset of X. () If there exists u P X such tht s u for ll s P S, thenu is clled n upper ound of S nd the set S is sid to e ounded ove (y u). Ex: S Q 0 Ñ Q doesn t hve mximl element, ut it does hve n upper ound: u 0. In fct, it hs lots of upper ounds: u 1, u 100, u 101{15, etc.but u 1 is not n upper ound since 1{2 P S nd 1{2 1. Ex: S Q 0 Ñ Q doesn t hve ny upper ounds. Ex: As suset of X Q 0, S Q 0 doesn t hve n upper ound. () Similrly, if there exists ` P X such tht s ` for ll s P S, then ` is clled lower ound of S nd the set S is sid to e ounded elow (y `). Note: Upper nd lower ounds do depend on the set X.
Consider X Q nd S tx P Q x 2 ` x 1 0u y y x 2 ` x 1 S 3 2 1 1 2 Mth Orcle: We cn compute x 2 ` x 1 0 ô x 1 2 p 1?5q RQ. So S doesn t hve minimum or mximum. But S is ounded ove nd elow, e.g. u 1 nd ` 2; or u.62 nd ` 1.62; or... But wht is the est ound? Does it even hve est ound? Orcle: In R, the est ounds re ` 1 2 p 1?5q nd u 1?5q. 2p 1 ` x Let X e n ordered set of numers (think N, Z, Q, nd, eventully, R), nd let S e nonempty suset of X. From efore: If there exists u P X such tht s u for ll s P S, thenu is clled n upper ound of S nd the set S is sid to e ounded ove (y u). Similrly, lower ound is numer ` P X such tht s ` for ll s P S; if` exists, we sy S is ounded elow. () If S is ounded ove, we cll n upper ound U stisfying U u for ll upper ounds u the lest upper ound or supremum of S, denoted y sup S. sup S min ptu P X u s for ll s P Suq () If S is ounded elow, we cll lower ound L stisfying L ` for ll lower ounds ` the gretest lower ound or infimum of S, denoted y inf S. inf S mx pt` P X ` s for ll s P Suq
Bck to X Q nd S tx P Q x 2 ` x 1 0u. y Orcle: In R, lower ounds Therefore, even though inf S S sup S y x 2 ` x 1 x upper ounds inf S 1 2 p 1?5q nd sup S 1?5q. p 1 ` 2 t` P X ` s for ll s P Su nd tu P X u s for ll s P Su re non-empty, inf S nd sup S don t exist in Q. Thm. The rtionl numers re incomplete in the sense tht there exist ounded susets tht do not hve infimums or supremums. Gol: Define the completion of Q the smllest set contining Q so tht every set tht s ounded ove/elow hs sup/inf. (R) Completeness Axiom: Every non-empty suset of R tht is ounded ove hs lest upper ound, i.e. for ll S Ñ R, ifs is ounded ove, then sup S exists nd is in R. The rel numers Let R e the set of susets of Q tht stisfy the following: C P R whenever 1. C à Q nd C H (C is proper, non-empty suset of Q); 2. for ll x P C, ify P Q stisfies y x, theny P C 3. mx C does not exist. (if x P C, theneverythinglessthnx is lso in C); (Recll, in contrst to upper ounds, mx C hs to e n element of C.) Orcle: R consists entirely of sets of the form tx P Q x u p 8,q Sets in R re clled Dedekind cuts, nd R R is the set of rel numers. for some fixed P R. Intuition: identify P R with the cut tx P Q x u PR. Thm. The completeness xiom holds. (This should feel uncomfortle... n xiom shouldn t hve to e proven! Whether this is n xiom or theorem depends on your perspective... )
Opertions Intuition: R consists entirely of sets of the form For, tx P Q x u p 8,q for some fixed P R. P R, define ` t` P,P u. Creful: wedo not wnt to define y t P,P u! Exmple: Consider p 1q p 1q. We certinly need this to e 1 tx P Q x 1u. Compre to t, Pp 1q u. This ltter set contins, for exmple, p 2qp 3q 6 R 1. Insted: for non-negtive, P R (i.e. 0, 0 H), define 1. t P 0,P 0 uyq 0 ; 2. t x P Q x for ll P u; 3. p q; nd 4. p q p q. x Addition: For, P R, define ` t ` P,P u. Multipliction: For non-negtive, P R (i.e. 0, 0 H), define t P 0,P 0 uyq 0 ; t x P Q x for ll P u; p q; nd p q p q. Agin... 1. Addition nd multipliction still stisfy commuttivity, ssocitivity, nd distriutivity. 2. We still hve ` 0 x (dditive identity) nd 1 (multiplictive identity) forll P R. 3. We lso hve tht `p q 0. (dditive inverses) 4. Dor ll P R with 0, thereexists 1 P R tht stisfies 1 1 (multiplictive inverses). So R field (gin, modern lger).
Intuition: R consists entirely of sets of the form tx P Q x u p 8,q for some fixed P R. For ll P Q, we cn concretely identify with tx P Q x u PR. Nmely, Q Ñ R defined y fiñ is n injective mp, which will respect ddition, multipliction, nd order (once we define them). Comprisons: For, P R, define whenever Ñ. Thm. (Archimeden property) If, 0, thenthereexists n P N such tht n. Thm. (Denseness of Q) If, thenthereexistsc P Q such tht c. x
Wrpping up 1. Context is king! You cn do most mth t ny level of strction. For exmple, there s huge di erence etween wht you cn ssume when working in set theory versus clculus. 2. Go slowly when reding/writing new mth. Py ttention to detils! For exmple, the simple inversion of quntifiers cn mess whole sttement up. 3. Alwys do exmples! Illustrtive exmples cn help you understnd the ig picture; extreme exmples help you understnd the detils. And if you don t know where to strt, then do n exmple! 4. Revise, revise, revise. When solving mth prolems, lot goes on ehind the scenes. Don t e frid to write down logicl fllcies (like strting with the conclusion, or proof y exmple ) in the privcy of your own home. Just don t stop there! Tip: Keep old ides or notes to yourself in your.tex file, commented out with % s, in cse you need them gin lter. 5. Mth isn t liner; mth is frctl. While logicl rgument needs to come in logicl order, there isn t just one good order to explin ll of mth. In prticulr, every good nswer spins o mny good questions! 6. You cn do it!! Be kind to yourself! If you don t get something right wy, tht doesn t men you re stupid, or tht you cn t get there. Mth is hrd, ut dole; nd the struggle is wht mkes the rekthroughs so fun!