MAT 215: Analysis in a single variable Course notes, Fall Michael Damron

Transcription

1 MAT 215: Anlysis in single vrible Course notes, Fll 2012 Michel Dmron Compiled from lectures nd exercises designed with Mrk McConnell following Principles of Mthemticl Anlysis, Rudin Princeton University 1

2 Contents 1 Fundmentls Sets Reltions nd functions Crdinlity Nturl numbers nd induction Crdinlity nd the nturl numbers Exercises The rel numbers Rtionls nd suprem Existence nd properties of rel numbers R n for n Exercises Metric spces Definitions Open nd closed sets Limit points Compctness Heine-Borel Theorem: compctness in R n The Cntor set Exercises Sequences Definitions Subsequences, Cuchy sequences nd completeness Specil sequences Exercises Series Definitions Rtio nd root tests Non non-negtive series Exercises Function limits nd continuity Function limits Continuity Reltions between continuity nd compctness Connectedness nd the IVT Discontinuities Exercises

3 7 Derivtives Introduction Properties Men vlue theorem L Hopitl s rule Power series Tylor s theorem Exercises Integrtion Definitions Properties of integrtion Fundmentl theorems Chnge of vribles, integrtion by prts Exercises A Rel powers 105 A.1 Nturl roots A.2 Rtionl powers A.3 Rel powers B Logrithm nd exponentil functions 110 B.1 Logrithm B.2 Exponentil function B.3 Sophomore s drem C Dimension of the Cntor set 115 C.1 Definitions C.2 The Cntor set C.3 Exercises

4 1 Fundmentls This chpter is primrily bsed on course notes by Prof. Gunning. 1.1 Sets We begin with the concepts of set, object nd set membership. We will leve these s primitive in sense; tht is, undefined. You cn think of set s collection of objects nd if is n object nd A is set then A mens is member of A. If A nd B re sets, we sy tht A is subset of B (written A B) whenever A we hve B. If A B nd B A we sy the sets re equl nd we write A = B. A is proper subset of B if A B but A B. Note tht, the set with no elements, is subset of every set. There re mny opertions we cn perform with sets. If A nd B re sets, A B is the union of A nd B nd is the set A B = { : A or B}. If A nd B re sets, A B is the intersection of A nd B nd is the set A B = { : A nd B}. Of course we cn generlize these to rbitrry numbers of sets. If C is (possibly infinite) collection of sets (tht is, set whose elements re themselves sets), we define A = { : A for some A C} A C A = { : A for ll A C}. A C The sets A nd B re clled disjoint if A B =. These opertions obey the following properties A (B C) = (A B) (A C) A (B C) = (A B) (A C). Let us give proof of the first. To show these sets re equl, we must show ech is contined in the other. So let A (B C). We would like to show tht (A B) (A C). We know A nd (B C). One possibility is tht A nd B, in which cse A B, giving (A B) (A C). The only other possibility is tht A nd C, since must be in either B or C. Then A C nd the sme conclusion holds. The other direction is n exercise. If A nd B re sets then define the difference A \ B s A \ B = { : A but / B}. 4

5 One cn verify the following s well. Finlly the symmetric difference is 1.2 Reltions nd functions A \ (B C) = (A \ B) (A \ C) A \ (B C) = (A \ B) (A \ C). A B = (A \ B) (B \ A). Our lst importnt wy to build set from other sets is the product. We write A B = {(, b) : A, b B}. Definition A reltion R between sets A nd B is ny subset of A B. If (, b) R we think of s being relted to b. We will first mention types of reltions on single set. A reltion R between A nd A is reflexive if (, ) R for ll A. It is symmetric if whenever ( 1, 2 ) R we hve ( 2, 1 ) R. It is trnsitive if whenever ( 1, 2 ) R nd ( 2, 3 ) R we hve ( 1, 3 ) R. Definition A reltion R on A which is reflexive, symmetric nd trnsitive is clled n equivlence reltion. Given A nd n equivlence reltion R on A we write for the equivlence clss of. [] R = { A : (, ) R} Sometimes the condition (, ) R is written (nd sometimes R is not even mentioned). An exmple is equlity of sets; tht is, defining the reltion A B if A = B gives n equivlence reltion. And here we hve not specified R or the lrger set on which R is reltion. You cn check set equlity is reflexive, symmetric nd trnsitive. Proposition If R is n equivlence reltion on nonempty set A nd 1, 2 R then either [ 1 ] R = [ 2 ] R or [ 1 ] R [ 2 ] R =. Proof. We will first show tht both conditions cnnot simultneously hold. Then we will show tht t lest one must hold. To show the first, note tht 1 [ 1 ] R nd 2 [ 2 ] R since R is reflexive. Therefore if [ 1 ] R = [ 2 ] R then 1 [ 1 ] R [ 2 ] R, giving nonempty intersection. For the second clim, suppose tht [ 1 ] R [ 2 ] R, giving some in the intersection. We clim tht [] R = [ 1 ] R. If [] R then (, ) R. But [ 1 ] R, so (, 1 ) R. By trnsitivity, (, 1 ) R, so [ 1 ] R. This proves [] R [ 1 ] R. To show the other continment, let [ 1 ] R so tht (, 1 ) R. Agin, (, 1 ) R, giving ( 1, ) R. Trnsitivity then implies (, ) R, so [] R. 5

6 The picture is tht the equivlence clsses of R prtition A. Definition A prtition of A is collection P of subsets of A such tht 1. A = S P S nd 2. S 1 S 2 = whenever S 1 nd S 2 in P re not equl. Using this definition, we cn sy tht if R is n equivlence reltion on set A then the collection C R = {[] R : A} of equivlence clsses form prtition of A. Just note to conclude. If we hve n equivlence reltion R on set A, it is stndrd nottion to write R/A = {[] R : A} for the set of equivlence clsses of A under R. This is known s tking the quotient by n equivlence reltion. At times the reltion R is written in n implied mnner using symbol like. For instnce, (, b) R would be written b. In this cse, the quotient is R/. We will spend much of the course tlking bout functions, which re specil kinds of reltions. Definition Let A nd B be sets nd f reltion between A nd B. We sy tht f is (well-defined) function from A to B, written f : A B if the following hold. 1. For ech A, there is t lest one b B such tht (, b) f. 2. For ech A, there is t most one b B such tht (, b) f. Tht is, if we ever hve (, b 1 ) f nd (, b 2 ) f for b 1, b 2 B, it follows tht b 1 = b 2. The set A is clled the domin of f nd the B is clled the codomin of f. Of course we will not continue to use this nottion for function, but the more fmilir nottion: if (, b) f then becuse of item 2 bove, we cn unmbiguously write f() = b. We will be interested in certin types of functions. Definition The function f : A B is clled one-to-one (injective) if whenever 1 2 then f( 1 ) f( 2 ). It is clled onto (surjective) if for ech b B there exists A such tht f() = b. Another wy to define onto is to first define the rnge of function f : A B by f(a) = {f() : A} nd sy tht f is onto if f(a) = B. Mny times we wnt to compose functions to build other ones. Suppose tht f : A B nd g : B C re functions. Then (g f) : A C is defined s (g f)() = g((f()). 6

7 Formlly speking we define g f A C by (, c) g f if (, b) f nd (b, c) g for some b B. You cn check tht this defines function. Proposition Let f : A B nd g : B C be functions. 1. If f nd g re one-to-one then so is g f. 2. If f nd g re onto then so is g f. Proof. We strt with the first sttement. Suppose tht f nd g re one-to-one; we will show tht g f must be one-to-one. Suppose then tht nd in A re such tht (g f)() = (g f)( ). Then by definition, g(f()) = g(f( )). But g is one-to-one, so f() = f( ). Now since f is one-to-one, we find =. This shows tht if (g f)() = (g f)( ) then =, proving g f is one-to-one. Suppose then tht f nd g re onto. To show tht g f is onto we must show tht for ech c C there exists A such tht (g f)() = c.this is the sme sttement s g(f()) = c. We know tht g is onto, so there exists b B such tht g(b) = c. Furthermore, f is onto, so for this specific b, there exists A such tht f() = b. Putting these together, This completes the proof. (g f)() = g(f()) = g(b) = c. If function is both one-to-one nd onto we cn define n inverse function. Definition If f : A B is both one-to-one nd onto we cll f bijection. Theorem Let f : A B. There exists function f 1 : B A such tht f 1 f = id A nd f f 1 = id B, (1) where id A : A A nd id B : B B re the identity functions id A () = nd id B (b) = b if nd only if f is bijection. The mening of the bove equtions is f 1 (f()) = nd f(f 1 (b)) = b for ll A nd b B. Proof. Suppose tht f : A B is bijection. Then define f 1 B A by f 1 = {(b, ) : (, b) f}. This is clerly reltion. We clim it is function. To show this we must prove tht for ll b B there exists A such tht (b, ) f 1 nd 7

8 for ll b B there exists t most one A such tht (b, ) f 1. Restted, these re for ll b B there exists A such tht f() = b nd for ll b B there exists t most one A such tht f() = b. These re exctly the conditions tht f be bijection, so f 1 is function. Now we must show tht f 1 f = id A nd f f 1 = id B. We show only the first; the second is n exercise. For ech A, there is b B such tht f() = b. By definition of f 1, we then hve (b, ) f 1 ; tht is, f 1 (b) =. Therefore (, b) f nd (b, ) f 1, giving (, ) f 1 f, or (f 1 f)() = = id A (). We hve now shown tht if f is bijection then there is function f 1 tht stisfies (1). For the other direction, suppose tht f : A B is function nd g : B A is function such tht g f = id A nd f g = id B. We must show then tht f is bijection. To show one-to-one, suppose tht f( 1 ) = f( 2 ). Then 1 = id A ( 1 ) = g(f( 1 )) = g(f( 2 )) = id A ( 2 ) = 2., giving tht f is one-to-one. To show onto, let b B; we clim tht f mps the element g(b) to b. To see this, compute b = id B (b) = f(g(b)). This shows tht f is onto nd completes the proof. Here re some more fcts bout inverses nd injectivity/surjectivity. If f : A B is bijection then so is f 1 : B A. If f : A B nd g : B C re bijections then so is g f. The identity mp id A : A A is bijection. If function f : A B is not bijection then there is no inverse function f 1 : B A. However we cn in ll cses consider the inverse imge. Definition Given f : A B nd C B we define the inverse imge of C s f 1 (C) = { A : f() C}. Note tht if we let C be singleton set {b} for some b B then we retrieve ll elements A mpped to b: f 1 ({b}) = { A : f() = b}. In the cse tht f is invertible, this just gives the singleton set consisting of the point f 1 (b). We note the following properties of inverse imges (proved in the homework). For f : A B nd C 1, C 2 B, f 1 (C 1 C 2 ) = f 1 (C 1 ) f 1 (C 2 ). f 1 (C 1 C 2 ) = f 1 (C 1 ) f 1 (C 2 ). 8

9 1.3 Crdinlity The results of the previous section llow us to define n equivlence reltion on sets: Definition If A nd B re sets, we sy tht A nd B re equivlent (A B or A nd B hve the sme crdinlity) if there exists bijection f : A B. The crdinlity of set A (written (A)) is defined s the equivlence clss of A under this reltion. Tht is (A) = {B : A B}. To compre crdinlities, we introduce new reltion on sets. Definition If A nd B re sets then we write (A) (B) if there exists one-to-one function f : A B. Write (A) < (B) if (A) (B) but (A) (B). The following properties follow. (Exercise: verify the first two.) 1. (reflexivity) For ech set A, (A) (A). 2. (trnsitivity) For ll sets A, B, C, if (A) (B) nd (B) (C) then (A) (C). 3. (ntisymmetry) For ll sets A nd B, if (A) (B) nd (B) (A) then (A) = (B). Any reltion on set tht stisfies these properties is clled prtil order. For crdinlity, estblishment of ntisymmetry is done by the Cntor-Bernstein theorem, which we will skip. Theorem (Cntor s Theorem). For ny set A let P(A) be the power set of A; tht is, the set whose elements re the subsets of A. Then (A) < (P(A)). Proof. We first show tht (A) (P(A)). We proceed by contrdiction. Suppose tht A is set but ssume tht (A) = (P(A)). Then there exists bijection f : A P(A). Using this function, define the set S = { A : / f()}. Since this is subset of A, it is n element of P(A). As f is bijection, it is onto nd therefore there exists s A such tht f(s) = S. There re now two possibilities; either s S or s / S. In either cse we will derive contrdiction, proving tht the ssumption we mde cnnot be true: no such f cn exist nd (A) (P(A)). In the first cse, s S. Then s S = f(s), we hve s f(s). But then by definition of S, it must ctully be tht s / S, contrdiction. In the second cse, s / S, giving by the definition of S tht s f(s). However f(s) = S so s S, nother contrdiction. Second we must show tht (A) (P(A)). To do this we define the function f : A P(A) by f() = {}. To prove injectivity, suppose tht f( 1 ) = f( 2 ). Then { 1 } = { 2 } nd therefore 1 = 2. 9

10 Let us now give n exmple of two sets with the sme crdinlity. If A nd B re sets we write B A for the set of functions f : A B. Let F 2 be set with two elements, which we cll 0 nd 1. We clim tht (P(A)) = (F A 2 ). To see this we must disply bijection between the two. Define f : P(A) F A 2 by the following. For ny subset S A ssocite the chrcteristic function χ S : A F 2 by { 1 if S χ S () = 0 if / S. Exercise: show tht the function f : P(A) F A 2 given by f(s) = χ S is bijection. 1.4 Nturl numbers nd induction To introduce the nturl numbers in n xiomtic wy we will use the Peno xioms. Assumption. We ssume the existence of set N, n element 1 N nd function s : N N with the following properties. 1. For ech n N, s(n) (the successor of n) is not equl to s is injective. 3. (Inductive xiom) If ny subset S N contins 1 nd hs the property tht whenever n S then s(n) S, it follows tht S = N. The third property seems bit weird t first, but ctully there re mny sets which stisfy the first two properties nd re not N. For instnce, the set {n/2 : n N} does. So we need it to relly pin down N. From these xioms mny properties follow. Here is one. for ll n N, s(n) n. Proof. Let S = {n N : s(n) n}. Clerly 1 S. Now suppose tht n S for some n. Then we clim tht s(n) S. To see this, note tht by injectivity of s, s(n) n implies tht s(s(n)) s(n). Thus s(n) S. By the inductive xiom, since 1 S nd whenever n S we hve s(n) S, we see tht S = N. In other words, s(n) n for ll n. Addition It is customry to cll s(1) = 2, s(2) = 3, nd so on. We define ddition on the nturl numbers in recursive mnner: for ny n N, define n + 1 to be s(n) nd 10

11 for ny n, m N, define n + s(m) to be s(n + m). Tht this indeed defines function + : N N N requires proof, but we will skip this nd ssume tht ddition is defined normlly. Of course, ddition stisfies the commuttive nd ssocitive lws. 1. For ny m, n, r N, m + (n + r) = (m + n) + r. Proof. First we show the sttement for r = 1 nd ll m, n. We hve m + (n + 1) = m + s(n) = s(m + n) = (m + n) + 1, where we hve used the inductive definition of ddition. Now suppose tht the formul holds for some r N; we will show it holds for s(r). Indeed, m + (n + s(r)) = m + (n + (r + 1)) = m + ((n + r) + 1) = m + s(n + r) In other words, the set = s(m + (n + r)) = s((m + n) + r) = (m + n) + s(r). S = {r N : m + (n + r) = (m + n) + r for ll m, n N} hs 1 S nd whenever r S, lso s(r) S. By the inductive xiom, S = N. 2. For ny m, n N, m + n = n + m. Proof. Agin we use n inductive rgument. Define S = {n N : n + m = m + n for ll m N}. The first step is to show tht 1 S; tht is, tht 1 + m = m + 1 for ll m N. For this we lso do n induction. Set T = {m N : 1 + m = m + 1}. First, 1 T since = Suppose then tht m T. We clim tht this implies m + 1 T. To see this, write By the induction, T = N. 1 + (m + 1) = (1 + m) + 1 = (m + 1) + 1. Now tht we hve shown 1 S, we ssume n S nd prove n + 1 S. For m N, (n + 1) + m = n + (1 + m) = n + (m + 1) = (n + m) + 1 = (m + n) + 1 = m + (n + 1). By the inductive xiom, S = N nd we re done. 11

12 3. For ll n, m N, n + m n. Proof. Define the set Then since by the Peno xioms, S = {n N : n + m nfor ll m N}. 1 + m = s(m) 1 for ll m N, so 1 N. Suppose then tht n S; tht is, n is such tht n + m n for ll m N. Then by injectivity, for m N, (n + 1) + m = (n + m) + 1 = s(n + m) s(n) = n + 1, giving n + 1 S. By the inductive xiom, S = N nd we re done. Lst, for proving fcts bout ordering we show s is bijection from N to N \ {1}. Proof. We know s does not mp ny element to 1 so s is in fct function to N \ {1}. Also it is injective. To show surjective, consider the set S = {1} {s(n) : n N}. Clerly 1 S. Supposing tht n S then n N, so s(n) S. Therefore S = N. Therefore if k 1 then k = s(n) for some n N. The bove lets us define n 1 for n 1. It is the element such tht (n 1) + 1 = n. Ordering We lso define n ordering on the nturl numbers. We sy tht m n for m, n N if either m = n or m + = n for some N. This defines totl ordering of N; tht is, it is prtil ordering tht lso stisfies for ll m, n N, m n or n m. In the cse tht m n but m n we write m < n. Note tht by item 3 bove, n < n + m for ll n, m N. In prticulr, n < s(n). Proposition is totl ordering of N. 12

13 Proof. First ech n n so it is reflexive. Next if n 1 n 2 nd n 2 n 3 then if n 1 = n 2 or n 2 = n 3, we clerly hve n 1 n 3. Otherwise there exists m 1, m 2 N such tht n 1 + m 1 = n 2 nd n 2 + m 2 = n 3. In this cse, n 3 = n 2 + m 2 = (n 1 + m 1 ) + m 2 = n 1 + (m 1 + m 2 ), giving n 1 n 3. For ntisymmetry, suppose tht m n nd n m. For contrdiction, if m n then there exists, b N such tht m = n+ nd n = m+b. Then m = (m+)+b = m+(+b), contrdiction with item 3 bove. Therefore m = n. So fr we hve proved tht is prtil order. We now prove is totl ordering. To begin with, we clim tht for ll n N, 1 n. Clerly this is true for n = 1. If we ssume it holds for some n then n + 1 = 1 + n 1, verifying the clim by induction. Now for ny m > 1 (tht is, m N with m 1), define the set S = {n N : n m} {n N : m n}. By the bove remrks, 1 S. Supposing now tht n S for some n N, we clim tht n + 1 S. To show this, we hve three cses. 1. Cse 1: n = m. In this cse, n + 1 = m + 1 m, giving n + 1 S. 2. Cse 2: n > m, so there exists N such tht n = m+. Then n+1 = m++1 m, giving n + 1 S. 3. Cse 3: n < m, so there exists N such tht m = n+. If = 1 then n+1 = m S. Otherwise > 1, implying tht 1 N (tht is, 1 is defined), so m = n + = n = (n + 1) + 1 > n + 1, so tht n + 1 S. By the inductive xiom, S = N nd therefore for ll n, we hve n m or m n. A consequence of these properties is trichotomy of the nturl numbers. For ny m, n N, exctly one of the following holds: m < n, m = n or n < m. A property tht reltes ddition nd ordering is if m, n, r N such tht m < n then m + r < n + r. Proof. There must be N such tht n = m +. Then n + r = m + + r = m + r +, giving m + r < n + r. Clerly then if m n nd r N we hve m + r n + r. 13

14 If n < k then n + 1 k. Proof. If n < k then there exists j N such tht n + j = k. Becuse 1 j we find n + 1 n + j = k. Multipliction. We define multipliction inductively by n 1 = n for ll n N n s(m) = n + (n m). One cn prove the following properties; (try it!) let m, n, r, s N: 1. for ll n, m, r N, n (m + r) = (n m) + (n r). 2. n m = m n. 3. (n m) r = n (m r). 4. if n < m nd r s then rn < sm. 1.5 Crdinlity nd the nturl numbers For ech n N we write the set Note tht J 1 = {1} nd for n 1, we hve J n = {m N : m n}. J n+1 = J n {n + 1}. To show this let k be in the right side. If k = n + 1 then k J n+1. Otherwise k n, giving by n n + 1 the inequlity k n + 1, or k J n+1. To prove the inclusion, suppose tht k J n+1. If k J n we re done, so suppose tht k / J n. Therefore k > n, so k n + 1. On the other hnd, k n + 1, so k = n + 1. Definition For n rbitrry set A we sy tht A hs crdinlity n if A J n. In this cse we sy A is finite nd we write (A) = n. If A is not equivlent to ny J n we sy A is infinite. In this definition, (A) is n equivlence clss of sets nd n is number, so wht we hve written here is purely symbolic: it mens A J n. Lemm If A nd B re sets such tht A B then (A) (B). Proof. Define f : A B by f() =. Then f is n injection. 14

15 Theorem For ll n N, (J n ) < (J n+1 ) < N. Proof. Ech set bove is subset of the next, so the proposition holds using insted of <. We must then prove in ech spot bove. Assume first tht we hve proved tht (J n ) (J n+1 ) for ll n N; we will show tht (J n ) N for ll n N. If we hd equlity, then we would find (J n+1 ) N = (J n ). This contrdicts the first inequlity. To prove the inequlity (J n ) (J n+1 ), we use induction. Clerly it holds for n = 1 since J 1 = {1} nd J 2 = {1, 2} nd ny function from J 1 to J 2 cn only hve one element in its rnge (cnnot be onto). Suppose then tht (J n ) (J n+1 ); we will prove tht (J n+1 ) (J n+2 ) by contrdiction. Assume tht there is bijection f : J n+1 J n+2. Then some element must be mpped to n + 2; cll this k J n+1. Define h : J n+1 J n+1 by m m k, n + 1 h(m) = n + 1 m = k. k m = n + 1 This function just swps k nd n + 1. It follows then tht ˆf = f h : J n+1 J n+2 is bijection tht mps n + 1 to n + 2. Now J n is just J n+1 \ {n + 1} nd J n+1 is just J n+2 \ {n + 2}, so define g : J n J n+1 to do exctly wht ˆf does: g(m) = ˆf(m). It follows tht g is bijection from J n to J n+1, giving J n J n+1, contrdiction. Becuse of the proposition, if set A hs A N it must be infinite. In this cse we sy tht A is countble. Otherwise, if A is infinite nd (A) N, we sy it is uncountble. From this point on, we will be more loose bout working with the nturl numbers. For exmple, we will use the terms finite nd infinite in the sme wy tht we normlly do set is finite if it hs finitely mny elements nd infinite otherwise. Of course every proof we write from now on could be done using the Peno xioms, but we will be spred tht. Theorem Let S be n infinite subset of N. Then S is countbly infinite. Proof. We must construct bijection from N to S. We cn ctully do this using the wellordering property: tht ech non-empty subset of N hs lest element. Define f : N S recursively: f(1) is the lest element of S nd, ssuming we hve defined f(1),..., f(n), define f(n + 1) to be the lest element of S \ {f(1),..., f(n)}. This is bijection. Definition We sy set A is countble if it is either finite or countbly infinite. Note tht A is countble if nd only if there is n injection f : A N; tht is, tht (A) N. Theorem Let C be countble collection of countble sets. Then A C A is countble. 15

16 Proof. To prove this we need to construct bijection from N. We will do this somewht non-rigorously, thinking of bijection from N s listing of elements of A C A in sequence. For exmple, given countbly infinite set S we my tke bijection f : N S nd list ll of the elements of S s f(1), f(2), f(3),... If S is finite then this corresponds to finite list. Since ech A C is countble, we my list its elements. The collection C itself is countble so we cn list the elements of A C A in n rry: 1 2 b 1 b 2 b 3 c 1 d 1 d 2 d 3 d 4 Note tht some rows re finite. We now list the elements ccording to digonls. Tht is, we write the list s 1, b 1, 2, c 1, b 2, d 1, b 3, d 2,... Becuse we wnt the list to correspond to bijection, we need to mke sure tht no element is repeted. So, for instnce, if b 1 nd 2 re equl we would only include the first. 1.6 Exercises 1. Let f : A B nd g : B C be functions. Show tht the reltion g f A C, defined by (, c) g f if (, b) f nd (b, c) g for some b B is function. 2. Show tht the function f : P(A) F A 2 mentioned t the end of Section 1.3 nd given by f(s) = χ S is bijection. 3. Prove the properties of multipliction listed t the end of Section Prove the following sttements by induction. () For ll n N, (b) For ll n N, n = n 2 = n(n + 1) 2 n(n + 1)(2n + 1)

17 5. Strong Induction. In this exercise we introduce strong mthemticl induction, which, lthough being referred to s strong, is ctully equivlent to mthemticl induction. Suppose we re given collection {P (n) : n N} of mthemticl sttements. To show P (n) is true for ll n, mthemticl induction dicttes tht we show two things hold: P (1) is true nd if P (n) is true for some n N then P (n + 1) is true. To rgue insted using strong induction we prove tht P (1) is true nd if n N is such tht P (k) is true for ll k n then P (n + 1) is true. () Define sequence ( n ) of rel numbers recursively by 1 = 1 nd n = [n/2] for n 2. (Here [n/2] is the lrgest integer no bigger thn n/2.) Prove by strong induction tht n 2 n 1 for n 2. Is it possible to find b < 2 such tht n b n 1 for ll n 2? (b) Why does strong induction follow from mthemticl induction? In other words in the second step of strong induction, why re we llowed to ssume tht P (k) is true for ll k n to prove tht P (n + 1) is true? 6. Prove tht ny non-empty subset S N hs lest element. Tht is, there is n s S such tht for ll t S we hve s t. This is mjor result bout N, expressed by sying tht N is well-ordered. Hint. Assume there is no lest element. Let M = {m N : t S, m t}. Use Peno s induction xiom to prove tht M = N. Does this led to contrdiction? 17

18 2 The rel numbers 2.1 Rtionls nd suprem From now on we will proceed through Rudin, using the stndrd nottions Z = {..., 1, 0, 1,...} Q = {m/n : m, n Z nd n 0}. When thinking bout the rtionl numbers, we quickly come to relize tht they do not cpture ll tht we wish to express using numbers. For instnce, Theorem There is no rtionl number whose squre is 2. Proof. We rgue by contrdiction, so ssume tht 2 = (m/n) 2 for some m, n Z with n 0. We my ssume tht m nd n re not both even; otherwise, we cn reduce the frction, removing enough fctors of 2 from the numertor nd denomintor. Then 2n 2 = m 2, so m 2 is even. This ctully implies tht m must be even, for otherwise m 2 would be odd (since the squre of n odd number is odd). Therefore we cn write m = 2s for some s Z. Plugging bck in, we find 2n 2 = 4s 2 or n 2 = 2s 2, so n 2 is lso even, giving tht n is even. This is contrdiction. From the previous theorem, wht we know s 2 is not rtionl number. Therefore if we were to construct theory from only rtionls, we would hve hole where we think 2 should be. Wht is even strnger is tht there re rtionl numbers rbitrrily close to this hole. Theorem If q Q stisfies 0 < q 2 < 2 then we cn find nother rtionl ˆq Q such tht q 2 < ˆq 2 < 2. Similrly, for ech r Q such tht r 2 > 2, there is nother rtionl ˆr such tht 2 < ˆr 2 < r 2. Proof. Suppose tht q > 0 stisfies q 2 < 2 nd define ˆq = q + 2 q2 q + 2. Then ˆq > q nd giving ˆq 2 < 2. ˆq 2 2 = 2(q2 2) (q + 2) 2, 18

19 We see from bove tht the set {q Q : q 2 < 2} does not hve lrgest element. This leds us to study lrgest elements of sets more crefully. Definition If A is set with prtil ordering we sy tht A is n upper bound for subset B A if b for ll b B. We sy tht is lest upper bound for B if whenever is n upper bound for B, we hve. We define lower bound nd gretest lower bound similrly. Note tht if is lest upper bound for B then is unique. Indeed, ssume tht nd re lest upper bounds. Since they re both upper bounds, we hve nd, so by ntisymmetry of prtil orderings, =. Becuse of this uniqueness, there is no hrm in writing = sup B when is the lest upper bound of B nd = inf B when is the gretest lower bound of B. Proposition Let A be totlly ordered set nd B subset. Define C to be the set of ll upper bounds for B. Then sup B = inf C. Proof. We re trying to show tht some element (inf C) is the supremum of B, so we must show two things: inf C is n upper bound for B nd ny other upper bound for B stisfies inf C. The second sttement is esy becuse if is n upper bound for B then C. As inf C is lower bound for C we then hve inf C. For the first, ssume tht inf C is not n upper bound for B, so there exists b B such tht inf C is not b. By trichotomy, inf C < b. We clim then tht b is lower bound for C which is lrger thn the gretest lower bound, contrdiction. Why is this? If c C then c is n upper bound for B, giving c b, or b c. Note tht the second sttement of Theorem sttes tht the set {q Q : q 2 > 2} does not hve supremum in Q. Indeed, if it did hve supremum r, then r would be rtionl upper bound for this set nd then we could find smller ˆr tht is still n upper bound, contrdiction. So one wy of formulting the fct tht there re holes in Q is to sy tht it does not hve the lest upper bound property. Definition Let A be totlly ordered set with order. We sy tht A hs the lest upper bound property if ech nonempty subset B A with n upper bound in A hs lest upper bound in A. 2.2 Existence nd properties of rel numbers Therefore we re led to extend the rtionl numbers to fill in the holes. This is ctully quite difficult procedure nd there re mny routes to its end. We will not discuss these, however, nd will insted stte the min theorem bout the existence of the rel numbers without proof. The min point of this course will be to understnd properties of the rel numbers, nd not its existence nd uniqueness. 19

20 For the sttement, one needs the definition of n ordered field, which is certin type of totlly ordered set with multipliction nd ddition (like the rtionls). Theorem (Existence nd uniqueness of R). There exists unique ordered field with the lest upper bound property. The sense in which uniqueness holds is somewht technicl; it is not tht ny two ordered fields s bove must be equl, but they must be isomorphic. Agin we defer to Rudin for these definitions. We will now ssume the existence of R, tht it contins Q nd Z, nd its usul properties. One extremely useful property of R tht follows from the lest upper bound property is Theorem (Archimeden property of R). Given x, y R with x 0, there exists n Z such tht nx > y. Proof. First let x, y R such tht x, y > 0 nd ssume tht there is no such n. Then the set {nx : n N} is bounded bove by y. As it is clerly nonempty, it hs supremum s. Then s x < s, so s x cnnot be n upper bound, giving the existence of some m N such tht s x < mx. However this implies tht s < (m+1)x, so s ws ctully not n upper bound, contrdiction. This proves the sttement for the cse x < y. The other cses cn be obtined from this one by insted considering x nd/or y. The Archimeden property implies Corollry (Density of Q in R). Let x, y R with x < y. There exists q Q such tht x < q < y. Proof. Apply the Archimeden property to y x nd 1 to find n Z such tht n(y x) > 1. We cn lso find m 1 > nx nd m 2 > nx, so m 2 < nx < m 1. It follows then tht there is n m Z such tht m 1 nx < m. Finlly, Dividing by n we get x < m/n < y. nx < m 1 + nx < ny. Now we return to countbility. Theorem The set Q is countble, wheres R is uncountble. 20

21 Proof. We lredy know tht N N is countble: this is from setting up the rry (1, 1) (2, 1) (3, 1) (1, 2) (2, 2) (3, 2) (1, 3) (2, 3) (3, 3) nd listing the elements long digonls. On the other hnd, there is n injection f : Q + N N, where Q + is the set of positive rtionls. One such f is given by f(m/n) = (m, n), where m/n is the reduced frction for the rtionl, expressed with m, n N. Therefore Q + is countble. Similrly, Q, the set of negtive rtionls, is countble. Lst, Q = Q + Q {0} is union of 3 countble sets nd is thus countble. To prove R is uncountble, we will use deciml expnsions for rel numbers. In other words, we write x = where i {0,..., 9} for ll i. Since we hve not proved nything bout deciml expnsions, we re certinly ssuming lot here, but this is how things go. Note tht ech rel number hs t most 2 deciml expnsions (for instnce, 1/4 = = ). Assume tht R is countble. Then s there re t most two deciml expnsions for ech rel number, the set of deciml expnsions is countble (check this!) Now write the set of ll expnsions in list: b 0 b 1 b c 0 c 1 c 2... We will show tht no mtter wht list we re given (s bove), there must be sequence tht is not in the list. This implies tht there cn be no such list, nd thus R is uncountble. Consider the digonl element of the list. Tht is, we tke 0 for the first digit, b 1 for the second, c 2 for the third nd so on:. 0 b 1 c 2 d 3... We now hve rule to trnsform this digonl element into new one. We cn use mny, but here is one: chnge ech digit to 0 if it is not 0, nd replce it with 9 if it is 0. For exmple, Note tht this procedure chnges the digonl number into new one tht differs from the digonl element in every deciml plce. Cll this new expnsion A =.â 0 â 1... Now our originl list contins ll expnsions, so it must contin A t some point; let us sy tht the n-th element of the list is A. Then consider the n-th digit â n of A. On the one hnd, by construction, â n is not equl to the n-th digit of the digonl element. On the other hnd, by the position in the list, â n equls the n-th digit of the digonl element. This is contrdiction. 21

22 2.3 R n for n 2 A very importnt extension of R is given by n-dimensionl Eucliden spce. Definition For n 2, the set R n is defined s Addition of elements is defined s R n = { = ( 1,..., n ) : i R for ll i}. + b = ( 1,..., n ) + (b 1,..., b n ) = ( 1 + b 1,..., n + b n ) nd multipliction of elements by numbers is c = c( 1,..., n ) = (c 1,..., c n ), c R. Note tht this definition gives us R for n = 1. On R n we plce distnce, but to do tht, we need the existence of squre roots. We will tke this for grnted now, since we will prove it lter using continuity. Lemm For ech x R with x 0 there exists unique y R such tht y 2 = x. This element is written y = x. Definition On the set R n we define the norm = ( 1,..., n ) = n nd inner product b = ( 1,..., n ) (b 1,..., b n ) = 1 b n b n. Theorem Suppose, b, c R n nd c R. Then 1. 0 with = 0 if nd only if = c = c. 3. (Cuchy-Schwrz inequlity) b b. 4. (Tringle inequlity) + b + b. 5. b c + c b. Proof. The first two follow esily; for instnce since 2 0 for ll R (this is ctully prt of the definition of ordered field), we get n 0 nd therefore 0. If = 0 then by uniqueness of squre roots, n = 0 nd so 0 2 i for ll i, giving i = 0 for ll i. For the third item, we first give lemm. 22

23 Lemm If x 2 + bx + c 0 for ll x R then b 2 4c. Proof. If = 0 then bx c for ll x. Then we clim b must be zero. If not, then plugging in either 2c/b or 2c/b will give bx < c, contrdiction. Therefore is = 0 we must hve b = 0 nd therefore b 2 4c s climed. Otherwise 0. First ssume tht > 0. Plug in x = b/(2) to get b 2 /(4) + c 0 giving b 2 4c. Lst, if < 0 then we hve ( )x 2 + ( b)x + ( c) 0 nd pplying wht we hve proved lredy to this polynomil, we find ( b) 2 4( )( c), or b 2 4c. To prove Cuchy-Schwrz, note tht for ll x R, 0 ( 1 x b 1 ) ( n x b n ) 2 = ( n)x 2 2( 1 b n b n )x + (b b 2 n) = 2 x 2 2( b)x + b 2. So using the lemm, ( b) 2 2 b 2. The lst two items follow directly from the Cuchy-Schwrz inequlity. Indeed, + b 2 = ( b) ( b) = + 2 b + b b b + b 2 = ( + b ) 2. The lst inequlity follows by tking c nd c b in the previous. 2.4 Exercises 1. For ech of the following exmples, find the supremum nd the infimum of the set S. Also stte whether or not they re elements of S. () S = {x [0, 5] : cos x = 0}. (b) S = {x : x 2 2x 3 < 0}. (c) S = {s n : s n = n i=1 2 i }. 2. Prove by induction tht for ll n N nd rel numbers x 1,..., x n, x x n x x n. 3. Let A, B R be nonempty nd bounded bove. 23

24 () Define the sum set A + B = { + b : A, b B}. Prove tht sup(a + B) = sup A + sup B. (b) Define the product set A B = { b : A, b B}. Is it true tht sup(a B) = sup A sup B? If so, provide proof; otherwise, provide counterexmple. 4. Let C be collection of open intervls (sets I = (, b) for < b) such tht for ll I C, I nd if I, J C stisfy I J then I J =. Prove tht C is countble. Hint. Define function f : C S for some countble set S R by setting f(i) equl to some crefully chosen number. 24

25 3 Metric spces 3.1 Definitions Definition A set X with function d : X X R is metric spce if for ll x, y, z X, 1. d(x, y) 0 nd equls 0 if nd only if x = y nd 2. d(x, y) d(x, z) + d(z, y). Then we cll d metric. Exmples. 1. A useful exmple of metric spce is R n with metric d(, b) = b. 2. If X is ny nonempty set we cn define the discrete metric by { 1 if x y d(x, y) = 0 if x = y. 3. The set F [0, 1] of bounded functions f : [0, 1] R is metric spce with metric 3.2 Open nd closed sets d(f, g) = sup{ f(x) g(x) : x [0, 1]}. Let (X, d) be metric spce. We re interested in the possible subsets of X nd in wht wys we cn describe these using the metric d. Let s strt with the simplest. Definition Let r > 0. The neighborhood of rdius r centered t x X is the set For exmple, B r (x) = {y X : d(x, y) < r} 1. in R using the metric d(x, y) = x y we hve the open intervl B r (x) = (x r, x + r) = {y R : x r < y < x + r}. 2. In R n using the metric d(x, y) = x y we hve the open bll B r (x) = {(y 1,..., y n ) : (x 1 y 1 ) (x n y n ) 2 < r 2 }. To describe tht these sets pper to be open (tht is, no point is on the boundry), we introduce forml definition of open. Definition Let (X, d) be metric spce. A set Y X is open if for ech y Y there exists r > 0 such tht B r (y) Y. For ech point y we must be ble to fit (possibly tiny) neighborhood round y so tht it still stys in the set Y. Thinking of Y s, for exmple, n open bll in R n, s our point y pproches the boundry of this set, the rdius we tke for the neighborhood round this point will hve to decrese. 25

26 Proposition Any neighborhood is open. Proof. Let x X nd r > 0. To show tht B r (x) is open we must choose y B r (x) nd show tht there exists some s > 0 such tht B s (y) B r (x). The rdius s will depend on how close y is to the boundry. Therefore, choose s = r d(x, y). To show tht for this s, we hve B s (y) B r (x) we tke z B s (y). Then d(x, z) d(x, y) + d(y, z) < d(x, y) + s = r. Some more exmples: 1. In R, the only intervls tht re open re the (surprise!) open intervls. For instnce, let s consider the hlf-open intervl (0, 1] = {x R : 0 < x 1}. If it were open, we would be ble to, given ny x (0, 1], find r > 0 such tht B r (x) (0, 1]. But clerly this is flse becuse B r (1) contins 1 + r/2. 2. In R 2, the set {(x, y) : y > 0} {(x, y) : y < 1} is open. 3. In R 3, the set {(x, y, z) : y > 0} {(0, 0, 0)} is not open. Proposition Let C be collection of open sets. 1. The union O C O is open. 2. If C is finite then O C O is open. This need not be true if the collection is infinite. Proof. Let x O C O. Then there exists O C such tht x O. Since O is open, there exists r > 0 such tht B r (x) O. This is lso subset of O C O so this set is open. To show tht we cnnot llow infinite intersections, consider the sets ( 1/n, 1 + 1/n) in R. We hve n=1( 1/n, 1 + 1/n) = [0, 1], which is not open (under the usul metric of R). For finite intersections, let O 1,..., O n be the open sets from C nd x n i=1o i. Then for ech i, we hve x O i nd therefore there exists r i > 0 such tht B ri (x) O i. Letting r = min{r 1,..., r n }, we hve B r (x) B ri (x) for ll i nd therefore B r (x) O i for ll i. This implies B r (x) is subset of the intersection nd we re done. Definition An interior point of Y X is point y Y such tht there exists r > 0 with B r (y) Y. Write Y for the set of interior points of Y. Directly by definition, Y is open if nd only if Y = Y. 26

27 Exmples: 1. The set of interior points of [0, 1] (under the usul metric) is (0, 1). 2. The set of interior points of is just {(x, y) : y > 0}. 3. Wht is the set of interior points of Q? {(x, y) : y > 0} {(x, y) : x = 1, y 0} 4. Define metric on R 2 by d(x, y) = 1 if x y nd 0 otherwise. This cn be shown to be metric. Given set Y R 2, wht is Y? Definition A set Y X is closed if its complement X \ Y is open. Sets cn be both open nd closed. Consider, whose complement is clerly open, mking closed. It is lso open. Proposition Let C be collection of closed sets. 1. C C C is closed. 2. If C is finite then C C C is closed. Proof. Just use X \ [ C C C] = C C (X \ C). 3.3 Limit points There is n lterntive chrcteriztion of closed sets in terms of limit points Definition Let Y X. A point x X is limit point of Y if for ech r > 0 there exists y Y such tht y x nd y B r (x). Write Y for the set of limit points of Y. Exmples: 1. 0 is limit point of {1, 1/2, 1/3,...}. 2. {1, 2, 3} hs no limit points. 3. In R 2, B 1 (0) {(0, y) : y R} {(10, 10)} hs limit points {(x, y) : x 2 + y 2 1} {(0, y) : y R}. Actully we could hve given different definition of limit point. Proposition x X is limit point of Y if nd only if for ech r > 0 there re infinitely mny points of Y in B r (x) 27

28 Proof. We need only show tht if x is limit point of Y nd r > 0 then there re infinitely mny points of Y in B r (x). We rgue by contrdiction; ssume there re only finitely mny nd lbel the ones tht re not equl to x s y 1,..., y n. Choosing r = min{d(x, y 1 ),..., d(x, y n )}, we then hve tht B r (x) contins no points of Y except possibly x. This contrdicts the fct tht x is limit point of Y. Here is yet nother definition of closed. Theorem Y is closed if nd only if Y Y. Proof. Suppose Y is closed nd let y be limit point of Y. If y / Y then becuse X \ Y is open, we cn find r > 0 such tht B r (y) (X \ Y ). But for this r, there is no x B r (y) tht is lso in Y, so tht y is not limit point of Y, contrdiction. Suppose conversely tht Y Y ; we will show tht Y is closed by showing tht X \ Y is open. To do this, let z X \ Y. Since z / Y nd Y Y, z cnnot be limit point of Y. Therefore there is n r > 0 such tht B r (z) contins no points p z such tht p Y. Since z is lso not in Y, we must hve B r (z) (X \ Y ), implying tht X \ Y is open. Exmples: 1. Agin the set {1, 2, 3} hs no limit points (becuse from the bove proposition, finite set cnnot hve limit points). However it is closed by the bove theorem. 2. Is Q closed in R? How bout Q 2 in R 2? 3. The set Z hs no limit points in R, so it is closed. Definition The closure of Y in X is the set Y = Y Y. Theorem Let C be the collection of ll sets C X such tht C is closed nd Y C. Then Y = C C C. Proof. We first show the inclusion. To do this we need to show tht ech y Y nd ech y Y must be in the intersection on the right (cll it J). First if y Y then becuse ech C C contins Y, we hve y J. Second, if y Y nd C C we lso clim tht y C. This is becuse y, being limit point of Y, is lso limit point of C (directly from the definition). However C is closed, so it contins its limit points, nd y C. For the inclusion, we will show tht Y C. This implies tht Y is one of the sets we re intersecting to form J, nd so J Y. Clerly Y Y, so we need to show tht Y is closed. If x / Y then x is not in Y nd x is not limit point of Y, so there exists r > 0 such tht B r (x) does not intersect Y. Since B r (x) is open, ech point in it hs neighborhood is contined in B r (x) nd therefore does not intersect Y. This mens tht ech point in B r (x) is not in Y nd is not limit point of Y, giving B r (x) ( Y ) c, so the complement of Y is open. Thus Y is closed. From the theorem bove, we hve couple of consequences: 28

29 1. For ll Y X, Y is closed. This is becuse the intersection of closed sets is closed. 2. Y = Y if nd only if Y is closed. One direction is cler: tht if Y = Y then Y is closed. For the other direction, if Y is closed then Y Y nd therefore Y = Y Y Y. Exmples: 1. Q = R. 2. R \ Q = R. 3. {1, 1/2, 1/3,...} = {1, 1/2, 1/3,...} {0}. For some prctice, we give Theorem 2.28 from Rudin: Theorem Let Y R be nonempty nd bounded bove. Then sup Y Y nd therefore sup Y Y if Y is closed. Proof. By the lest upper bound property, s = sup Y exists. To show s Y we need to show tht s Y or s Y. If s Y we re done, so we ssume s / Y nd prove tht s Y. Since s is the lest upper bound, given r > 0 there must exist y Y such tht s r < y s. If this were not true, then s r would be n upper bound for Y. But now we hve found y Y such tht y s nd y B r (s), proving tht s is limit point for Y. Note tht sup Y is not lwys limit point of Y. Indeed, consider the set Y = {0}. This set hs sup Y = 0 but hs no limit points. The set Y cn even hve limit points but just not with sup Y limit point. Consider Y = {0} [ 2, 1]. 3.4 Compctness It will be very importnt for us, during the study of continuity for instnce, to understnd exctly which sets Y R hve the following property: for ech infinite subset E Y, E hs limit point in Y. We will soon see tht the intervl [0, 1] hs this property, wheres (0, 1) does not (tke for exmple the subset {1, 1/2, 1/3,...}). The reson is tht we will mny times find ourselves exctly in this sitution: with n infinite subset E of some set Y nd we will wnt to find limit point for E (nd hope tht it is lso in E). This property is wht we will cll on the problem set limit point compctness. Limit point compctness ws pprently one of the originl notions of compctness (see the discussion in Munkres topology book t the beginning of the compctness section thnks Prof. McConnell). However over time it becme pprent tht there ws stronger nd more generl version of compctness (equivlent in metric spces, but not in ll topologicl spces) which could be formulted only in terms of open sets. We give this definition, now tken to be the stndrd one, below. 29

30 Definition A subset K of metric spce X is compct if for every collection C of open sets such tht K C C C, there re finitely mny sets C 1,..., C n C such tht K n i=1c i. The collection C is clled n open cover for K nd {C 1,..., C n } is finite subcover. The process of choosing this finite number of sets from C is referred to s extrcting finite subcover. The definition, in this lnguge, sttes tht K is compct if from every open cover of K we cn extrct finite subcover of K. It is quite difficult to gin intuition bout the bove definition, but it will develop s we go on nd use compctness in vrious circumstnces. The min point is tht finite collections re much more useful thn infinite collections. This is true for exmple with numbers: we lredy know tht set of finitely mny numbers hs min nd mx, wheres n infinite set does not necessrily. As we go through the course, to develop clerer view of compctness, you should revisit the following phrse: often times, compctness llows us to pss from locl informtion (vlid in ech open set from the cover) to globl informtion (vlid on the whole spce), by ptching together the sets in the finite subcover. Let us now give some properties of compct sets nd try to emphsize where the bility to extrct finite subcovers comes into the proofs. Theorem Any compct set is limit point compct. Proof. Let K X be compct nd let E K be n infinite set. Assume for contrdiction tht E hs no limit point in K, so for ech x K we cn find r x > 0 such tht B rx (x) intersects E only possibly t x. The collection C = {B rx (x) : x K} is n open cover of K, so by compctness it cn be reduced to finite subcover of K (nd thus of E). But this mens tht E must hve been finite, contrdiction. Definition A set E X is bounded if there exists x X nd R > 0 such tht E B R (x). Theorem Any compct K X is bounded. Proof. Pick x X nd define collection C of open sets by C = {B R (x) : R N}. We clim tht C is n open cover of K. We need just to show tht ech point of X is in t lest one of the sets of C. So let y X nd choose R > d(y, x). Then y B R (x). Since K is compct, there exist C 1,..., C n C such tht K n i=1c i. By definition of the sets in C we cn then find R 1,..., R n such tht K n i=1b Ri (x). Tking R = mx{r 1,..., R n }, we then hve K B R (x), completing the proof. In the proof it ws essentil to extrct finite subcover becuse we wnted to tke R to be the mximum of rdii of sets in C. This is clerly infinity if we hve n infinite subcover, nd so in this cse the proof would brek down (tht is, if K we were not ble to extrct finite subcover). Exmples. 30

31 1. The set {1/2, 1/3,...} is not compct. This is becuse we cn find n open cover tht dmits no finite subcover. Indeed, consider {( 1 C = n 1 2n, 1 n + 1 ) } : n 2. 2n Ech one of the sets in the bove collection covers only finitely mny elements from {1/2, 1/3,...}, nd so ny finite sub collection cnnot cover the whole set. 2. However if we dd 0, by considering the set {1/2, 1/3,...} {0}, it becomes compct. To prove this, let C be ny open cover; we will show tht there re finitely mny sets from C tht still cover our set. To do this, note first tht there must be some C C such tht 0 C. Since C is open, it contins some intervl ( r, r) for r > 0. Then for n > 1/r, ll points 1/n re in this intervl, nd thus C contins ll but finitely mny of the points from our set. Now we just need to cover the other points, of which there re finitely mny. Writing 1/2,..., 1/N for these points, choose for ech i set C i from C such tht 1/i C i. Then {C, C 2,..., C N } is finite subcover. The min problem in exmple 1 ws ctully tht the set ws not closed. It is not immeditely pprent how tht ws mnifested in our inbility to produce finite subcover, but it is generl fct: Theorem Any compct K X is closed. Proof. We will show tht K c = X \ K is open. Therefore pick x K c ; we will produce n r > 0 such tht B r (x) K c. We first produce n open cover of K. For ech y K, the distnce d(x, y) must be positive, since x y (s x / K). Therefore define the bll We now define the collection B y = {B(y, d(x, y)/2)}. C = {B y : y K}. Since ech y B y this is n open cover of K. By compctness, we cn extrct finite subcover {B y1,..., B yn }. Choosing r = min{d(x, y i )/2 : i = 1,..., n}, we clim then tht B r (x) K c. To show this, let z B r (x). Then d(z, x) < r nd by the tringle inequlity, d(y i, x) < d(z, y i ) + d(z, x) < d(z, y i ) + r, 31