Basic Geometry and Topology

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1 Bsic Geometry nd Topology Stephn Stolz Septemer 7, 2015 Contents 1 Pointset Topology Metric spces Topologicl spces Constructions with topologicl spces Suspce topology Product topology Quotient topology Properties of topologicl spces Husdorff spces Compct spces Connected spces Pointset Topology 1.1 Metric spces We recll tht mp f : R m R n etween Eucliden spces is continuous if nd only if where x X ɛ > 0 δ > 0 y X d(x, y) < δ d(f(x), f(y)) < ɛ, (1.1) d(x, y) = x y = (x 1 y 1 ) (x n y n ) 2 R 0 is the Eucliden distnce etween two points x, y in R n. Exmple 1.2. (Exmples of continuous mps.) 1. The ddition mp : R 2 R, x = (x 1, x 2 ) x 1 + x 2 ; 2. The multipliction mp m: R 2 R, x = (x 1, x 2 ) x 1 x 2 ; The proofs tht these mps re continuous re simple estimtes tht you proly rememer from clculus. Since the continuity of ll the mps we ll look t in these notes is proved y expressing them in terms of the mps nd m, we include the proofs of continuity of nd m for completeness. 1

2 1 POINTSET TOPOLOGY 2 Proof. To prove tht the ddition mp is continuous, suppose x = (x 1, x 2 ) R 2 nd ɛ > 0 re given. We clim tht for δ := ɛ/2 nd y = (y 1, y 2 ) R 2 with d(x, y) < δ we hve d((x), (y)) < ɛ nd hence is continuous function. To prove the clim, we note tht d(x, y) = x 1 y x 2 y 2 2 nd hence x 1 y 1 d(x, y), x 1 y 1 d(x, y). It follows tht d((x), (y)) = (x) (y) = x 1 + x 2 y 1 y 2 x 1 y 1 + x 2 y 2 2d(x, y) < 2δ = ɛ. To prove tht the multipliction mp m is continuous, we clim tht for δ := min{1, ɛ/( x 1 + x 2 + 1)} nd y = (y 1, y 2 ) R 2 with d(x, y) < δ we hve d(m(x), m(y)) < ɛ nd hence m is continuous function. The clim follows from the following estimtes: d(m(y), m(x)) = y 1 y 2 x 1 x 2 = y 1 y 2 x 1 y 2 + x 1 y 2 x 1 x 2 y 1 y 2 x 1 y 2 + x 1 y 2 x 1 x 2 = y 1 x 1 y 2 + x 1 y 2 x 2 d(x, y)( y 2 + x 1 ) d(x, y)( x 2 + y 2 x 2 + x 1 ) d(x, y)( x 1 + x 2 + 1) < δ( x 1 + x 2 + 1) ɛ Lemm 1.3. The function d: R n R n R 0 hs the following properties: 1. d(x, y) = 0 if nd only if x = y; 2. d(x, y) = d(y, x) (symmetry); 3. d(x, y) d(x, z) + d(z, y) (tringle inequlity) Definition 1.4. A metric spce is set X equipped with mp d: X X R 0 with properties (1)-(3) ove. A mp f : X Y etween metric spces X, Y is continuous if condition (1.1) is stisfied. n isometry if d(f(x), f(y)) = d(x, y) for ll x, y X; Two metric spces X, Y re homeomorphic (resp. isometric) if there re continuous mps (resp. isometries) f : X Y nd g : Y X which re inverses of ech other. Exmple 1.5. An importnt clss of exmples of metric spces re susets of R n. Here re prticulr exmples we will e tlking out during the semester:

3 1 POINTSET TOPOLOGY 3 1. The n-disk D n := {x R n x 1} R n, nd D n r := {x R n x r}, the n-disk of rdius r > 0. The diltion mp D n D n r x rx is homeomorphism etween D n nd D n r with inverse given y multipliction y 1/r. However, these two metric spces re not isometric for r 1. To see this, define the dimeter dim(x) of metric spce X y dim(x) := sup{d(x, y) x, y X} R 0 { }. For exmple, dim(dr n ) = 2r. It is esy to see tht if two metric spces X, Y re isometric, then their dimeters gree. In prticulr, the disks Dr n nd Dr n re not isometric unless r = r. 2. The n-sphere S n := {x R n+1 x = 1} R n The torus T = {v R 3 d(v, C) = r} for 0 < r < 1. Here C = {(x, y, 0) x 2 + y 2 = 1} R 3 is the unit circle in the xy-plne, nd d(v, C) = inf w C d(v, w) is the distnce etween v nd C. 4. The generl liner group GL n (R) = {vector spce isomorphisms f : R n R n } {(v 1,..., v n ) v i R n, det(v 1,..., v n ) 0} = {invertile n n-mtrices} R } n {{ R n } = R n2 n Here we think of (v 1,..., v n ) s n n n-mtrix with column vectors v i, nd the ijection is the usul one in liner lger tht sends liner mp f : R n R n to the mtrix (f(e 1 ),..., f(e n )) whose column vectors re the imges of the stndrd sis elements e i R n. 5. The specil liner group SL n (R) = {(v 1,..., v n ) v i R n, det(v 1,..., v n ) = 1} R n2 6. The orthogonl group O(n) = {liner isometries f : R n R n } = {(v 1,..., v n ) v i R n, v i s re orthonorml} R n2 We recll tht collection of vectors v i R n is orthonorml if v i = 1 for ll i, nd v i is perpendiculr to v j for i j.

4 1 POINTSET TOPOLOGY 4 7. The specil orthogonl group SO(n) = {(v 1,..., v n ) O(n) det(v 1,..., v n ) = 1} R n2 8. The Stiefel mnifold V k (R n ) = {liner isometries f : R k R n } = {(v 1,..., v k ) v i R n, v i s re orthonorml} R kn Exmple 1.6. The following mps etween metric spces re continuous. While it is possile to prove their continuity using the definition of continuity, it will e much simpler to prove their continuity y uilding these mps using compositions nd products from the continuous mps nd m of Exmple 1.2. We will do this elow in Lemm Every polynomil function f : R n R is continuous. We recll tht polynomil function is of the form f(x 1,..., x n ) = i 1,...,i n i1,...,i n x i 1 1 x in n for i1,...,i n R. 2. Let M n n (R) = R n2 e the set of n n mtrices. Then the mp M n n (R) M n n (R) M n n (R) (A, B) AB given y mtrix multipliction is continuous. Here we use the fct tht mp to the product M n n (R) = R n2 = R R is continuous if nd only if ech component mp is continuous (see Lemm 1.21), nd ech mtrix entry of AB is polynomil nd hence continuous function of the mtrix entries of A nd B. Restricting to the invertile mtrices GL n (R) M n n (R), we see tht the multipliction mp GL n (R) GL n (R) GL n (R) is continuous. The sme holds for the sugroups SO(n) O(n) GL n (R). 3. The mp GL n (R) GL n (R), A A 1 is continuous (this is homework prolem). The sme sttement follows for the sugroups of GL n (R). The Eucliden metric on R n given y d(x, y) = (x 1 y 1 ) (x n y n ) 2 for x, y R n is not the only resonle metric on R n. Another metric on R n is given y d 1 (x, y) = n x i y i. (1.7) i=1 The question rises whether it cn hppen tht mp f : R n R n is continuous with respect to one of these metrics, ut not with respect to the other. To see tht this doen t hppen, it is useful to chrcterize continuity of mp f : X Y etween metric spces X, Y in wy tht involves the metrics on X nd Y less directly thn Definition 1.4 does. This lterntive chrcteriztion will e sed on the following notion of open susets of metric spce.

5 1 POINTSET TOPOLOGY 5 Definition 1.8. Let X e metric spce. A suset U X is open if for every point x U there is some ɛ > 0 such tht B ɛ (x) U. Here B ɛ (x) = {y X d(y, x) < ɛ} is the ll of rdius ɛ round x. To illustrte this, lets look t exmples of susets of R n equipped with the Eucliden metric. The suset D n r = {v R n v r} R n is not open, since for for point v D n r with v = r ny open ll B ɛ (v) with center v will contin points not in D n r. By contrst, the suset B r (0) R n is open, since for ny x B r (0) the ll B δ (x) of rdius δ = r x is contined in B r (0), since for y B δ (x) y the tringle inequlity we hve d(y, 0) d(y, x) + d(x, 0) < δ + x = (r x ) + x = r. Lemm 1.9. A mp f : X Y etween metric spces is continuous if nd only if f 1 (V ) is n open suset of X for every open suset V Y. Corollry If f : X Y nd g : Y Z re continuous mps, then so it their composition g f : X Z. Exercise () Prove Lemm 1.9 () Assume tht d, d re two metrics on set X which re equivlent in the sense tht there re constnts C, C > 0 such tht d(x, y) Cd 1 (x, y) nd d 1 (x, y) C d(x, y) for ll x, y X. Show tht suset U X is open with respect to d if nd only if it is open with respect to d. (c) Show tht the Eucliden metric d nd the metric (1.7) on R n re equivlent. This shows in prticulr tht mp f : R n R n is continuous w.r.t. d if nd only if it is continuous w.r.t. d Topologicl spces Lemm 1.9 nd Exercise () ove shows tht it is etter to define continuity of mps etween metric spces in terms of the open susets of these metric spce insted of the originl ɛ-δ-definition. In fct, we cn go one step further, forget out the metric on set X ltogether, nd just consider collection T of susets of X tht we declre to e open. The next result summrizes the sic properties of open susets of metric spce X, which then motivtes the restrictions tht we wish to put on such collections T. Lemm Open susets of metric spce X hve the following properties. (i) X nd re open. (ii) Any union of open sets is open. (iii) The intersection of ny finite numer of open sets is open. Definition A topologicl spce is set X together with collection T of susets of X, clled open sets which re required to stisfy conditions (i), (ii) nd (iii) of the lemm ove. The collection T is clled topology on X. The sets in T re clled the open sets, nd their complements in X re clled closed sets. A suset of X my e neither closed nor open, either closed or open, or oth. A mp f : X Y etween topologicl spces X, Y is continuous if the inverse imge f 1 (V ) of every open suset V Y is n open suset of X.

6 1 POINTSET TOPOLOGY 6 It is esy to see tht the composition of continuous mps is gin continuous. Exmples of topologicl spces. 1. Let X e metric spce, nd T the collection of those susets of X tht re unions of lls B ɛ (x) in X (i.e., the susets which re open in the sense of Definition 1.8). Then T is topology on X, the metric topology. 2. Let X e set. Then T = {ll susets of X} is topology, the discrete topology. We note tht ny mp f : X Y to topologicl spce Y is continuous. We will see lter tht the only continuous mps R n X re the constnt mps. 3. Let X e set. Then T = {, X} is topology, the indiscrete topology. Sometimes it is convenient to define topology U on set X y first descriing smller collection B of susets of X, nd then defining U to e those susets of X tht cn e written s unions of susets elonging to B. We ve done this lredy when defining the metric topology: Let X e metric spce nd let B e the collection of susets of X of the form B ɛ (x) := {y X d(y, x) < ɛ} (the lls in X). Then the metric topology U on X consists of those susets U which re unions of susets elonging to B. Lemm Let B e collection of susets of set X stisfying the following conditions 1. Every point x X elongs to some suset B B. 2. If B 1, B 2 B, then for every x B 1 B 2 there is some B B with x B nd B B 1 B 2. Then T := {unions of susets elonging to B} is topology on X. Definition If the ove conditions re stisfied, we cll the collection B is clled sis for the topology T or we sy tht B genertes the topology T. It is esy to check tht the collection of lls in metric spce stisfies the ove conditions nd hence the collection of open susets is topology s climed y Lemm Constructions with topologicl spces Suspce topology Definition Let X e topologicl spce, nd A X suset. Then T = {A U U open X} is topology on A clled the suspce topology. Lemm Let X e metric spce nd A X. Then the metric topology on A grees with the suspce topology on A (s suset of X equipped with the metric topology). Lemm Let X, Y e topologicl spces nd let A e suset of X equipped with the suspce topology. Then the inclusion mp i: A X is continuous nd mp f : Y A is continuous if nd only if the composition i f : Y X is continuous.

7 1 POINTSET TOPOLOGY Product topology Definition The product topology on the Crtesin product X Y = {(x, y) x X, y Y } of topologicl spces X, Y is the topology with sis B = {U V U open X, V Y } open The collection B oviously stisfies property (1) of sis; property (2) holds since (U V ) (U V ) = (U U ) (V V ). We note tht the collection B is not topology since the union of U V nd U V is typiclly not Crtesin product (e.g., drw picture for the cse where X = Y = R nd U, U, V, V re open intervls). Lemm The product topology on R m R n (with ech fctor equipped with the metric topology) grees with the metric topology on R m+n = R m R n. Proof: homework. Lemm Let X, Y 1, Y 2 e topologicl spces. Then the projection mps p i : Y 1 Y 2 Y i is continuous nd mp f : X Y 1 Y 2 is continuous if nd only if the component mps re continuous for i = 1, 2. Proof: homework X f Y 1 Y 2 p i Yi Lemm Let X e topologicl spce nd let f, g : X R e continuous mps. Then f + g nd f g continuous mps from X to R. If g(x) 0 for ll x X, then lso f/g is continuous. 2. Any polynomil function f : R n R is continuous. 3. The multipliction mp µ: GL n (R) GL n (R) GL n (R) is continuous. Proof. To prove prt (1) we note tht the mp f + g : X R cn e fctored in the form X f g R R R The mp f g is continuous y Lemm 1.21 since its component mps f, g re continuous; the mp is continuous y Exmple 1.2, nd hence the composition f + g is continuous. The rgument for f g is the sme, with replced y m. To prove tht f/g is continuous, we fctor it in the form X f g R R p 1 (I p 2 ) R R m R, where R = {t R t 0}, p 1 (resp. p 2 ) is the projection to the first (resp. second) fctor of R R, nd I : R R is the inversion mp t t 1. By Lemm 1.21 the p i s re continuous, in clculus we lerned tht I is continuous, nd hence gin y Lemm 1.21 the mp p 1 (I p 2 ) is continuous.

8 1 POINTSET TOPOLOGY 8 To prove prt (2), we note tht the constnt mp R n R, x = (x 1,..., x n ) is oviously continuous, nd tht the projection mp p i : R n R, x = (x 1,..., x n ) x i is continuous y Lemm Hence y prt (1) of this lemm, the monomil function x x i 1 1 x in n is continuous. Any polynomil function is sum of monomil functions nd hence continuous. For the proof of (3), let M n n (R) = R n2 e the set of n n mtrices nd let µ: M n n (R) M n n (R) M n n (R) (A, B) AB e the mp given y mtrix multipliction. By Lemm 1.21 the mp µ is continuous if nd only if the composition M n n (R) M n n (R) µ M n n (R) p ij R is continuous for ll 1 i, j n, where p ij is the projection mp tht sends mtrix A to its entry A ij R. Since the p ij (µ(a, B)) = (A B) ij is polynomil in the entries of the mtrices A nd B, this is continuous mp y prt (2) nd hence µ is continuous. Restricting µ to invertile mtrices, we otin the multipliction mp µ : GL n (R) GL n (R) GL n (R) tht we wnt to show is continuous. We will rgue tht in generl if f : X Y is continuous mp with f(a) B for susets A X, B Y, then the restriction f A : A B is continuous. To prove this, consider the commuttive digrm A f A B i X f Y where i, j re the ovious inclusion mps. These inclusion mps re continuous w.r.t. the suspce topology on A, B y Lemm The continuity of f nd i implies the continuity of f i = j f A which gin y Lemm 1.18 implies the continuity of f A. j Quotient topology. Definition Let X e topologicl spce nd let e n equivlence reltion on X. We denote y X/ e the set of equivlence clsses nd y p: X X/ x [x] e the projection mp tht sends point x X to its equivlence clss [x]. The quotient topology on X/ is given y the collection of susets U = {U X/ p 1 (U) is n open suset of X}. The set X/ equipped with the quotient topology is clled the quotient spce.

9 1 POINTSET TOPOLOGY 9 The quotient topology is often used to construct topology on set Y which is not suset of some Eucliden spce R n, or for which it is not cler how to construct metric. If there is surjective mp p: X Y from topologicl spce X, then Y cn e identified with the quotient spce X/, where the equivlence reltion is given y x x if nd only if p(x) = p(x ). In prticulr, Y = X/ cn e equipped with the quotient topology. Here re importnt exmples. Exmple The rel projective spce of dimension n is the set The mp RP n := {1-dimensionl suspces of R n+1 }. S n RP n is surjective, leding to the identifiction nd the quotient topology on RP n. R n+1 v suspce generted y v RP n = S n /(v ±v), 2. Similrly, working with complex vector spces, we otin quotient topology on the the complex projective spce CP n := {1-dimensionl suspces of C n+1 } = S 2n+1 /(v zv), z S 1 3. Generlizing, we cn consider the Grssmnn mnifold There is surjective mp G k (R n+k ) := {k-dimensionl suspces of R n+k }. V k (R n+k ) = {(v 1,..., v k ) v i R n+k, v i s re orthonorml} G k (R n+k ) given y sending (v 1,..., v k ) V k (R n+k ) to the k-dimensionl suspce of R n+k spnned y the v i s. Hence the suspce topology on the Stiefel mnifold V k (R n+k ) R (n+k)k gives quotient topology on the Grssmnn mnifold G k (R n+k ) = V k (R n+k )/. The sme construction works for the complex Grssmnn mnifold G k (C n+k ). As the exmples elow will show, sometimes quotient spce X/ is homeomorphic to topologicl spce Z constructed in different wy. To estlish the homeomorphism etween X/ nd Z, we need to construct continuous mps f : X/ Z g : Z X/ tht re inverse to ech other. The next lemm shows tht it is esy to check continuity of the mp f, the mp out of the quotient spce. Lemm The projection mp p: X X/ is continuous nd mp f : X/ Z to topologicl spce Z is continuous if nd only if the composition f p: X Z is continuous.

10 1 POINTSET TOPOLOGY 10 As we will see in the next section, there re mny situtions where the continuity of the inverse mp for continuous ijection f is utomtic. So in the exmples elow, nd for the exercises in this section, we will defer checking the continuity of f 1 to tht section. Nottion. Let A e suset of topologicl spce X. Define equivlence reltion on X y x y if x = y or x, y A. We use the nottion X/A for the quotient spce X/. Exmple (1) We clim tht the quotient spce [ 1, +1]/{±1} is homeomorphic to S 1 vi the mp f : [ 1, +1]/{±1} S 1 given y [t] e πit. Geometriclly speking, the mp f wrps the intervl [ 1, +1] once round the circle. Here is picture. glue 1 +1 It is esy to check tht the mp f is ijection. To see tht f is continuous, consider the composition [ 1, +1] p [ 1, +1]/{±1} f S 1 i C = R 2, where p is the projection mp nd i the inclusion mp. This composition sends t [ 1, +1] to e πit = (sin πt, cos πt) R 2. By Lemm 1.21 it is continuous function, since its component functions sin πt nd cos πt re continuous functions. By Lemm 1.25 the continuity of i f p implies the continuity of i f, which y Lemm 1.18 implies the continuity of f. As mentioned ove, we ll postpone the proof of the continuity of the inverse mp f 1 to the next section. (2) More generlly, D n /S n 1 is homeomorphic to S n. (proof: homework) (3) Consider the quotient spce of the squre [ 1, +1] [ 1, +1] given y identifying (s, 1) with (s, 1) for ll s [ 1, 1]. It cn e visulized s squre whose top edge is to e glued with its ottom edge. In the picture elow we indicte tht identifiction y leling those two edges y the sme letter. glue

11 1 POINTSET TOPOLOGY 11 The quotient ([ 1, +1] [ 1, +1]) /(s, 1) (s, +1) is homeomorphic to the cylinder C = {(x, y, z) R 3 x [ 1, +1], y 2 + z 2 = 1}. The proof is essentilly the sme s in (1). A homeomorphism from the quotient spce to C is given y f([s, t]) = (s, sin πt, cos πt). The picture elow shows the cylinder C with the imge of the edge indicted. (4) Consider gin the squre, ut this time using n equivlence reltions tht identifies more points thn the one in the previous exmple. As efore we identify (s, 1) nd (s, 1) for s [ 1, 1], nd in ddition we identify ( 1, t) with (1, t) for t [ 1, 1]. Here is the picture, where gin corresponding points of edges leled y the sme letter re to e identified. We clim tht the quotient spce is homeomorphic to the torus T := {x R 3 d(x, K) = d}, where K = {(x 1, x 2, 0) x x 2 2 = 1} is the unit circle in the xy-plne nd 0 < d < 1 is rel numer (see ) vi homeomorphism tht mps the edges of the squre to the loops in T indicted in the following picture elow. Exercise: prove this y writing down n explicity mp from the quotient spce to T, nd rguing tht this mp is continuous ijection (s lwys in this section, we defer the proof of the continuity of the inverse to the next section). (5) We clim tht the quotient spce D n / with equivlence reltion generted y v v for v S n 1 D n is homeomorphic to the rel projective spce RP n. Proof: exercise. In prticulr, RP 1 = S 1 /v v is homeomorphic to D 1 / = [ 1, 1]/ 1 1, which y exmple (1) is homeomorphic to S 1.

12 1 POINTSET TOPOLOGY 12 (6) The quotient spce [ 1, 1] [ 1, 1]/ with the equivlence reltion generted y ( 1, t) (1, t) is represented grphiclly y the following picture. This topologicl spce is clled the Möius nd. It is homeomorphic to suspce of R 3 shown y the following picture (7) The quotient spce of the squre y edge identifictions given y the picture is the Klein ottle. It is hrder to visulize, since it is not homeomorphic to suspce of R 3 (which cn e proved y the methods of lgeric topology). (8) The quotient spce of the squre given y the picture is homeomorphic to the rel projective plne RP 2. Exercise: prove this (hint: use the sttement of exmple (5)). Like the Klein ottle, it is chllenging to visulize the rel projective plne, since it is not homeomorphic to suspce of R Properties of topologicl spces In the previous susection we descried numer of exmples of topologicl spces X, Y tht we climed to e homeomorphic. We typiclly constructed ijection f : X Y nd rgued tht f is continuous. However, we did not finish the proof tht f is homeomorphism, since we defered the rgument tht the inverse mp f 1 : Y X is continuous. We note tht not every continuous ijection is homeomorphism. For exmple if X is set, X δ (resp. X ind ) is the topologicl spce given y equipping the set X with the discrete (resp. indiscrete) topology, then the identity mp is continuous ijection from X δ to X ind. However its inverse, the identity mp X ind X δ is not continuous if X contins t lest two points. Fortuntely, there re situtions where the continuity of the inverse mp is utomtic s the following proposition shows.

13 1 POINTSET TOPOLOGY 13 Proposition Let f : X Y e continuous ijection. Then f is homeomorphism provided X is compct nd Y is Husdorff. The gol of this section is to define these notions, prove the proposition ove, nd to give tools to recognize tht topologicl spce is compct nd/or Husdorff Husdorff spces Definition Let X e topologicl spce, x i X, i = 1, 2,... sequence in X nd x X. Then x is the limit of the x i s if for ny open suset U X contining x there is some N such tht x i U for ll i N. Cvet: If X is topologicl spce with the indiscrete topology, every point is the limit of every sequence. The limit is unique if the topologicl spce hs the following property: Definition A topologicl spce X is Husdorff if for every x, y X, x y, there re disjoint open susets U, V X with x U, y V. Note: if X is metric spce, then the metric topology on X is Husdorff (since for x y nd ɛ = d(x, y)/2, the lls B ɛ (x), B ɛ (y) re disjoint open susets). In prticulr, ny suset of R n, equipped with the suspce topology, is Husdorff. Wrning: The notion of Cuchy sequences cn e defined in metric spces, ut not in generl for topologicl spces (even when they re Husdorff). Lemm Let X e topologicl spce nd A closed suspce of X. If x n A is sequence with limit x, then x A. Proof. Assume x / A. Then x is point in the open suset X \ A nd hence y the definition of limit, ll ut finitely mny elements x n must elong to X \ A, contrdicting our ssumptions Compct spces Definition An open cover of topologicl spce X is collection of open susets of X whose union is X. If for every open cover of X there is finite sucollection which lso covers X, then X is clled compct. Some ooks (like Munkres Topology) refer to open covers s open coverings, while newer ooks (nd wikipedi) seem to prefer to ove terminology, proly for the sme resons s me: to void confusions with covering spces, notion we ll introduce soon. Now we ll prove some useful properties of compct spces nd mps etween them, which will led to the importnt Corollries?? nd Lemm If f : X Y is continuous mp nd X is compct, then the imge f(x) is compct. In prticulr, if X is compct, then ny quotient spce X/ is compct, since the projection mp X X/ is continuous with imge X/.

14 1 POINTSET TOPOLOGY 14 Proof. To show tht f(x) is compct ssume tht {U }, A is n open cover of the suspce f(x). Then ech U is of the form U = V f(x) for some open suset V Y. Then {f 1 (V )}, A is n open cover of X. Since X is compct, there is finite suset A of A such tht {f 1 (V )}, A is cover of X. This implies tht {U }, A is finite cover of f(x), nd hence f(x) is compct. Lemm If K is closed suspce of compct spce X, then K is compct. 2. If K is compct suspce of Husdorff spce X, then K is closed. Proof. To prove (1), ssume tht {U }, A is n open covering of K. Since the U s re open w.r.t. the suspce topology of K, there re open susets V of X such tht U = V K. Then the V s together with the open suset X \ K form n open covering of X. The compctness of X implies tht there is finite suset A A such tht the susets V for A, together with X \ K still cover X. It follows tht U, A is finite cover of K, showing tht K is compct. The proof of prt (2) is homework prolem. Corollry If f : X Y is continuous ijection with X compct nd Y Husdorff, then f is homeomorphism. Proof. We need to show tht the mp g : Y X inverse to f is continuous, i.e., tht g 1 (U) = f(u) is n open suset of Y for ny open suset U of X. Equivlently (y pssing to complements), it suffices to show tht g 1 (C) = f(c) is closed suset of Y for ny closed suset C of C. Now the ssumption tht X is compct implies tht the closed suset C X is compct y prt (1) of Lemm 1.33 nd hence f(c) Y is compct y Lemm The ssumption tht Y is Husdorff then implies y prt (2) of Lemm 1.33 tht f(c) is closed. Lemm Let K e compct suset of R n. Then K is ounded, mening tht there is some r > 0 such tht K is contined in the open ll B r (0) := {x R n d(x, 0) < r}. Proof. The collection B r (0) K, r (0, ), is n open cover of K. By compctness, K is covered y finite numer of these lls; if R is the mximum of the rdii of these finitely mny lls, this implies K B R (0) s desired. Corollry If f : X R is continuous function on compct spce X, then f hs mximum nd minimum. Proof. K = f(x) is compct suset of R. Hence K is ounded, nd thus K hs n infimum := inf K R nd supremum := sup K R. The infimum (resp. supremum) of K is the limit of sequence of elements in K; since K is closed (y Lemm 1.33 (2)), the limit points nd elong to K y Lemm In other words, there re elements x min, x mx X with f(x min ) = f(x) for ll x X nd f(x mx ) = f(x) for ll x X.

15 1 POINTSET TOPOLOGY 15 In order to use Corollries 1.34 nd 1.36, we need to e le to show tht topologicl spces we re interested in, re in fct compct. Note tht this is quite difficult just working from the definition of compctness: you need to ensure tht every open cover hs finite sucover. Tht sounds like lot of work... Fortuntely, there is very simple clssicl chrcteriztion of compct suspces of Eucliden spces: Theorem (Heine-Borel Theorem) A suspce X R n is compct if nd only if X is closed nd ounded. We note tht we ve lredy proved tht if K R n is compct, then K is closed suset of R n (Lemm 1.33(2)), nd K is ounded (Lemm 1.35). There two importnt ingredients to the proof of the converse, nmely the following two results: Lemm A closed intervl [, ] is compct. This lemm hs short proof tht cn e found in ny pointset topology ook, e.g., [?]. Theorem If X 1,..., X n re compct topologicl spces, then their product X 1 X n is compct. For proof see e.g. [?, Ch. 3, Thm. 5.7]. The sttement is true more generlly for product of infinitely mny compct spce (s discussed in [?, p. 113], the correct definition of the product topology for infinite products requires some cre), nd this result is clled Tychonoff s Theorem, see [?, Ch. 5, Thm. 1.1]. Proof of the Heine-Borel Theorem. Let K R n e closed nd ounded, sy K B r (0). We note tht B r (0) is contined in the n-fold product P := [ r, r] [ r, r] R n which is compct y Theorem So K is closed suset of P nd hence compct y Lemm 1.33(1) Connected spces Definition A topologicl spce X is connected if it cn t e written s decomposed in the form X = U V, where U, V re two non-empty disjoint open susets of X. For exmple, if,, c, d re rel numers with < < c < d, consider the suspce X = (, ) (c, d) R. The topologicl spce X is not connected, since U = (, ), V = (c, d) re open disjoint susets of X whose union is X. This remins true if we replce the open intervls y closed intervls. The spce X = [, ] [c, d] is not connected, since it is the disjoint union of the susets U = [, ], V = [c, d]. We wnt to emphsize tht while U nd V re not open s susets of R, they re open susets of X, since they cn e written s U = (, c) X V = (, ) X, showing tht they re open susets for the suspce topology of X R.

16 1 POINTSET TOPOLOGY 16 Lemm Any intervl I in R (open, closed, hlf-open, ounded or not) is connected. Proof. Using proof y contrdiction, let us ssume tht I hs decomposition I = U V s the union of two non-empty disjoint open susets. Pick points u U nd v V, nd let us ssume u < v without loss of generlity. Then [u, v] = U V with U := U [u, v] V := U [u, v] is decomposition of [u, v] s the disjoint union of non-empty disjoint open susets U, V of [u, v]. We clim tht the supremum c := sup U elongs to oth, U nd V, thus leding to the desired contrdiction. Here is the rgument. Assuming tht c doesn t elong to U, for ny ɛ > 0, there must e some element of U elonging to the intervl (c ɛ, c), llowing us to construct sequence of elements u i U converging to c. This implies c U y Lemm 1.30, since U is closed suspce of [u, v] (its complement V is open). By construction, every x [u, v] with x > c = sup U elongs to V. So we cn construct sequence v i V converging to c. Since V is closed suset of [u, v], we conclude c V. Theorem (Intermedite Vlue Theorem) Let X e connected topologicl spce, nd f : X R continuous mp. If elements, R elong to the imge of f, then lso ny rel numer c etween nd elongs to the imge of f. Proof. Assume tht c is not in the imge of f. Then X = f 1 (, c) f 1 (c, ) is decomposion of X s union of non-empty disjoint open susets. There is nother notion, closely relted to the notion of connected topologicl spce, which might e esier to think of geometriclly. Definition A topologicl spce X is pth connected if for ny points x, y X there is pth connecting them. In other words, there is continuous mp γ : [, ] X from some intervl to X with γ() = x, γ() = y. Lemm Any pth connected topologicl spce is connected. Proof. Using proof y contrdiction, let us ssume tht the topologicl spce X is pth connected, ut not connected. So there is decomposition X = U V of X s the union of non-empty open susets U, V X. The ssumption tht X is pth connected llows us to find pth γ : [, ] X with γ() U nd γ() V. Then we otin the decomposition [, ] = f 1 (U) f 1 (V ) of the intervl [, ] s the disjoint union of open susets. These re non-empty since f 1 (U) nd f 1 (V ). This implies tht [, ] is not connected, the desired contrdiction.

17 1 POINTSET TOPOLOGY 17 For typicl topologicl spces we will consider, the properties connected nd pth connected re equivlent. But here is n exmple known s the topologist s sine curve which is connected, ut not pth connected, see [?, Exmple 7, p. 156]. It is the following suspce of R 2 : X = {(x, sin 1 x ) R2 0 < x < 1} {(0, y) R 2 1 y 1}.