Question 1: Question 2: Topology Exercise Sheet 3

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1 Topology Exercise Shee 3 Prof. Dr. Alessandro Siso Due o 14 March Quesions 1 and 6 are more concepual and should have prioriy. Quesions 4 and 5 admi a relaively shor soluion. Quesion 7 is harder, and you should leave i as he las one. Quesion 1: Le X and Y be opological spaces, and le A be a subse of X and B be a subse of Y. Show ha A B = A B. In paricular, conclude ha he produc of wo closed ses is a closed se. We sar by showing A B A B. In order o do ha, we will show ha X Y A B X Y A B. Le (x, y) (X Y A B). Then we have x A or y B (or boh). Assume ha he firs one holds (he oher case is analogous). Then here exiss an open se O of X such ha x O and O A =. By definiion of produc opology, we have ha O Y is an open in X Y. By consrucion, (x, y) O Y and O Y A B =. Thus (x, y) X Y A B. For he oher inclusion, le (x, y) A B, and le O be an open of X Y such ha (x, y) O. We wan o show ha O A B. By definiion of produc opology, we have ha O = i V i U i, where he V i are open ses of X and he U i are open ses of Y. Since x A, we have ha for each i, V i A. Similarly for y. Thus, for each i we have ha V i U i A B. In paricular, O A B. Quesion 2: Show ha he following are homeomorphic: (i) The inerval [0, 1] and he inerval [2, 5]; 1

2 Consider he map f : [0, 1] [2, 5] defined as f() = 2+3. I is clear ha f is coninuous and i s inverse is coninuous. Indeed, we have f 1 () = 2 3. Since f 1 is coninuous, f is a homeomorphism. (ii) The inerval ( 1, 1) and he real line R; Consider he map f : ( 1, 1) R defined as { if 0 1 f() = if 0 +1 Since = if = 0, he map is coninuous. I is no hard o see ha 1 +1 f is bijecive. The limi of f for 1, resp. 1, is, resp., and f(0) = 0. A simple exercise is derivaives ells us ha f is sricly increasing in f (0,1) and in f ( 1,0). Thus f is bijecive. The inverse of f is f 1 () = { 1+ if 0 1 which is a coninuous funcion. Thus f is an homeomorphism. (iii) The closed disk or radius one in R 2 and he closed square [ 1, 1] [ 1, 1] is R 2 ; We wan o produce an homeomorphism from he closed uni disk B = {x R 2 d(x, (0, 0)) 1} o he square Q = [ 1, 1] [ 1, 1]. The idea is he following: we will srech he disk o he square. To o his, i is convenien o work in polar coordinaes. Le θ be he angular coordinae and ρ be he radial one. Noe ha for each value of θ, here only one value ρ B such ha (θ, ρ B ) is a poin of he boundary of he disk (and he value is ρ B = 1), and only one value ρ Q such ha (θ, ρ Q ) is a poin of he boundary of he square. This amoun o say ha he ray saring a (0, 0) wih angle θ inersecs each of he boundaries of B and Q in exacly 2

3 one poin. Le g be he funcion defined as g(θ) = ρ Q ρ B = ρ Q. We claim ha g is coninuous. Indeed, we can explicily wrie g as: 1 if θ [ π cos(θ), π] if θ [ π sin(θ) g(θ) =, 3π] if θ [ 3 cos(θ) 4 π, 5π] 4 1 if θ [ 5π, 7π] sin(θ) 4 4 Le f : B Q be he funcion defined as f(θ, ρ) = (θ, g(θ)ρ). The funcion f is a bijecion from B o Q. Indeed, f sreches each ray of he disk {(θ, ρ) ρ 1} o he ray of he square ( {(θ, ) ρ): ρ g(θ)}. The inverse of f is clearly defined as f 1 (θ, ρ) = θ, (noe ha g is ρ g(θ) always posiive). Since g is coninuous, we have ha f and f 1 are also coninuous. Quesion 3: Le p = ( 1, 0) and q = (2, 0) be poins in R 2, and le D 1 = {z R 2 d(z, p) < 1} and D 2 = {z R 2 d(z, q) < 2}. Which of he following are conneced? (i) D 1 D 2 ; (ii) D 1 D 2 ; (iii) D 1 D 2 ; You may use he fac ha D 1 = {z R 2 d(z, p) 1} and D 2 = {z R 2 d(z, q) 2}. Noe ha, by definiion of he opology on R 2, he ses D 1 and D 2 are open. I is easy o see ha D 1 D 2 =. Indeed by riangular inequaliy, for each poin z R 2 we have d(z, p) + d(z, q) d(p, q) = 3. If d(z, p) < 1, hen d(z, q) 3 d(z, p) > 2, and similarly for he oher case. Thus he firs se is he disjoin union of open ses, hence no conneced. The second and he hird are conneced. To see his, we will show ha hey are pah conneced. Clearly for each poin z D 1, here exiss a pah conained in D 1 ha connecs z o p. This can be easily believed by saring a a picure, or wriing an explici 3

4 formula in polar coordinaes. Similarly, each poin z D 2 can be conneced o q inside D 2. Now we claim ha here exiss a pah ha connecs p and q ha is conained in boh D 1 D 2 and D 1 D 2. This implies ha boh of he unions above are pah conneced. Le γ : [0, 3] R 2 be he pah defined as γ() = ( 1 +, 0). Clearly γ is a coninuous pah ha connecs p and q. We claim ha he image of γ is conained in D 1 D 2 (and hence in D 1 D 2 ). For each, if 1, hen d(γ(), p) = d(( 1, 0), ( 1, 0)) = ( 1 + 1) 2 =, hus γ() D 1. Oherwise, if > 1, hen d(γ(), q) = d(( 1, 0), (2, 0)) = ( 1 2) 2 = 3. Since 1 < 3, we have ha d(γ(), q) < 2, and hus γ() D 2. Quesion 4: Le X be a se equipped wih he discree opology. conneced? Which subses of X are The only conneced subses are he poins of X, i.e. he subses {x}, for x X. Since he se {x} canno be wrien as he disjoin union of any pair of non-empy ses, he poins of X are rivially conneced. On he oher hand, le Y X be a subse ha is no a poin, ha is, Y conains a leas wo elemens. Le x Y. Then {x} and Y {x} are disjoin open ses, because in he discree opology every se is open. In paricular, Y is no conneced. Quesion 5: Le X and Y be pah conneced spaces. Show ha X Y is pah conneced. Le (x 1, y 1 ) and (x 2, y 2 ) be poins in X Y. We wan o find a coninuous map f : [0, 1] X Y such ha f(0) = (x 1, y 1 ) and f(1) = (x 2, y 2 ). Since X and Y are boh pah conneced, here exis coninuous funcions γ : [0, 1] X, δ : [0, 1] Y such ha γ(0) = x 1, γ(1) = x 2, δ(0) = y 1 4

5 and δ(1) = y 2. Le f = γ δ. Since he produc of coninuous funcions is coninuous, hen f is coninuous. Moreover f(0) = (γ(0), δ(0)) = (x 1, y 1 ) and f(1) = (γ(1), δ(1)) = (x 2, y 2 ). Quesion 6: From he fac ha an inerval [a, b] is conneced, deduce he inermediae value heorem. Tha is, prove ha for a coninuous funcion f : [a, b] R and for each c such ha f(a) < c < f(b), here exiss x [a, b] such ha f(x) = c. Suppose ha his was no he case. Then [a, b] = f 1 ((, c)) ((c, )). Since f(a) < c and f(b) > c, he wo ses f 1 ((, c)) and ((c, )) are non empy, and since f is coninuous hey are open. Moreover, he union has o be disjoin: for each poin y [a, b], eiher f(y) > c or f(y) < c. Thus, we wroe [a, b] as a disjoin union of non-empy, open ses, which is a conradicion. Quesion 7: Le S 1 = {z R 2 d(z, (0, 0)) = 1} be he uni circle in R 2, and le f : S 1 R be a coninuous funcion. Show ha here exiss z S 1 such ha f(z) = f( z). In paricular, f is no injecive. Noe: if z = (x 0, y 0 ) is a poin in S 1, hen z is he poin ( x 0, y 0 ). To simplify noaion, le s assume ha S 1 is paramerized in polar coordinaes by he angle θ R, wih he usual convenion ha θ and θ represen he same poin if θ θ is a muliple of 2π. Le s consider f(0) and f(π). If hey are he same, hen we are done. Oherwise, consider he funcion g : [0, π] R defined as: g(θ) = f(θ) f( θ). 5

6 I is clear ha g(0) and g(π) have differen sign, and ha he funcion g is coninuous. Thus, by he inermediae value heorem, here is θ such ha g(θ) = 0, which concludes he proof. 6