RealAnalysis
Question58
PartA
Bydefinition, A set which can be separated into two sets U and Vwhich have no points in common.which are such that no accumulationpoint of U belongs to V, and no accumulation point of V belongs to U.A set is totally disconnected if there exixts no subsetcontaining more than one point that are connected,for example,the setof rational numbers is totally disconnected.
Aset S is called disconnected if there are two open sets, U and V suchthat

(U S) # 0 and (V S) # 0

(U S) (V S) = 0

(U S) (V S) = S
Therefore,the closure of a disconnected set is NOT disconnected.
Proof:Let X be a connected subset of R, and let Z be the closure of X.Suppose (to get a contradiction) that Z is disconnected. That meansthere are two nonempty open subsets of R covering Z, call them U andV, such that U∩Z and V∩Z are nonempty while U∩V∩Z is empty.ThenU and V cover X also. And since U∩V∩X ⊆U∩V∩Z, then U∩V∩X is empty. Pick any points u∈U∩Z and v∈V∩Z(which exist since those sets are nonempty). That means u is in theclosure of X and U is an open neighborhood of u, so U∩X isnonempty. Similarly, V∩X is nonempty.
PartB
Takinga point s in a set S. If there is some open neighborhood of s thatlie entirely in S, then s is in the interior. For example, 1/2 in (0,1) is in the neighborhood (1/4, 3/4).
Alimit point of a set S is a point of which every neighborhood of s,no matter how small the neighborhood is, contains some point of Sother than s. Take an element x not in S and not a limit point of S.Then there`s some neighborhood of x that doesn`t intersect S. So x isin the interior of the complement of S. If all limit points of S arein S, then the complement of S is open, since every point x not in Sis in the interior of the complement of S.
Question63
PartA
InEuclidean space, a convexsetis a region, such that, for every pair of points within the region,every point on the straight line segment that joins the pair ofpoints is also within the region.
Welet Ebea set in a real or complex vector space. Eisstarconvexif there exists an x_{0}in Esuchthat the line segment from x_{0}to any point yin Eis contained in E.Hence a nonempty convex set is always starconvex but a starconvexset is not always convex.
PartB
Theconverse is false because a nonempty convex set is alwaysstarconvex, but a starconvex set is not always convex.
PartC
Yes,every star domain is a contractible set, via a straightlinehomotopy. In particular, any star domain is a simply connected set,that is, every star domain, and only a star domain, can be `shrunkeninto itself.`
Part D
No,because aconnectedspaceis a topological space that cannot be represented as the union of twoor more disjoint nonempty open subsets which is not the case forstarlike.
Question71
PartA
Ametric space M{displaystyle M} Mis connected if the only subsets that are both open and closed arethe empty set and M{displaystyle M} Mitself. Therefore, M and N having a bijective isometry between them,In this case, the two metric spaces are substantially identical.
Proof:If ( M 1 , d 1 ) , … , ( M n , d n ){displaystyle (M_{1},d_{1}),ldots ,(M_{n},d_{n})} (M_{1}d_{1})………(M_{n} d_{n}) are metricspaces, and N is the Euclidean norm on R^{n},then (M_{1X} …….X M_{n}, N (d_{1},…,d_{n})(M 1 × … × M n , N ( d 1 , … , d n ) ) {displaystyle {Big(}M_{1} imes ldots imes M_{n},N(d_{1},ldots ,d_{n}){Big )}} is a metric space, where the product metric is defined by:
N(d_{1},….,d_{n}){(x_{1},…x_{n}),(y_{1},…..y_{n})} =N{d_{i}(_{x1y1}),…d_{n}(X_{n},Y_{n})
AAndthe induced topology agrees with the product topology. By theequivalence of norms in finite dimensions, an equivalent metric isobtained if Nis the taxicabnorm, a pnorm,the maxnorm, or any other norm which is nondecreasing as thecoordinates of a positive ntupleincrease (yielding the triangle inequality).
PartB
Theconverse:If Mis a metric space with metric d,and ~is an equivalence relation on M,then we can endow the quotient set M/~with the following (pseudo)metric. Given two equivalence classes [x]and [y],we define
d’([x],[y]= inf{d(p1,q1) + d(p2,q2) +……+ d(pn,qn)} d′ ( [ x ] , [ y ] ) = inf { d ( p 1 , q 1 ) + d ( p 2 , q 2 ) + ⋯+ d ( p n , q n ) } {displaystyled`([x],[y])=inf{d(p_{1},q_{1})+d(p_{2},q_{2})+dotsb+d(p_{n},q_{n})}} wherethe infimumis taken over all finite sequences (p 1 , p 2 , … , p n ) {displaystyle (p_{1},p_{2},dots ,p_{n})}(p1,p2,…pn)and (q 1 , q 2 , … , q n ) {displaystyle (q_{1},q_{2},dots ,q_{n})}(q1,q2, …..qn)with [p 1 ] = [ x ] {displaystyle [p_{1}]=[x]} [p1]= [x]. [qn] =[y], [q n ] = [ y ] {displaystyle [q_{n}]=[y]} [ q i ] = [ p i + 1 ] , i= 1 , 2 , … , n − 1 {displaystyle [q_{i}]=[p_{i+1}],i=1,2,dots,n1}Ingeneral this will only define a pseudo metric,d′ ( [ x ] , [ y ] ) = 0 {displaystyle d`([x],[y])=0} [x ] = [ y ] {displaystyle [x]=[y]}However,for nice equivalence relations (for instance, those given by gluingtogether polyhedra along faces), it is a metric.
Question76
PartA
Theintersection of connected sets is not necessarily connected. Example, Consider a collection { X i }{displaystyle {X_{i}}} {Xi} of connected sets whose unionis X = ∪i X i {displaystyle X=cup _{i}{X_{i}}} X= Ui Xi. If X{displaystyle X} X is disconnected and U∪V {displaystyle Ucup V} U Ụ V is a separation of X{displaystyle X} X(with U , V{displaystyle U,V} U,V disjoint and open in X{displaystyle X} X), then each Xi {displaystyle X_{i}} Xi must be entirely contained ineither U {displaystyle U} U or V{displaystyle V} V, since otherwise, Xi ∩ U {displaystyle X_{i}cap U} Xi ∩ U and Xi ∩ V {displaystyle X_{i}cap V} Xi ∩V (which aredisjoint and open in X i {displaystyleX_{i}} Xi ) would be a separation of Xi {displaystyle X_{i}} Xi, contradicting the assumption thatit is connected.
Thismeans that, if the union X{displaystyle X} Xis disconnected, then the collection {X i } {displaystyle {X_{i}}} {Xi}canbe partitioned to two subcollections, such that the unions of thesubcollections are disjoint and open in X{displaystyle X} X. It implies that in several cases, a union of connected sets isnecessarily connected.
PartB
LetS be a closed set, and let X_{n}be a sequence in S (i.e., 8n 2 N : xn 2 S) that converges to _x 2 X.We must show that _x 2 S. Suppose not  i.e., _x 2 Sc. Since Sc isopen, there is an ^_ > 0 for which B(_x ^_) _ Sc. Since xn ! _x,let _n be such that n > _n ) d(xn _x) < ^_ > 0. Then for n> _n we have both xn 2 S and xn 2 B(_x ^_) _ Sc, a contradiction.((): Suppose S is not closed. We must show that S does not containall its limit points. Since S is not closed, Sc is not open.Therefore there is at least one element _x of Sc such that every ballB(_x _) contains at least one element of (Sc)c = S. For every n 2 N,let xn 2 B(_x 1n) S. Then we have a sequence xn in S whichconverges to _x 62 S  i.e., _x is a limit point of S but is not inS, so S does not contain all its limit points.
PartC
Theanswer does not change in the case it is compact since it will bestill be closed going by the definition of compactness.∀i , j : X i ∩ X j ≠ ∅{displaystyle forall i,j:X_{i}cap X_{j}
eq emptyset }∀i : X i ∩ X i + 1 ≠ ∅{displaystyle forall i:X_{i}cap X_{i+1}
eq emptyset } X / { X i} {displaystyle X/{X_{i}}} X {displaystyle X} q ( U ) ∪q ( V ) {displaystyle q(U)cup q(V)} q ( U ) , q ( V ){displaystyle q(U),q(V)} f : ( M , d ) ⟶( X , δ ) {displaystyle f:(M,d)longrightarrow (X,delta )} δ ( f( x ) , f ( y ) ) ≤ d ( x , y ) {displaystyle delta(f(x),f(y))leq d(x,y)} x ∼y , {displaystyle xsim y,} f ¯ : M / ∼⟶X {displaystyle {overline {f}}colon M/sim longrightarrow X} f ¯( [ x ] ) = f ( x ) {displaystyle {overline {f}}([x])=f(x)} f ¯ 🙁 M / ∼, d ′ ) ⟶( X , δ ) . {displaystyle {overline {f}}colon (M/sim,d`)longrightarrow (X,delta ).}