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Solutions to Problem Set 3: Limits and closures

Problem 1. Let X be a topological space and A, B ⊂ X .

a. Show that A ∪B = A ∪B .

b. Show that A ∩ B ⊂ A ∩ B .

c. Give an example of X , A, and B such that A ∩ B = A ∩ B .

d. Let Y be a subset of X such that A ⊂ Y . Denote by A the closure of Ain X , and equip Y with the subspace topology. Describe the closure of Ain Y in terms of A and Y .

Solution 1. a. Taking closures preserve inclusion relation and A and B aresubsets of A∪B , so A and B are subsets of A ∪B . Hence, A∪B ⊂ A ∪B .On the other hand, A∪B ⊂ A∪B and the latter is closed; hence, A ∪B ⊂A ∪B . Thus, the equality holds.

b. A ∩ B ⊂ A and this implies A ∩ B ⊂ A. Similarly, A ∩ B ⊂ B; thus,A ∩ B ⊂ A ∩ B .

c. Let X = R , with the standard topology, A = R < 0 and B = R > 0 . Then,clearly A ∩ B = ∅, but A ∩ B = R ≤ 0 ∩ R ≥ 0 = {0}. So the equality fails.

d. The closure of A inside of Y is equal to A⊂F ⊂Y, closed in subspace topology F .But the set of closed subsets of Y , with respect to subspace topology, isexactly {F ∩Y : F is closed in X} and the set over which we take inter-section is {F ∩Y : F is closed in X, A ⊂ F }. Hence the above intersectionis equal to Y ∩ A⊂F, F is closed in X F = Y ∩ A. This is the closure in Y with respect to subspace topology.

Problem 2. Let X be a set and let

τ = {U ∈ P (X ) : X \ U is nite, or U = ∅}.

a. Show that τ is a topology on X . This topology is called the conite

topology (or nite complement topology).

b. Let X be a set equipped with the conite topology. Let A ⊂ X . Describethe boundary ∂A of A.

c. Suppose X = N . To which points does the sequence ( n)n ∈N converge?

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Solution 2. a. By assumption ∅ ∈ τ and X ∈ τ as X \ X = ∅ is nite. Allsupersets of sets with nite complement, also have nite complement. Inparticular unions of open sets have nite complement(unless they are allempty, in which case the union is empty as well). Given two non-empty,conite sets U, V ∈ τ , X \ (U ∩V ) = ( X \ U )∪(X \ V ) is nite, so U ∩V ∈ τ .

b. • First the trivial case: If X is nite then the topology is the discretetopology, so everything is open and closed and boundaries are empty.

• If X is innite but A is nite, it is closed, so its closure is A. Butthere is no non-empty open set in A, so its interior is empty and itsboundary is A.

• The complement of the last case is also similar: If A is innite witha nite complement, it is open, so its interior is itself, but the onlyclosed set containing it is X , so its boundary is equal to X \ A.

• If both A and its complement is innite, then arguing as above we seethat it has empty interior and its closure is X . Thus, its boundaryis also X .

c. To every point: Given x ∈N and an open neighborhood U , all but nitelymany natural numbers lie in U . Thus, the sequence eventually falls intoU as well and this implies the sequence converges to x for any x ∈N .

Problem 3. Let (X, d ) be a metric space. Prove that the metric topology onX is Hausdorff.

Solution 3. Given x, y ∈ X , x = y, the distance d(x, y ) > 0. Let r = d(x, y )/ 2.Clearly B (x, r ) and B(y, r ) are open subsets containing x and y and which doesnot intersect each other. Thus, every such pair is seperated by open sets and X is Hausdorff.

Problem 4. Let X and Y be topological spaces. A map f : X → Y is calledopen if for every open U ⊂ X , the image f (U ) is open in Y .

a. Consider X × Y equipped with the product topology. Show that the map p1 : X × Y → X, (x, y) → x is both continuous and open.

b. Consider X Y equipped with the sum topology. Show that the mapi1 : X → X Y, x → (x, 0) is both continuous and open.

Solution 4. a. Given U ⊂ X open, the preimage p− 11 (U ) is equal to U × Y which is one of the basic open subsets. Hence, it is open and so, p1 iscontinuous. Also as any open set in the product topology a union of setsof type U × V , where U and V are open and as taking images commuteswith taking unions, it is enough to check p1 (U × V ) is open. But this setis equal to either U or ∅ hence it is open. Thus, p1 is open.

b. The images of open sets U ⊂ X are U × { 0}; hence are among the basiselements for the sum topology introduced last week. Hence they are open,

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Solution 5. a. The preimages of ∅ and X/ ∼ are ∅ and X respectively, hencethey are open. Taking preimages commute with unions and intersections;hence, respective closedness properties on X under these operations implythe same for X . For instance, if U and V are open in X/ ∼, then π− 1 (U ∩V ) = π− 1 (U ) ∩π− 1 (V ) is open in X . Hence, U ∩V is open in X/ ∼. It issimilar for unions. Hence, this denes a topology. Preimages of opens areopen by denition, hence π is continuous.

b. No. For instance, let X = [0, 1] and let x ∼ y if and only if x = y orx, y ∈ {0, 1}, i.e. we are identifying 0 and 1. Let U = (1 / 2, 1]. It is openin X , but π(U ) is not: π− 1 (π(U )) = {0}∪(1/ 2, 1] is not open in X = [0, 1].

c. If [x1 ] = [x2 ], i.e. x1 ∼ x2 , then f (x1 ) = f (x2 ), hence such a map iswell dened. Uniqueness is also an easy set theoretic fact, it follows fromsurjectivity of π : Any a ∈ X/ ∼ can be written as π (x) and f (a) = f (x)

is uniquely determined by f . To show continuity, let V ⊂ Y be open,then f

− 1(V ) has preimage π − 1 (f

− 1(V )) = f − 1 (V ), which is open as f is

continuous. Hence, f is also continuous.

d. For points [( x, i )] with x = 0, and for 0 < < |x | , B (x, ) × { 0, 1} maps toan open subset(whose preimage is itself), that contains only nitely manyof the points of the sequence. On the other hand, every neighborhood of [(0, 0)] and [(0, 1)] pulls back to an open subset of R R that contains(U × { 0}) \ { (0, 0), (0, 1)}, where U is a neighborhood of 0 ∈R . It clearlycontains all but nitely many of the points from the sequence; hence,the sequence has limits [(0 , 0)] and [(0, 1)], i.e. the two origins. Non-uniqueness of the limit implies it is not Hausdorff.

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