Topology Homework 1: Chapter 2. Continuity 1. Page 31, question no. 1 :
(a) A ∪ B = A ∪ B ∪ A ∪ B0= A ∪ B ∪ A0∪ B0 = A ∪ A0 ∪ B ∪ B0 = A ∪ B, where A ∪ B0
, A0and B0denotes the set of limit points of A ∪ B, A and B, respectively.
(b) A ∩ B = A ∩ B ∪ A ∩ B0⊆ A ∩ B ∪ A0∩ B0 = A ∪ A0 ∩ B ∪ B0 = A ∩ B.
(c) Since the closure of A is defined to be A = ∩{C ⊂ X | A ⊆ C, C is closed in X }, A is the smallest closed set containing A. Hence, A = (A) = A.
(d) Since the interior of A is defined to be IntA = ∪{U ⊂ X | U ⊂ A, U is open in X }, IntA is the largest open set contained in A. Hence, IntA ∪ IntB ⊆ Int A ∪ B since IntA ∪ IntB is an open subset of A ∪ B.
(e) Since Int(A) ∩ Int(B) is an open set contained in A ∩ B, we have Int(A) ∩ Int(B) ⊆ Int A ∩ B.
Conversely, since Int A ∩ B is an open set contained in both A and B, we have Int A ∩ B ⊆ Int(A) ∩ Int(B). Hence, we have Int A ∩ B = Int(A) ∩ Int(B).
(f) Since IntA is open, it is the largest open subset contained in IntA and Int IntA.
2. Page 31, question no. 2 : For n ∈ N, let An= − 2+1
n , 2 −1
n. It is easy to see that ∪∞n=1An= − 2, 2, an open interval in R.
3. Page 31, question no. 3(c) : Let A = R2\ {(x, sin(1/x)) | x > 0}. Then (a) the interior Int(A) = R2\{(x, sin(1/x)) | x > 0} ∪ {(0, y) | 0 ≤ y ≤ 1}.
(b) the closure A = R2.
(c) the frontier (or the boundary) ∂ A ={(x, sin(1/x)) | x > 0} ∪ {(0, y) | 0 ≤ y ≤ 1}.
4. Page 31, question no. 4 :
(a) The set of limit points is {1/n | n = 1, 2, . . .} ∪ {0}.
(b) The set of limit points is {0}.
5. Page 31, question no. 5 : By using the result in 1(b), the assumption A is dense and X \ O is closed, we have
A∩ (X \ O) ⊆ A ∩ X \ O = X ∩ X \ O = X \ O = X \ O. (∗) Also, by using the assumption that A is dense, the results in 1(a) and (∗), we have
O∪ (X \ O) = X = A = (A ∩ O) ∪ (A ∩ (X \ O)) = (A ∩ O) ∪ (A ∩ (X \ O)) ⊆ (A ∩ O) ∪ (X \ O).
Thus, we have
O∪ (X \ O) ⊆ (A ∩ O) ∪ (X \ O), and
O⊆ (A ∩ O).
6. Page 31, question no. 9 : If A is open (closed) in Y then A = U ∩Y, where U is an open (closed) subset of X . Therefore, if Y is open (closed) in X then U ∩Y = A is open (closed) in X .
7. Page 35, question no. 13 :
(a) Since {0} is closed in R and g : X → R is continuous, the set {x | g(x) = 0} = g−1({0}) is closed in X .
Topology Homework (Continued)
(b) If f : R → R is continuous, then the function g : R → R defined by g(x) = f (x) − x is also contin- uous. Thus, the set {x ∈ R | f (x) = x} = {x ∈ R | g(x) = 0} is closed in R.
8. Page 35, question no. 15 :
(a) Let W be a basic open set in R2of the form W = (a, b) × (c, d). Then W ∩ graph( f ) = {(x, f (x)) | x∈ (a, b)} is open in graph( f ) = {(x, f (x) | x ∈ R}. Since Γ−1f (W ∩ graph( f )) = (a, b) is open in R, the map Γf : R → R2is continuous.
(b) For each x ∈ R, define the inverse map g : graph( f ) → R by g(x, f (x)) = x. Since g−1((a, b)) = W∩ graph( f ), g is continuous. Hence, the image of Γf, i.e. the graph( f ), is homeomorphic to R.
9. Page 36, question no. 18 : Let V be an open set in Y and let M ∈ N such that f−1(V ) = ( f |AM)−1(V ) ⊆ AM.
The continuity of ( f |AM) on AM and AM⊆ Int(AM+1) implies that there exists an open set U ⊆ X such that
f−1(V ) = ( f |AM)−1(V ) = U ∩ AM= U ∩ IntAM+1. This proves that f : X → Y is continuous.
10. Page 36, question no. 21 : Let B(1) = {x ∈ Rn| ∑ni=1xi2≤ 1} and let C(1) = {x ∈ Rn| |xi| ≤ 1, 1 ≤ i ≤ n}
denote the unit ball and unit cube in Rn. Define f : C(1) → B(1) by
f(x) =
(0 if x = 0
t(x)x ∀x 6= 0 ∈ C(1)
where t(x) is the distance from 0 to the boundary ∂C(1) in the direction of x. It is easy to see that f is a homeomorphism from C(1) onto B(1).
11. Page 41, question no. 27 : Suppose that 0 = d(x, A) = inf{d(x, a) | a ∈ A}, there exists a sequence {an} of points in A such that limn→∞d(x, an) = 0. This implies that limn→∞an= x and x ∈ A.
Conversely, suppose that x ∈ A, there exists an∈ A ∩ Bx(1/n) for each n ∈ N, where Bx(1/n) = {y ∈ X | d(y, x) < 1/n}. This implies that limn→∞d(x, an) = 0 and d(x, A) = 0.
12. Page 41, question no. 28 : By the Lemma (2.14) on page 40 (i.e. the Urysohn Lemma), there exists a continuous function f : X → R such that
f(x) =
−1 if x ∈ A
1 if x ∈ B
r∈ (−1, 1) if x ∈ X \ (A ∪ B)
Then U = f−1((−∞, 0)) and V = f−1((0, ∞)) are disjoint open subsets satisfying A ⊆ U and B ⊆ V.
13. Page 41, question no. 30 : Let A be a closed subset of X . If A = X then A = ∩∞n=1An, where An= X for all n ∈ N.
If A 6= X , then A = A = ∩∞n=1An, where n ∈ N and An= {x ∈ X | d(x, A) < 1/n} is open in X . 14. Page 41, question no. 35 : Let h : Rn\ {0} → Snbe defined by
h(x) = x
kxk ∀x ∈ Rn\ {0}.
Then h(x) = x for all x ∈ Snand the map g : X → Sn defined by g(x) = h( f (x)) for each x ∈ Rn\ {0}
agrees with f on f−1(Sn).
