,Tej Bahadur Singh
Solutions Manual to
Introduction to Topology
(by the author)
,Chapter 1
Topological Spaces
Section 1.1
1. For x, y, z ∈ X, if d(x, z) = 1, then x 6= z so that d(x, y) + d(y, z) is at
least 1.
2. (a) Positivity and symmetry are clear. For triangle inequality, let
x, y, z ∈ X. Then d(x, z) ≤ d(x, y) + d(y, z). If d(x, z) > 0, then
(1 + d(x, z))/d(x, z) ≥ [1 + d(x, y) + d(y, z)]/[d(x, y) + d(y, z)] ⇒
d(x, z)/(1 + d(x, z)) ≤ [d(x, y) + d(y, z)] /[1 + d(x, y) + d(y, z)] ≤
d(x, y)/(1 + d(x, y)) + d(y, z)/(1 + d(y, z)), since d(x, y) ≥ 0 and
d(y, z) ≥ 0. If d(x, z) = 0, the inequality still holds good, so
d′ (x, z) ≤ d′ (x, y) + d′ (y, z). Clearly, d′ (x, y) ≤ 1 for all x, y ∈ X
and, therefore, diam(X) ≤ 1.
(b) d1 (x, y) = min{1, d(x, y)} ≥ 0 and d1 (x, y) = 0 ⇔ d(x, y) =
0 ⇔ x = y. Symmetry is obvious. For triangle inequality, con-
sider points x, y, z ∈ X. If d(y, z) < 1 and d(x, y) < 1, then
d(x, z) ≤ d(x, y) + d(y, z) implies d1 (x, z) ≤ d1 (x, y) + d1 (y, z).
If d(x, y) ≥ 1 or d(y, z) ≥ 1, then d1 (x, y) + d1 (y, z) is at least 1,
while d1 (x, z) ≤ 1. Also diam(X) ≤ 1.
3. (a) If x 6= 0, then xi 6= 0 for some i. So k x k= max1≤i≤n | xi |> 0.
(b) k xa k= maxi | xi a |=| a | maxi | xi |=| a |k x k.
(c) k x + y k= maxi | xi + yi |≤ maxi (| xi | + | yi |) ≤ maxi | xi |
+maxi | yi |≤k x k + k y k.
4. (a) d∗ (x, y) = max | xi − yi | is obviously nonnegative and d∗ (x, y) =
i
0 ⇔ x = y. The symmetry is also obvious. For triangle inequality,
let x = (xi ), y = (yi ), z = (zi ) be any three elements of Rn . Then,
for each i, | xi − zi |≤| xi − yi | + | yi − zi |≤ d∗ (x, y) + d∗ (y, z). So
d∗ (x, z) ≤ d∗ (x, y) + d∗ (y, z) and d∗ is a metric.
(b) Let x = (xi ), y = (yi ) and z = (zi ) be elements of Rn . Clearly
d+ (x, y) ≥ 0 and d+ (x, y) = 0 ⇔ x = y. Also, the equality
d+ (x, y) = d+ (y, x) is obvious. For the triangle inequality, we have
Pn Pn
1 | xi − zi | ≤ P1 (| xi − yi | + P | yi − zi |)
n n
= 1 | x i − y i | + 1 | y i − zi | .
1
,2 Introduction to Topology
5. B(x, r) in (R2 , d∗ ) is the square without its boundary centered at x
and side 2r, B(x, r) in (R2 , d+ ) is the rhombus without its boundary
and B(x, r) in (R2 , d) is the disc without its boundary centered at x
with radius r. In R3 , Bd∗ (x, r) is a cube without its faces, Bd+ (x, r) is
a tetragonal bipyramid without its faces and Bd (x, r) is a (solid) ball
without its surface.
6. The properties of a metric for the function d in Ex 1.1.4 follow from the
properties of Riemann integral. For the function d∗ in Ex 1.1.5, we see
that the set {| f (x) − g(x) |: x ∈ X} bounded for any f, g ∈ B(X), for
| f (x) − g(x) |≤| f (x) | + | g(x) | ∀ x ∈ X. So d∗ (f, g) is a real number
for all f, g ∈ B(X). The verification of the properties of a metric are
routine.
7. Choose a fixed point x0 ∈ X. Then for every x ∈ X,
d(f (x), g(x)) ≤ d(f (x), f (x0 )) + d(f (x0 ), g(x0 ))+
d(g(x0 ), g(x))
≤ diam (f (X)) + d(f (x0 ), g(x0 )) + diam (g(X)).
∗
So d (f, g) ≤ diam (f (X)) + d (f (x0 ), g(x0 )) + diam (g(X)) < ∞.
∗ ∗
Since d(y, y ′ ) ≥ 0 ∀ y, y ′ ∈ Y , d (f, g) ≥ 0 and d (f, g) = 0 ⇔
d(f (x), g(x)) = 0 ∀ x ∈ X ⇔ f = g. If f, g, h ∈ B(X, Y ), then
∗ ∗
d(f (x), g(x)) ≤ d(f (x), h(x)) + d(h(x), g(x)) ≤ d (f, h) + d (h, g). This
∗ ∗ ∗
implies d (f, g) ≤ d (f, h) + d (h, g). Symmetry is obvious.
8. (a) If f (x) = x + v, then kf (x) − f (y)k = kx − yk. So f sends B(x; ǫ)
into B(f (x); ǫ).
(b) Suppose g(x) = rx. If r = 0, then g is a constant map. If r 6= 0, then
kx − yk < ǫ/r ⇒ kg(x) − g(y)k < ǫ.
9. This is immediate from the inequality kxk − kyk ≤ kx − yk.
10. Let x, y ∈ X be arbitrary. Then, for every a ∈ A, d(x, a) ≤ d(x, y) +
d(y, a) ⇒ inf a∈A d(x, a) ≤ d(x, y) + inf a∈A d(y, a) ⇒ d(x, A) − d(y, A) ≤
d(x, y). Interchanging the roles of x and y, we obtain d(y, A) − d(x, A) ≤
d(x, y), and hence d(x, A) − d(y, A) ≤ d(x, y). This clearly implies that
f is continuous.
11. If there exists an x ∈ X and a real K such that d(x, a) ≤ K for all
a ∈ A, then d(a, b) ≤ 2K for all a, b ∈ A. So diam(A) ≤ 2K.
12. (a) Choose x0 ∈ A ∩ B. Let x, y ∈ A ∪ B. If x ∈ A, y ∈ B, then
d(x, y) ≤ d(x, x0 ) + d(x0 , y) ≤ diam(A) + diam(B). If both x, y are
in A, then d(x, y) ≤ diam(A), and if both of them are in B, then
d(x, y) ≤ diam(B). So diam(A ∪ B) ≤ diam(A) + diam(B).
, Topological Spaces 3
(b) Fix a ∈ A, b ∈ B. Then for any x, y ∈ A ∪ B, d(x, y) ≤ diam(A) +
d(a, b) + diam(B). So diam(A ∪ B) ≤ diam(A) + d(a, b) + diam(B).
Taking infimum over a ∈ A and b ∈ B, we get diam(A ∪ B) ≤
diam(A) + dist(A, B) + diam(B).
Section 1.2
1. (a) No.
(b) The discrete topology is the smallest topology on X containing both
T1 and T2 and the indiscrete topology is the largest topology contained
in both T1 and T2 .
2. Verification is routine.
3. T = {G : X − G is finite or contains x0 }. Clearly ∅, X ∈ T . Let
G1 , G2 ∈ T and G1 ∩ G2 6= ∅. If both X − G1 and X − G2 are finite,
then X − (G1 ∩ G2 ) = (X − G1 ) ∪ (X − G2 ) is finite. And, if X − G1 or
X − G2 is infinite, then it contains x0 . So X − (G1 ∩ G2 ) contains x0 ,
and G1 ∩ G2 ∈ T . Next, let {G Sα } be aTfamily of sets in T . If X − Gα
is finite for some α, then X − GαS= (X − GS α ) is finite. Otherwise,
x0 ∈ X − Gα for all α so that x0 ∈/ Gα . Thus Gα ∈ T .
If x 6= x0 , then x0 ∈ X − {x} so that {x} is open. Also {x} is closed,
being a finite subset of X.
4. The sets [−1, − 12 ) ∪ ( 12 , 1] in (a), (−1, − 12 ] ∪ [ 12 , 1) 1
in 1(b)1 and [−1, − 2 ] ∪
1 1
[ 2 , 1] in (c) are not open. The set (−1, 0) ∪ 2 , 1 ∪ 3 , 2 ∪ · · · in (d) is
obviously open. Except the set in (c), none are closed in R.
5. In the topology {∅, R, {p}, R − {p}}, every open set is closed and vice
versa.
6. (a) If (x, y) ∈
/ A, then x < 0 or y < 0. If x < 0, put r =| x | /2, and if
y < 0, put r =| y | /2. Then B((x, y); r) ⊆ R2 − A. So R2 − A is open
⇒ A is closed.
(b) For any r > 0, B((x, 0); r) is not contained in the set S = {(x, 0) :
−1 < x < 1}, so it is not open. The points -1 and 1 are in R2 − S, but
it is not a nbd of either of them. So S is not closed.
(c) As in (a), one sees that R2 − {(x, 0) : −1 ≤ x ≤ 1} is nbd of each of
its points.
7. Let F = Rn × {0} ⊂ Rn+m = X. If x = (xi ) ∈ / F , then xi 6= 0 for some
index i > n. Put r =| xi | /2. If y ∈ B(x; r) ∩ F , then | xi |≤ d(x, y) <
,4 Introduction to Topology
r =| xi | /2, a contradiction. So B(x; r) ∩ F = ∅ and B(x; r) ⊆ X − F .
Thus X − F is nbd of each of its points and hence F is closed.
8. Suppose that f ∈ / C(I). Then there exists a point t0 ∈ I and a real ǫ > 0
s.t for each δ > 0, there is a point tδ ∈ I satisfying | tδ − t0 |< δ but
| f (tδ ) − f (t0 ) |≥ ǫ. Put r = ǫ/3 and consider the open ball B(f ; r).
If g ∈ B(f ; r), then | f (t) − g(t) |< r for all t ∈ I. Consequently,
| g(tδ ) − g(t0 ) |≥| f (tδ ) − f (t0 ) | − | f (tδ ) − g(tδ ) | − | g(t0 ) − f (t0 ) |> r
and g is discontinuous at t0 . So B(f ; r) ⊆ B(I) − C(I).
9. Let A = {(x, y) : xy = 1, x > 0}, and B = {(x, y) : xy = −1, x < 0}.
Then A ∩ B = ∅. Since the first and second quadrants are closed
in R2 and the multiplication R2 → R1 is continuous, A and B are
closed subsets of R2 . For (1/y, y) ∈ A and (y, 1/y) ∈ B, we have
d((1/y, y), (−1/y, y)) = 2/y → 0. So d(A, B) = 0. Clearly, B can be
taken to be y-axis.
10. Let B[x; r] = {y ∈ X : d(y, x) ≤ r}. If p ∈ X − B[x; r], then d(p, x) > r.
Put ǫ = d(p; x) − r > 0. Then B(p, ǫ) ∩ B[x; r] = ∅, for y ∈ B(p; ǫ) ⇒
r < d(x, p) − d(y, p) ≤ d(x, y).
11. (a) Discrete topology, for B(x; 1/2) = {x} is open for every x ∈ X.
(b) No, for every one point set in a metric space is closed.
12. No. On the set R, the cocountable topology is not discrete, since the set
(−∞, 0) is not open.
S
13. (a) Q = q∈Q {q}. Since Q is countable and each {q} is closed in R, Q
is Fσ .
T
(b) The set Gq = R − {q} is open ∀ q ∈ Q and R − Q = q∈Q Gq .
(c) Since [a, b] is closed subset of R, it is Fσ . Also, [a, b] = ∩n∈N (a −
1 1
n , b + n ), so [a, b] is Gδ .
14. S
Let F be a closed subset of (X, d). For each n = 1, 2, . . . , put Gn =
{B(x, 1/n) : x ∈ F }. Then each Gn is an open set containing F . If
y∈/ F , then ∃ a real number r > 0 such that B(y,Tr) ⊆ X − F . For n
sufficiently large so that 1/n < r, y ∈
/ Gn . So F = n Gn .
In the cofinite space Rf , the closed set {p}, p ∈ R, is not a Gδ -set.
For, otherwise,
T we have open sets Gn , n S= 1, 2, . . ., in Rf such that
{p} = n Gn . This implies that R − {p} = n (R − Gn ), a contradiction,
since R − Gn is finite ∀ n.
15. Apply De-Morgan’s rule.
, Topological Spaces 5
16. (a) Suppose Ai , i T ∈ N are Gδ -sets. T∞ Then for T each i, ∃ open sets Gij ,
∞
j ∈ N, s.t. Ai = j=1 Gij . So i=1 Ai = i,j∈N Gij . Since N × N is
countable,
T the r.h.s. is countable intersection of open sets, and therefore
i Ai is a Gδ -set. Also,
Sn T
i=1 Ai = {G1j1 ∪ · · · ∪ Gnjn : (j1 , . . . , jn ) ∈ N × · · · × N}.
Sn
Clearly, each G1j1 ∪ · · · ∪ Gnjn is open, and thus i=1 Ai is a countable
intersection of open sets, for N × · · · × N is countable.
S∞
(b) Suppose for each i = 1, 2, . . ., Ai = j=1 Fij , where each Fij is closed
S∞ S
in X. Then i=1 Ai = i,j∈N Fij is countable union of closed sets. So
S∞
i=1 Ai is Fσ . Also,
Tn S
i=1 Ai = {F1j1 ∩ · · · ∩ Fnjn : (j1 , . . . , jn ) ∈ N × · · · × N}.
Tn Tn
Since, for each n-tuple (j1 , . . . , jn ), 1 Fiji is closed, 1 Ai is Fσ .
Section 1.3
1. No point of X is a limit point of A.
2. {a}′ = {b, c}, {b}′ = {c}, {c}′ = ∅ and {a, c}′ = {b, c}.
3. (a) If A = {1, 21 , 31 , ...}, then A◦ = ∅, A = A ∪ {0}, ∂A = {0}, A′ = {0}.
(b) If A = (−1, 0) ∪ (0, 1), then A◦ = A, A = [−1, 1], ∂A = {−1, 0, 1}
and A′ = A.
(c) For A = {(m + n)/mn : m, n ∈ N}, A◦ = ∅, A′ = {0} ∪ {1, 21 , . . .},
A = A ∪ A′ and ∂A = A.
A′ : Clearly, 0, 1, 1/2, 1/3, . . . are all limit points of A. On the other hand,
no point of R − [0, 2] is in A′ . Further, since A ∩ (1, 2] = {2, 3/2, 4/3, . . .},
no point of (1, 2] too is a limit point of A. Now, suppose that 0 < x < 1
and x 6= 1/n for all n. Then ∃ an integer m > 0 such that 1/m <
x < 1/(m − 1). If 1/m < (1/k + 1/l) < 1/(m − 1), then both k, l ≥ m
and k or l is greater than m2 + m only when the other integer is m.
Choose a positive real ǫ < min{x − 1/m, 1/(m − 1) − x}. Then we find
an integer n0 such that 1/m + 1/n < x − ǫ for all n > n0 . It follows
that U = (x − ǫ, x + ǫ) contains only finitely many points of A. Since
each singleton in R is closed, (U − A) ∪ {x} is an open nbd of x, which
contains no point of A other than possibly x. Accordingly, x is not a
limit point of A.
(d) For A = {sin n/n : n ∈ N}, A◦ = ∅, A′ = {0}, A = A∪{0}, ∂A = A.
,6 Introduction to Topology
4. (a) For A = R × {0}, A◦ = ∅, A′ = A, A = A, ∂A = A.
(b) For A = {(x, 0) : 0 < x < 1}, A◦ = ∅, A′ = {(0, 0), (1, 0)}, A =
A ∪ A′ , ∂A = A′ .
(c) For A = {(x, y) : x ∈ Q}, A◦ = ∅, A′ = R2 , A = A′ , ∂A = R2 , for
R2 − A = R2 also.
(d) Same as in (c).
(e) For A = {(x, y) : 1 < x2 + y 2 ≤ 2}, A◦ = {(x, y) : 1 < x2 + y 2 < 2},
A′ = A ∪ {(x, y) : x2 + y 2 = 1}, A = A′ , ∂A = {(x, y) : x2 + y 2 = 1 or 2}.
(f) For A = {(x, y) : x ≥ 0, y > 0}, A◦ = {(x, y) : x > 0, y > 0},
A′ = A∪{(x, 0) : x ≥ 0}, A = A′ , ∂A = {(x, 0) : x ≥ 0}∪{(0, y) : y ≥ 0}.
(g) For A = {(x, y) : x 6= 0, y ≤ 1/x}, A◦ = {(x, y) : y < 1/x},
′
A′ = A ∪ {(0, y) : y ∈ R}, A = A , ∂A = {(x, y) : xy = 1 or x = 0}.
(h) For A = {(x, y) : x ≥ y 2 }, A◦ = {(x, y) : x > y 2 }, A′ = A, A = A′ ,
∂A = {(x, y) : x = y 2 }.
T T
5. (a): (i)T Aα ⊆ ATβ for every β. So ( Aα )◦ ⊆ A◦β for all β, which implies
that ( Aα )◦ ⊆ A◦α .
S S S
(ii) Each Aβ ⊆ Aα ⇒ A◦α ⊆ ( Aα )◦ .
T T T T
(iii) As Aα ⊆ Aβ , Aα ⊆ Aβ for every β. So Aα ⊆ Aα .
S S S S
(iv) Since Aβ ⊆ Aα , Aβ ⊆ Aα , for every β. So Aα ⊆ Aα .
(b) (i) In R, consider ◦
T 1/n] for n = 1, 2, . . .. Then An =
T ◦ An = [−1/n,
(−1/n, 1/n). So An = {0}, but An = {0} and int{0} = ∅.
(ii) Take A1 = [−1, 0), A2 = [0, 1], then A◦1 ∪ A◦2 = (−1, 0) ∪ (0, 1) while
(A1 ∪ A2 )◦ = (−1, 1).
(iii) Let A = (−1, 0) and B = (0, 1). Then A ∩ B = ∅ while A∩B = {0}.
S
(iv) For each n = 1, 2, . . ., put An = (1/n, 2). Then An = (0, 2) so
S S
that An = [0, 2]. On the other hand, An = (0, 2] for An = [1/n, 2].
S S S S
(c) If Aα is closed, then Aα ⊆ Aα = Aα .
6. (a) X − A is open and contained in X − A, so X − A ⊆ (X − A)◦ . If
G is an open set contained in X − A, then A ⊆ X − G ⇒ G ⊆ X − A.
Thus X − A = (X − A)◦ .
(b) X − A◦ is a closed set containing X − A, so X − A◦ ⊇ X − A.
Conversely, if F is closed and X − A ⊆ F , then A ⊇ X − F ⇒ X − F ⊆
A◦ ⇒ F ⊇ X − A◦ . So X − A◦ ⊆ X − A and the equality holds.
(c) ∂A = A ∩ X − A = A ∩ (X − A◦ ) = A − A◦ .
(d) Obviously, A◦ ⊆ A and ∂A ⊆ A so that A◦ ∪ ∂A ⊆ A. Conversely, if
x ∈ A and x ∈ / X − A ⇒ x ∈ A◦ .
/ ∂A, then x ∈
(e) A◦ ∩ ∂A = A◦ ∩ (A − A◦ ) = ∅.
, Topological Spaces 7
(f) We have A◦ = X −X − A = A−X − A = A−((X −A)∪∂(X −A)) =
A − ∂(X − A) = A − ∂A.
7. If A is clopen, then ∂A = A ∩ X − A = A ∩ (X − A) = ∅. Conversely,
∂A = ∅ ⇒ A is closed. Also, X − A ⊆ X − A ⇒ X − A = X − A ⇒
X − A is closed, and hence A is open.
8. It is obvious that A′ and B ′ are contained in (A ∪ B)′ . Furthermore, if
x∈ / A′ ∪ B ′ , then ∃ a nbd U of x s.t U ∩ (A ∪ B) ⊆ {x}. So x ∈
/ (A ∪ B)′ .
Regarding the second question, we have
∂(A ∪ B) ⊆ A ∪ B ∩ X − A ∩ X − B
= A∩X −A∩X −B ∪ B∩X −A∩X −B
⊆ ∂A ∪ ∂B.
The inclusion may be proper. Let A = [0, 1], B = [1, 2]. Then ∂(A∪B) =
{0, 2} in R, but ∂A ∪ ∂B = {0, 1, 2}.
9. U is open ⇒ U = U ◦ .
◦ ◦ ◦
(a) Since U ⊆ U , U ⊆ U . Also, U ⊆ U = U .
(b) ∂U = U − U ◦ = U − U .
◦
(c) Take U = (0, 1) ∪ (1, 2) in R, then U = (0, 2) 6= U .
(d) If x ∈ U ∩ A and V is any nbd of x, then U ∩ V ∩ A 6= ∅. This
implies x ∈ U ∩ A.
10. Suppose that G is open. Then G ∩ A ⊆ G ∩ A ⇒ G ∩ A ⊆ G ∩ A. In
the opposite direction, we have G ∩ A ⊆ G ∩ A, by Exercise 9(d). To see
the converse, take A = X − G. Then G ∩ X − G = G ∩ (X − G) = ∅ ⇒
G ∩ X − G = ∅ ⇒ X − G ⊆ X − G ⇒ X − G is closed and hence G is
open.
11. If A is infinite and U 6= ∅ is open in X, then U ∩ A is infinite, for X − U
is finite. If A is finite and x ∈
/ A, then U = X − A is a nbd of x with
U ∩ A = ∅. If x ∈ A, then X − (A − {x}) is a nbd of x not containing
any other point of A.
12. (a) x ∈ A◦ ⇔ there exists an open set G such that x ∈ G ⊆ A ⇔ ∃
r > 0 s.t B(x, r) ⊆ A.
(b) x is a limit point of A ⇐⇒ for every nbd U of x, U ∩ A − {x} =
6
∅ ⇐⇒ ∀ r > 0, B(x, r) ∩ A − {x} 6= ∅.
/ A ⇒ ∃ r > 0 s.t B(x, r) ∩ A = ∅ ⇒ d(x, a) ≥ r ∀ a ∈ A. Con-
(c) x ∈
versely, if x ∈ A, then for every r > 0, B(x, r)∩A 6= ∅. So dist(x, A) < r.
Since r > 0 is arbitrary, dist(x, A) = 0.
, 8 Introduction to Topology
13. (a) By definition, B(x, r) ⊆ B[x, r] and, by Exercise 1.2.10, B[x, r] is
closed. So B(x, r) ⊆ B[x, r]. To see the reverse inclusion, let y ∈ B[x, r]
with k y − x k= r and consider an open ball B(y, δ). Choose 0 < ǫ <
min{δ, r}. Then z = y + (ǫ/r)(x − y) belongs to B(y, δ) ∩ B(x, r). Thus
every nbd of y intersects B(x, r) and y ∈ B(x, r). So B[x, r] ⊆ B(x, r),
and the equality holds.
(b) Let X be a set having more than one point and d be the discrete
metric. Then B(x, 1) = {x}, B[x, 1] = X, and ∂B(x, 1) = ∅ while
{y : d(y, x) = 1} = X − {x}.
(c) In general, ∂B(x, r) ⊆ {y : d(y, x) = r}.
14. (a) Let x ∈ A be an isolated point. Then ∃ an open nbd U of x such that
U ∩ A = {x}. Since A has no isolated point, x ∈ / A. As x ∈ A, U ∩ A 6=
∅ ⇒ U contains more than one point of A, a contradiction. So A also
has no isolated points.
(b) Let G ⊆ X be an open set and x ∈ G be a isolated point. Then ∃
an open set U such that U ∩ G = {x}. Thus {x} is open in X and so an
isolated point of X.
15. (a) Let X be a trivial space and ∅ 6= A ⊆ X. Since X is the only closed
set in X, which contains A, A = X. If X is discrete space and A is
proper subset of X, then A = A 6= X. So A is not dense in X.
(b) Yes, for each x ∈ X, X − {x} = X − {x} ⇒ {x} is open.
(c) Let X be a discrete space and A ⊆ X. Then both A and X − A are
closed sets, so ∂A = A ∩ (X − A) = ∅. Further, if Y is trivial space and
A is a non empty proper subset of Y , then A = Y and Y − A = Y so
that ∂A = Y . If A = ∅ or Y , then ∂A = ∅.
16. (a) D is dense in X ⇐⇒ D = X ⇐⇒ F = X for any closed set
F ⊇ D ⇐⇒ G = ∅ for every open set G ⊆ X − D.
(b) Obviously G ∩ D ⊆ G for every set G of X. Suppose that G is open
in X and x ∈ G. If D is dense in X and U is an open nbd of x, then
U ∩ G ∩ D 6= ∅, since G ∩ U is a non empty open set. So x ∈ G ∩ D.
(c) By Exercise 10, G ∩ H = G ∩ H = G = X.
Section 1.4
1. (a) For n ∈ N, {n} = (n−1, n+1), if n > 1 and (−∞, 2) = [1, 2) = {1}.
So each point of N is open in the order topology.
(b) The one-point set {(2, 1)} is not open in the dictionary order topol-
ogy on {1, 2} × N.
Solutions Manual to
Introduction to Topology
(by the author)
,Chapter 1
Topological Spaces
Section 1.1
1. For x, y, z ∈ X, if d(x, z) = 1, then x 6= z so that d(x, y) + d(y, z) is at
least 1.
2. (a) Positivity and symmetry are clear. For triangle inequality, let
x, y, z ∈ X. Then d(x, z) ≤ d(x, y) + d(y, z). If d(x, z) > 0, then
(1 + d(x, z))/d(x, z) ≥ [1 + d(x, y) + d(y, z)]/[d(x, y) + d(y, z)] ⇒
d(x, z)/(1 + d(x, z)) ≤ [d(x, y) + d(y, z)] /[1 + d(x, y) + d(y, z)] ≤
d(x, y)/(1 + d(x, y)) + d(y, z)/(1 + d(y, z)), since d(x, y) ≥ 0 and
d(y, z) ≥ 0. If d(x, z) = 0, the inequality still holds good, so
d′ (x, z) ≤ d′ (x, y) + d′ (y, z). Clearly, d′ (x, y) ≤ 1 for all x, y ∈ X
and, therefore, diam(X) ≤ 1.
(b) d1 (x, y) = min{1, d(x, y)} ≥ 0 and d1 (x, y) = 0 ⇔ d(x, y) =
0 ⇔ x = y. Symmetry is obvious. For triangle inequality, con-
sider points x, y, z ∈ X. If d(y, z) < 1 and d(x, y) < 1, then
d(x, z) ≤ d(x, y) + d(y, z) implies d1 (x, z) ≤ d1 (x, y) + d1 (y, z).
If d(x, y) ≥ 1 or d(y, z) ≥ 1, then d1 (x, y) + d1 (y, z) is at least 1,
while d1 (x, z) ≤ 1. Also diam(X) ≤ 1.
3. (a) If x 6= 0, then xi 6= 0 for some i. So k x k= max1≤i≤n | xi |> 0.
(b) k xa k= maxi | xi a |=| a | maxi | xi |=| a |k x k.
(c) k x + y k= maxi | xi + yi |≤ maxi (| xi | + | yi |) ≤ maxi | xi |
+maxi | yi |≤k x k + k y k.
4. (a) d∗ (x, y) = max | xi − yi | is obviously nonnegative and d∗ (x, y) =
i
0 ⇔ x = y. The symmetry is also obvious. For triangle inequality,
let x = (xi ), y = (yi ), z = (zi ) be any three elements of Rn . Then,
for each i, | xi − zi |≤| xi − yi | + | yi − zi |≤ d∗ (x, y) + d∗ (y, z). So
d∗ (x, z) ≤ d∗ (x, y) + d∗ (y, z) and d∗ is a metric.
(b) Let x = (xi ), y = (yi ) and z = (zi ) be elements of Rn . Clearly
d+ (x, y) ≥ 0 and d+ (x, y) = 0 ⇔ x = y. Also, the equality
d+ (x, y) = d+ (y, x) is obvious. For the triangle inequality, we have
Pn Pn
1 | xi − zi | ≤ P1 (| xi − yi | + P | yi − zi |)
n n
= 1 | x i − y i | + 1 | y i − zi | .
1
,2 Introduction to Topology
5. B(x, r) in (R2 , d∗ ) is the square without its boundary centered at x
and side 2r, B(x, r) in (R2 , d+ ) is the rhombus without its boundary
and B(x, r) in (R2 , d) is the disc without its boundary centered at x
with radius r. In R3 , Bd∗ (x, r) is a cube without its faces, Bd+ (x, r) is
a tetragonal bipyramid without its faces and Bd (x, r) is a (solid) ball
without its surface.
6. The properties of a metric for the function d in Ex 1.1.4 follow from the
properties of Riemann integral. For the function d∗ in Ex 1.1.5, we see
that the set {| f (x) − g(x) |: x ∈ X} bounded for any f, g ∈ B(X), for
| f (x) − g(x) |≤| f (x) | + | g(x) | ∀ x ∈ X. So d∗ (f, g) is a real number
for all f, g ∈ B(X). The verification of the properties of a metric are
routine.
7. Choose a fixed point x0 ∈ X. Then for every x ∈ X,
d(f (x), g(x)) ≤ d(f (x), f (x0 )) + d(f (x0 ), g(x0 ))+
d(g(x0 ), g(x))
≤ diam (f (X)) + d(f (x0 ), g(x0 )) + diam (g(X)).
∗
So d (f, g) ≤ diam (f (X)) + d (f (x0 ), g(x0 )) + diam (g(X)) < ∞.
∗ ∗
Since d(y, y ′ ) ≥ 0 ∀ y, y ′ ∈ Y , d (f, g) ≥ 0 and d (f, g) = 0 ⇔
d(f (x), g(x)) = 0 ∀ x ∈ X ⇔ f = g. If f, g, h ∈ B(X, Y ), then
∗ ∗
d(f (x), g(x)) ≤ d(f (x), h(x)) + d(h(x), g(x)) ≤ d (f, h) + d (h, g). This
∗ ∗ ∗
implies d (f, g) ≤ d (f, h) + d (h, g). Symmetry is obvious.
8. (a) If f (x) = x + v, then kf (x) − f (y)k = kx − yk. So f sends B(x; ǫ)
into B(f (x); ǫ).
(b) Suppose g(x) = rx. If r = 0, then g is a constant map. If r 6= 0, then
kx − yk < ǫ/r ⇒ kg(x) − g(y)k < ǫ.
9. This is immediate from the inequality kxk − kyk ≤ kx − yk.
10. Let x, y ∈ X be arbitrary. Then, for every a ∈ A, d(x, a) ≤ d(x, y) +
d(y, a) ⇒ inf a∈A d(x, a) ≤ d(x, y) + inf a∈A d(y, a) ⇒ d(x, A) − d(y, A) ≤
d(x, y). Interchanging the roles of x and y, we obtain d(y, A) − d(x, A) ≤
d(x, y), and hence d(x, A) − d(y, A) ≤ d(x, y). This clearly implies that
f is continuous.
11. If there exists an x ∈ X and a real K such that d(x, a) ≤ K for all
a ∈ A, then d(a, b) ≤ 2K for all a, b ∈ A. So diam(A) ≤ 2K.
12. (a) Choose x0 ∈ A ∩ B. Let x, y ∈ A ∪ B. If x ∈ A, y ∈ B, then
d(x, y) ≤ d(x, x0 ) + d(x0 , y) ≤ diam(A) + diam(B). If both x, y are
in A, then d(x, y) ≤ diam(A), and if both of them are in B, then
d(x, y) ≤ diam(B). So diam(A ∪ B) ≤ diam(A) + diam(B).
, Topological Spaces 3
(b) Fix a ∈ A, b ∈ B. Then for any x, y ∈ A ∪ B, d(x, y) ≤ diam(A) +
d(a, b) + diam(B). So diam(A ∪ B) ≤ diam(A) + d(a, b) + diam(B).
Taking infimum over a ∈ A and b ∈ B, we get diam(A ∪ B) ≤
diam(A) + dist(A, B) + diam(B).
Section 1.2
1. (a) No.
(b) The discrete topology is the smallest topology on X containing both
T1 and T2 and the indiscrete topology is the largest topology contained
in both T1 and T2 .
2. Verification is routine.
3. T = {G : X − G is finite or contains x0 }. Clearly ∅, X ∈ T . Let
G1 , G2 ∈ T and G1 ∩ G2 6= ∅. If both X − G1 and X − G2 are finite,
then X − (G1 ∩ G2 ) = (X − G1 ) ∪ (X − G2 ) is finite. And, if X − G1 or
X − G2 is infinite, then it contains x0 . So X − (G1 ∩ G2 ) contains x0 ,
and G1 ∩ G2 ∈ T . Next, let {G Sα } be aTfamily of sets in T . If X − Gα
is finite for some α, then X − GαS= (X − GS α ) is finite. Otherwise,
x0 ∈ X − Gα for all α so that x0 ∈/ Gα . Thus Gα ∈ T .
If x 6= x0 , then x0 ∈ X − {x} so that {x} is open. Also {x} is closed,
being a finite subset of X.
4. The sets [−1, − 12 ) ∪ ( 12 , 1] in (a), (−1, − 12 ] ∪ [ 12 , 1) 1
in 1(b)1 and [−1, − 2 ] ∪
1 1
[ 2 , 1] in (c) are not open. The set (−1, 0) ∪ 2 , 1 ∪ 3 , 2 ∪ · · · in (d) is
obviously open. Except the set in (c), none are closed in R.
5. In the topology {∅, R, {p}, R − {p}}, every open set is closed and vice
versa.
6. (a) If (x, y) ∈
/ A, then x < 0 or y < 0. If x < 0, put r =| x | /2, and if
y < 0, put r =| y | /2. Then B((x, y); r) ⊆ R2 − A. So R2 − A is open
⇒ A is closed.
(b) For any r > 0, B((x, 0); r) is not contained in the set S = {(x, 0) :
−1 < x < 1}, so it is not open. The points -1 and 1 are in R2 − S, but
it is not a nbd of either of them. So S is not closed.
(c) As in (a), one sees that R2 − {(x, 0) : −1 ≤ x ≤ 1} is nbd of each of
its points.
7. Let F = Rn × {0} ⊂ Rn+m = X. If x = (xi ) ∈ / F , then xi 6= 0 for some
index i > n. Put r =| xi | /2. If y ∈ B(x; r) ∩ F , then | xi |≤ d(x, y) <
,4 Introduction to Topology
r =| xi | /2, a contradiction. So B(x; r) ∩ F = ∅ and B(x; r) ⊆ X − F .
Thus X − F is nbd of each of its points and hence F is closed.
8. Suppose that f ∈ / C(I). Then there exists a point t0 ∈ I and a real ǫ > 0
s.t for each δ > 0, there is a point tδ ∈ I satisfying | tδ − t0 |< δ but
| f (tδ ) − f (t0 ) |≥ ǫ. Put r = ǫ/3 and consider the open ball B(f ; r).
If g ∈ B(f ; r), then | f (t) − g(t) |< r for all t ∈ I. Consequently,
| g(tδ ) − g(t0 ) |≥| f (tδ ) − f (t0 ) | − | f (tδ ) − g(tδ ) | − | g(t0 ) − f (t0 ) |> r
and g is discontinuous at t0 . So B(f ; r) ⊆ B(I) − C(I).
9. Let A = {(x, y) : xy = 1, x > 0}, and B = {(x, y) : xy = −1, x < 0}.
Then A ∩ B = ∅. Since the first and second quadrants are closed
in R2 and the multiplication R2 → R1 is continuous, A and B are
closed subsets of R2 . For (1/y, y) ∈ A and (y, 1/y) ∈ B, we have
d((1/y, y), (−1/y, y)) = 2/y → 0. So d(A, B) = 0. Clearly, B can be
taken to be y-axis.
10. Let B[x; r] = {y ∈ X : d(y, x) ≤ r}. If p ∈ X − B[x; r], then d(p, x) > r.
Put ǫ = d(p; x) − r > 0. Then B(p, ǫ) ∩ B[x; r] = ∅, for y ∈ B(p; ǫ) ⇒
r < d(x, p) − d(y, p) ≤ d(x, y).
11. (a) Discrete topology, for B(x; 1/2) = {x} is open for every x ∈ X.
(b) No, for every one point set in a metric space is closed.
12. No. On the set R, the cocountable topology is not discrete, since the set
(−∞, 0) is not open.
S
13. (a) Q = q∈Q {q}. Since Q is countable and each {q} is closed in R, Q
is Fσ .
T
(b) The set Gq = R − {q} is open ∀ q ∈ Q and R − Q = q∈Q Gq .
(c) Since [a, b] is closed subset of R, it is Fσ . Also, [a, b] = ∩n∈N (a −
1 1
n , b + n ), so [a, b] is Gδ .
14. S
Let F be a closed subset of (X, d). For each n = 1, 2, . . . , put Gn =
{B(x, 1/n) : x ∈ F }. Then each Gn is an open set containing F . If
y∈/ F , then ∃ a real number r > 0 such that B(y,Tr) ⊆ X − F . For n
sufficiently large so that 1/n < r, y ∈
/ Gn . So F = n Gn .
In the cofinite space Rf , the closed set {p}, p ∈ R, is not a Gδ -set.
For, otherwise,
T we have open sets Gn , n S= 1, 2, . . ., in Rf such that
{p} = n Gn . This implies that R − {p} = n (R − Gn ), a contradiction,
since R − Gn is finite ∀ n.
15. Apply De-Morgan’s rule.
, Topological Spaces 5
16. (a) Suppose Ai , i T ∈ N are Gδ -sets. T∞ Then for T each i, ∃ open sets Gij ,
∞
j ∈ N, s.t. Ai = j=1 Gij . So i=1 Ai = i,j∈N Gij . Since N × N is
countable,
T the r.h.s. is countable intersection of open sets, and therefore
i Ai is a Gδ -set. Also,
Sn T
i=1 Ai = {G1j1 ∪ · · · ∪ Gnjn : (j1 , . . . , jn ) ∈ N × · · · × N}.
Sn
Clearly, each G1j1 ∪ · · · ∪ Gnjn is open, and thus i=1 Ai is a countable
intersection of open sets, for N × · · · × N is countable.
S∞
(b) Suppose for each i = 1, 2, . . ., Ai = j=1 Fij , where each Fij is closed
S∞ S
in X. Then i=1 Ai = i,j∈N Fij is countable union of closed sets. So
S∞
i=1 Ai is Fσ . Also,
Tn S
i=1 Ai = {F1j1 ∩ · · · ∩ Fnjn : (j1 , . . . , jn ) ∈ N × · · · × N}.
Tn Tn
Since, for each n-tuple (j1 , . . . , jn ), 1 Fiji is closed, 1 Ai is Fσ .
Section 1.3
1. No point of X is a limit point of A.
2. {a}′ = {b, c}, {b}′ = {c}, {c}′ = ∅ and {a, c}′ = {b, c}.
3. (a) If A = {1, 21 , 31 , ...}, then A◦ = ∅, A = A ∪ {0}, ∂A = {0}, A′ = {0}.
(b) If A = (−1, 0) ∪ (0, 1), then A◦ = A, A = [−1, 1], ∂A = {−1, 0, 1}
and A′ = A.
(c) For A = {(m + n)/mn : m, n ∈ N}, A◦ = ∅, A′ = {0} ∪ {1, 21 , . . .},
A = A ∪ A′ and ∂A = A.
A′ : Clearly, 0, 1, 1/2, 1/3, . . . are all limit points of A. On the other hand,
no point of R − [0, 2] is in A′ . Further, since A ∩ (1, 2] = {2, 3/2, 4/3, . . .},
no point of (1, 2] too is a limit point of A. Now, suppose that 0 < x < 1
and x 6= 1/n for all n. Then ∃ an integer m > 0 such that 1/m <
x < 1/(m − 1). If 1/m < (1/k + 1/l) < 1/(m − 1), then both k, l ≥ m
and k or l is greater than m2 + m only when the other integer is m.
Choose a positive real ǫ < min{x − 1/m, 1/(m − 1) − x}. Then we find
an integer n0 such that 1/m + 1/n < x − ǫ for all n > n0 . It follows
that U = (x − ǫ, x + ǫ) contains only finitely many points of A. Since
each singleton in R is closed, (U − A) ∪ {x} is an open nbd of x, which
contains no point of A other than possibly x. Accordingly, x is not a
limit point of A.
(d) For A = {sin n/n : n ∈ N}, A◦ = ∅, A′ = {0}, A = A∪{0}, ∂A = A.
,6 Introduction to Topology
4. (a) For A = R × {0}, A◦ = ∅, A′ = A, A = A, ∂A = A.
(b) For A = {(x, 0) : 0 < x < 1}, A◦ = ∅, A′ = {(0, 0), (1, 0)}, A =
A ∪ A′ , ∂A = A′ .
(c) For A = {(x, y) : x ∈ Q}, A◦ = ∅, A′ = R2 , A = A′ , ∂A = R2 , for
R2 − A = R2 also.
(d) Same as in (c).
(e) For A = {(x, y) : 1 < x2 + y 2 ≤ 2}, A◦ = {(x, y) : 1 < x2 + y 2 < 2},
A′ = A ∪ {(x, y) : x2 + y 2 = 1}, A = A′ , ∂A = {(x, y) : x2 + y 2 = 1 or 2}.
(f) For A = {(x, y) : x ≥ 0, y > 0}, A◦ = {(x, y) : x > 0, y > 0},
A′ = A∪{(x, 0) : x ≥ 0}, A = A′ , ∂A = {(x, 0) : x ≥ 0}∪{(0, y) : y ≥ 0}.
(g) For A = {(x, y) : x 6= 0, y ≤ 1/x}, A◦ = {(x, y) : y < 1/x},
′
A′ = A ∪ {(0, y) : y ∈ R}, A = A , ∂A = {(x, y) : xy = 1 or x = 0}.
(h) For A = {(x, y) : x ≥ y 2 }, A◦ = {(x, y) : x > y 2 }, A′ = A, A = A′ ,
∂A = {(x, y) : x = y 2 }.
T T
5. (a): (i)T Aα ⊆ ATβ for every β. So ( Aα )◦ ⊆ A◦β for all β, which implies
that ( Aα )◦ ⊆ A◦α .
S S S
(ii) Each Aβ ⊆ Aα ⇒ A◦α ⊆ ( Aα )◦ .
T T T T
(iii) As Aα ⊆ Aβ , Aα ⊆ Aβ for every β. So Aα ⊆ Aα .
S S S S
(iv) Since Aβ ⊆ Aα , Aβ ⊆ Aα , for every β. So Aα ⊆ Aα .
(b) (i) In R, consider ◦
T 1/n] for n = 1, 2, . . .. Then An =
T ◦ An = [−1/n,
(−1/n, 1/n). So An = {0}, but An = {0} and int{0} = ∅.
(ii) Take A1 = [−1, 0), A2 = [0, 1], then A◦1 ∪ A◦2 = (−1, 0) ∪ (0, 1) while
(A1 ∪ A2 )◦ = (−1, 1).
(iii) Let A = (−1, 0) and B = (0, 1). Then A ∩ B = ∅ while A∩B = {0}.
S
(iv) For each n = 1, 2, . . ., put An = (1/n, 2). Then An = (0, 2) so
S S
that An = [0, 2]. On the other hand, An = (0, 2] for An = [1/n, 2].
S S S S
(c) If Aα is closed, then Aα ⊆ Aα = Aα .
6. (a) X − A is open and contained in X − A, so X − A ⊆ (X − A)◦ . If
G is an open set contained in X − A, then A ⊆ X − G ⇒ G ⊆ X − A.
Thus X − A = (X − A)◦ .
(b) X − A◦ is a closed set containing X − A, so X − A◦ ⊇ X − A.
Conversely, if F is closed and X − A ⊆ F , then A ⊇ X − F ⇒ X − F ⊆
A◦ ⇒ F ⊇ X − A◦ . So X − A◦ ⊆ X − A and the equality holds.
(c) ∂A = A ∩ X − A = A ∩ (X − A◦ ) = A − A◦ .
(d) Obviously, A◦ ⊆ A and ∂A ⊆ A so that A◦ ∪ ∂A ⊆ A. Conversely, if
x ∈ A and x ∈ / X − A ⇒ x ∈ A◦ .
/ ∂A, then x ∈
(e) A◦ ∩ ∂A = A◦ ∩ (A − A◦ ) = ∅.
, Topological Spaces 7
(f) We have A◦ = X −X − A = A−X − A = A−((X −A)∪∂(X −A)) =
A − ∂(X − A) = A − ∂A.
7. If A is clopen, then ∂A = A ∩ X − A = A ∩ (X − A) = ∅. Conversely,
∂A = ∅ ⇒ A is closed. Also, X − A ⊆ X − A ⇒ X − A = X − A ⇒
X − A is closed, and hence A is open.
8. It is obvious that A′ and B ′ are contained in (A ∪ B)′ . Furthermore, if
x∈ / A′ ∪ B ′ , then ∃ a nbd U of x s.t U ∩ (A ∪ B) ⊆ {x}. So x ∈
/ (A ∪ B)′ .
Regarding the second question, we have
∂(A ∪ B) ⊆ A ∪ B ∩ X − A ∩ X − B
= A∩X −A∩X −B ∪ B∩X −A∩X −B
⊆ ∂A ∪ ∂B.
The inclusion may be proper. Let A = [0, 1], B = [1, 2]. Then ∂(A∪B) =
{0, 2} in R, but ∂A ∪ ∂B = {0, 1, 2}.
9. U is open ⇒ U = U ◦ .
◦ ◦ ◦
(a) Since U ⊆ U , U ⊆ U . Also, U ⊆ U = U .
(b) ∂U = U − U ◦ = U − U .
◦
(c) Take U = (0, 1) ∪ (1, 2) in R, then U = (0, 2) 6= U .
(d) If x ∈ U ∩ A and V is any nbd of x, then U ∩ V ∩ A 6= ∅. This
implies x ∈ U ∩ A.
10. Suppose that G is open. Then G ∩ A ⊆ G ∩ A ⇒ G ∩ A ⊆ G ∩ A. In
the opposite direction, we have G ∩ A ⊆ G ∩ A, by Exercise 9(d). To see
the converse, take A = X − G. Then G ∩ X − G = G ∩ (X − G) = ∅ ⇒
G ∩ X − G = ∅ ⇒ X − G ⊆ X − G ⇒ X − G is closed and hence G is
open.
11. If A is infinite and U 6= ∅ is open in X, then U ∩ A is infinite, for X − U
is finite. If A is finite and x ∈
/ A, then U = X − A is a nbd of x with
U ∩ A = ∅. If x ∈ A, then X − (A − {x}) is a nbd of x not containing
any other point of A.
12. (a) x ∈ A◦ ⇔ there exists an open set G such that x ∈ G ⊆ A ⇔ ∃
r > 0 s.t B(x, r) ⊆ A.
(b) x is a limit point of A ⇐⇒ for every nbd U of x, U ∩ A − {x} =
6
∅ ⇐⇒ ∀ r > 0, B(x, r) ∩ A − {x} 6= ∅.
/ A ⇒ ∃ r > 0 s.t B(x, r) ∩ A = ∅ ⇒ d(x, a) ≥ r ∀ a ∈ A. Con-
(c) x ∈
versely, if x ∈ A, then for every r > 0, B(x, r)∩A 6= ∅. So dist(x, A) < r.
Since r > 0 is arbitrary, dist(x, A) = 0.
, 8 Introduction to Topology
13. (a) By definition, B(x, r) ⊆ B[x, r] and, by Exercise 1.2.10, B[x, r] is
closed. So B(x, r) ⊆ B[x, r]. To see the reverse inclusion, let y ∈ B[x, r]
with k y − x k= r and consider an open ball B(y, δ). Choose 0 < ǫ <
min{δ, r}. Then z = y + (ǫ/r)(x − y) belongs to B(y, δ) ∩ B(x, r). Thus
every nbd of y intersects B(x, r) and y ∈ B(x, r). So B[x, r] ⊆ B(x, r),
and the equality holds.
(b) Let X be a set having more than one point and d be the discrete
metric. Then B(x, 1) = {x}, B[x, 1] = X, and ∂B(x, 1) = ∅ while
{y : d(y, x) = 1} = X − {x}.
(c) In general, ∂B(x, r) ⊆ {y : d(y, x) = r}.
14. (a) Let x ∈ A be an isolated point. Then ∃ an open nbd U of x such that
U ∩ A = {x}. Since A has no isolated point, x ∈ / A. As x ∈ A, U ∩ A 6=
∅ ⇒ U contains more than one point of A, a contradiction. So A also
has no isolated points.
(b) Let G ⊆ X be an open set and x ∈ G be a isolated point. Then ∃
an open set U such that U ∩ G = {x}. Thus {x} is open in X and so an
isolated point of X.
15. (a) Let X be a trivial space and ∅ 6= A ⊆ X. Since X is the only closed
set in X, which contains A, A = X. If X is discrete space and A is
proper subset of X, then A = A 6= X. So A is not dense in X.
(b) Yes, for each x ∈ X, X − {x} = X − {x} ⇒ {x} is open.
(c) Let X be a discrete space and A ⊆ X. Then both A and X − A are
closed sets, so ∂A = A ∩ (X − A) = ∅. Further, if Y is trivial space and
A is a non empty proper subset of Y , then A = Y and Y − A = Y so
that ∂A = Y . If A = ∅ or Y , then ∂A = ∅.
16. (a) D is dense in X ⇐⇒ D = X ⇐⇒ F = X for any closed set
F ⊇ D ⇐⇒ G = ∅ for every open set G ⊆ X − D.
(b) Obviously G ∩ D ⊆ G for every set G of X. Suppose that G is open
in X and x ∈ G. If D is dense in X and U is an open nbd of x, then
U ∩ G ∩ D 6= ∅, since G ∩ U is a non empty open set. So x ∈ G ∩ D.
(c) By Exercise 10, G ∩ H = G ∩ H = G = X.
Section 1.4
1. (a) For n ∈ N, {n} = (n−1, n+1), if n > 1 and (−∞, 2) = [1, 2) = {1}.
So each point of N is open in the order topology.
(b) The one-point set {(2, 1)} is not open in the dictionary order topol-
ogy on {1, 2} × N.
