[MA 351 — Homework Problems 4 — Solutions]
2.1 Determine Int(A) and Cl(A) in each case.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
A = (0, 1] in the lower limit topology on R.
A = {a} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = {a, c} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = {b} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = (−1, 1) ∪ {2} in the standard topology on R.
A = (−1, 1) ∪ {2} in the lower limit topology on R.
A = {(x, 0) ∈ R2 | x ∈ R} in R2 with standard topology.
A = {(0, y) ∈ R2 | y ∈ R} in R2 with vertical interval topology. (See Exercise
1.19)
(i) A = {(x, 0) ∈ R2 | x ∈ R} in R2 with vertical interval topology. (See Exercise
1.19)
Answers: (a) Int(A) = (0, 1), Cl(A) = [0, 1].
(b) Int(A) = {a}, Cl(A) = X.
(c) Int(A) = {a}, Cl(A) = X.
(d) Int(A) = ∅, Cl(A) = {b, c}.
(e) Int(A) = (−1, 1), Cl(A) = [−1, 1] ∪ {2}.
(f) Int(A) = (−1, 1), Cl(A) = [−1, 1) ∪ {2}.
(g) Int(A) = ∅, Cl(A) = A.
(h) Int(A) = A, Cl(A) = A.
(i) Int(A) = ∅, Cl(A) = A.
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2.6 Prove that Cl(Q) = R in the standard topology on R.
Solution: Let x ∈ R be any real number and U any open neighborhood of x.
Then, by the definition of the topology on R, there are real numbers a, b such that
a < x < b and (a, b) ⊂ U . Since between every two real numbers there is a rational
number, one can find a q ∈ Q such that a < q < b. Then q ∈ U ∩ Q and U ∩ Q 6= ∅.
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2.13 Determine the set of limit points of A in each case.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
A = (0, 1] in the lower limit topology on R.
A = {a} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = {a, c} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = {b} in X = {a, b, c} with topology {X, ∅, {a} , {a, b}}.
A = (−1, 1) ∪ {2} in the standard topology on R.
A = (−1, 1) ∪ {2} in the lower limit topology on R.
A = {(x, 0) ∈ R2 | x ∈ R} in R2 with standard topology.
A = {(0, y) ∈ R2 | y ∈ R} in R2 with vertical interval topology. (See Exercise
1.19)
2
(i) A = {(x, 0) ∈ R2 | x ∈ R} in R2 with vertical interval topology. (See Exercise
1.19)
Answers: (a) A0 = [0, 1).
(b) A0 = {b, c}.
(c) A0 = {b, c}.
(d) A0 = {c}.
(e) A0 = [−1, 1].
(f) A0 = [−1, 1). (g) A0 = A.
(h) A0 = A.
(i) A0 = ∅.
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2.14 For each n ∈ N, let Bn = {n, n + 1, n + 2, . . .}, and consider the collection
B = {Bn | n ∈ N}.
(a) Show that B is a basis for a topology on N.
(b) Show that the topology on N generated by B is not Hausdorff.
(c) Show that the sequence (2, 4, 6, 8, . . .) converges to every point in N with the
topology generated by B.
(d) Prove that every injective sequence converges to every point in N with the topology generated by B.
Solution: (a) Note that for every n ∈ N, n ∈ Bn . Also, Bm ∩ Bn = Bk , where
k = min {m, n}. This shows that B is a basis.
(b) Since B1 = N is the only basis open set containing 1, in is also the only open
set containing 1. Therefore 1 cannot be separated for, say, 2 with open sets, and N
is not Hausdorff.
(c) Our sequence is (xk ), where xk = 2k. Let n ∈ N be any number, and U be any
open set containing n. Then for some t ∈ N, we have n ∈ Bt ⊂ U . In particular,
t ≤ n. For any integer k ≥ n/2, xk = 2k ≥ n ≥ t and xk ∈ Bt ⊂ U . Thus (xk ) → n.
(d) Let, as in (c), n ∈ N be any number, and U be any open set containing n.
Chose Bt so that n ∈ Bt ⊂ U . If (xk ) is injective (i.e. all the terms of the sequence
are different), there may be no more than t − 1 terms of the sequence that are not in
Bt . Thus for some N and all k ≥ N we have xk ∈ Bt ⊂ U . Thus (xk ) → n.
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2.15 Determine the set of limit points of [0, 1] in the finite complement topology on
R
Solution: Let x ∈ R be any real number and U and open neighborhood of x (in
the finite complement topology). Since U 6= ∅, U = R − F for some finite set F .
Then
U ∩ [0, 1] = (R − F ) ∩ [0, 1] = [0, 1] − (F ∩ [0, 1]) = [0, 1] − finite set
is infinite and therefore contains points of [0, 1] different from x. Thus x is a limit
point. Finally [0, 1]0 = R.
F
3
2.21 Determine the set of limit points of the set
1
2
S=
x, sin
∈R |0<x≤1 .
x
Solution: The rectangle R = [0, 1]×[−1, 1] is a closed set containing A. Therefore
A0 ⊂ R and we only need to examine the points P = (a, b) ∈ R. We consider three
cases:
(1) a = 0 and −1 ≤ b ≤ 1. Let z = arcsin b and let
1
xn =
,
n ≥ 1.
z + 2πn
The corresponding point Pn is
1
= (xn , b)
Pn = xn , sin
xn
is a point of A and clearly lim Pn = P . This implies that P ∈ Cl(A). Since P 6∈ A,
P ∈ A0 .
The geometric meaning of this is as follows: we constructed a sequence (Pn ) of
points of A by taking intersection points of A with the horizontal line y = b. This
sequence converges to P = (0, b) implying that P ∈ A0 .
(2) 0 < a ≤ 1 and b = f (a), i.e. P ∈ A. It is geometrically obvious that since
A is a continuous curve, any open disk centered at P intersects A at infinitely many
points. Thus P ∈ A0 .
(3) 0 < a ≤ 1 and b 6= f (a). Then one can find a small disk around P that does
not intersect A. (Draw a picture!)
In conclusion, A0 = A ∪ ({0} × [−1, 1]). Note that (2) and (3) can be made more
formal using the continuity of f (x) = sin(1/x). I will provide such a proof if anybody
is interested.
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