Solutions to Assignment 3, Math 220
1 Classify each of the following sets as open, closed, neither or both.
a: n12 : n ∈ N
b: Z
c: R \ Q
d: {x : |x − π| < 1/2}
e: {sin x : x > 0}
Solution
a: neither
b: closed
c: neither
d: open
e: closed
2 Let A be a nonempty open subset of R and let Q be the set of
rationals. Prove that the intersection of A and Q is not the empty set.
Solution Since A is nonempty and open, it contains an interval. Since
Q is dense in R, this interval contains a rational number.
3 Find the set of accumulation points of each of the following sets:
a: (−1)k (1 − 1/k) : k ∈ Z and k 6= 0
b: Z
c: R \ Q
d: {x ∈ R : 4k < x < 4k + 1 for some integer k}
e: {x ∈ R : 1/(4n + 1) < x < 1/4n for some n ∈ N}
Solution
a: S 0 = {−1, 1}.
b: S 0 = ∅.
c: S 0 = R.
d: S 0 = ∪∞
k=−∞ [4k, 4k + 1].
e: S 0 = ∪∞
k=1 (1/(4k + 1), 1/4k) ∪ {0}.
4 Let B = [0, ∞). Give an example of an open cover of B that has
no finite subcover.
1
2
Solution Let Bn = (−1, n). Then Bn is open and the union ∪∞
n=1 Bn =
(−1, ∞). It follows that this provides an open cover of B. The union
of any finite collection of the Bn ’s is a bounded set and hence cannot
cover B.
5 If A and B are nonempty sets, we define the distance between A
and B to be
d(A, B) = inf |a − b| .
a∈A,b∈B
Prove that if d(A, B) = 0, where A is compact and B is closed, then
A ∩ B 6= ∅. Give an example to show that if A and B are closed, it is
possible to have d(A, B) = 0 with A ∩ B = ∅.
Solution Suppose that d(A, B) = 0, where A is compact and B is
closed. Since d(A, B) = 0, for each n ∈ N, we can find an ∈ A and
bn ∈ B with |an − bn | < 1/n. Since A is compact, A is bounded (say
|a| < M) and so the set
B1 = {b1 , b2 , b3 , . . .}
is bounded by M + 1. Now B1 is either finite or infinite. In the first
case, there is a b ∈ B1 with bn = b for infinitely many different n ∈ N.
We claim that b ∈ A ∩ B. Since B1 ⊆ B, we have b ∈ B. Now
|an − b| < 1/n for infinitely many n implies that b ∈ A0 . Since A is
compact, A is closed and hence A0 ⊆ A whereby b ∈ A (and hence
b ∈ A ∩ B).
Next suppose that B1 is infinite. Then since B1 is bounded, it has
an accumulation point, say b. Again, given > 0, choosing n so large
that n > 2/ and |bn − b| < /2, we have that
|an − b| ≤ |an − bn | + |bn − b| < 1/n + /2 < for infinitely many n. It follows that b is an accumulation point for
both A and B and, since both sets are closed, b ∈ A ∩ B.
If A = N and B = {n + 1/n : n ∈ N, n > 1}, then d(A, B) = 0, but
A ∩ B = ∅.