MA2223: SOLUTIONS TO PROBLEM SHEET 2
1. Let (X, d) be a metric space and let A be a subset of X.
(a) Prove that the closure Ā is the intersection of all closed sets
which contain A.
(b) Using (a) show that the closure of A is closed.
Solution: (a) Suppose x ∈ Ā and let C be a closed set which contains
A. If x ∈ A then clearly x ∈ C. If x ∈
/ A then x is a limit of a
sequence of points in A. In particular, x is a limit point of C and
so x ∈ C. This shows the inclusion
\
Ā ⊆
C
C∈Λ
where Λ is the collection of all closed sets which contain A.
Now suppose x ∈
/ Ā. We claim that there exists a closed set C
which contains A but does not contain x. Note that there must exist
an open ball B(x, r) which does not contain points in A. (Otherwise
we could construct a sequence (xn ) of points in A with xn ∈ B(x, n1 ).
But this would mean x is limit point of A which contradicts our
hypothesis). Now B(x, r) is open and so C = X\B(x, r) is a closed
set which contains A but not x. This proves the reverse inclusion so
we have
\
Ā =
C
C∈Λ
(b) We know that the intersection of any collection of closed sets
is a closed set. By (a) we see that Ā is a closed set.
2. Let (X, d) be a metric space and let a ∈ X. The closed ball with
centre a and radius r > 0 is defined as the set {x ∈ X : d(a, x) ≤ r}.
(a) Prove that this is a closed set.
(b) Prove that B(a, r) ⊆ {x ∈ X : d(a, x) ≤ r}.
(c) Prove that in Euclidean space, B(a, r) = {x ∈ X : d(a, x) ≤ r}.
Solution: Let A = {x ∈ X : d(a, x) ≤ r} be the closed ball.
(a) To prove that A is closed we will show that its complement
. If
X\A is open. Let x ∈ X\A. Then d(a, x) > r. Let = d(a,x)−r
2
1
2
MA2223: SOLUTIONS TO PROBLEM SHEET 2
y ∈ B(x, ) then by the triangle inequality
d(a, x) ≤ d(a, y) + d(y, x) < d(a, y) + which gives us
r = d(a, x) − 2 ≤ d(a, x) − < d(a, y)
This shows that B(x, ) is contained in X\A and so x is an interior
point of X\A. We conclude that X\A is open and so A is closed.
(b) Note that A is a closed set which contains B(a, r). By Q.1
B(a, r) is contained in A.
(c) In Rn we can make use of the vector space structure. Suppose
x ∈ A. Consider the sequence (xk ) in Rn where
1
1
xk = (1 − ) x + a
k
k
Note that for each k the Euclidean distance gives
1
1
d(a, xk ) = (1 − ) d(a, x) ≤ (1 − ) r < r
k
k
This shows that xk ∈ B(a, r). Also note that (xk ) converges to x
and so x is a limit point of B(a, r). We conclude that A ⊆ B(a, r).
The reverse inclusion is shown in (b).