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MATH343
Topological Properties of Metric Spaces
3. Topological Properties of Metric Spaces
3.1
open sets
Defintion 3.1. Let (X, ρ) be a metric space. Let x ∈ X and r > 0. The set
B (x, r) = { y ∈ X | ρ(x, y) < r }
is called an open ball of radius r and center at x. The set
Bc (x, r) = { y ∈ X | ρ(x, y) 6 r }
is called the closed ball of radius r and center at x.
Proposition 3.2.
1. In (R, d) (where d is the usual Euclidean metric),
B (x, r) = { y ∈ R | |x − y| < r }
= { y ∈ R | −r < y − x < r }
= {y ∈ R|x − r < y < x + r}
= (x − r, x + r) .
Similarly, Bc (x, r) = [x − r, x + r].
2. In (R2 , d)
p
o
(x1 , x2 ) ∈ R2 x21 + x22 < 1
= (x1 , x2 ) ∈ R2 x21 + x22 < 1 ,
B ((0, 0) , 1) =
n
which is the interior of a circle of radius 1 centered at the origin.
3. Let ρ be the discrete metric on a non-empty set X. If r 6 1, then
B (x, r) = { y ∈ X | ρ(x, y) < r } = {x} ,
and if r > 1, then
B (x, r) = { y ∈ X | ρ(x, y) < r } = X .
4. Note that ρ(x, y) = max{|x1 − y1 |, |x2 − y2 |} is a metric on R2 . In (R, ρ)
B ((0, 0) , 1) = (x1 , x2 ) ∈ R2 max{|x1 |, |x2 |} < 1
= (x1 , x2 ) ∈ R2 −1 < x1 < 1 and − 1 < x2 < 1
which is the interior of a square centered at the origin with sides of length 2.
Defintion 3.3. A subset U in a metric space (X, ρ) is called open if for every x ∈ X, there exists r > 0,
(depending on x and U ) such that
B (x, r) ⊆ U .
Intuitively an open set does not contain any points on its boundary.
MATH343
Topological Properties of Metric Spaces
Theorem 3.4. In a metric space (X, ρ) each open ball is an open set.
Proof. Consider an open ball B (x, r), x ∈ X, and r > 0.
Let y ∈ B (x, r). Then ρ(x, y) < r, so
δ = r − ρ(x, y) > 0 .
For any z ∈ B (y, δ),
ρ(x, z) 6 ρ(x, y) + ρ(y, z)
< ρ(x, y) + δ
= ρ(x, y) + r − ρ(x, y) = r .
That means that z ∈ B (x, r).
Thus for each y ∈ B (x, r), there exists an open ball B (y, δ), such that
B (y, δ) ⊆ B (x, r) ,
which means B (x, r) is open.
Example 3.5.
1. In (R, d), an open interval is an open set. This is because we can write
a+b b−2
(a, b) = B
,
2
2
Also (−∞, a) and (a, ∞) are open.
2. But (a, b] is not open in (R, d), because there is no r > 0 such that B (b, r) = (b − r, b + r) is contained in
(a, b].
3. Similarly neither {a} nor [a, b] or [a, b) is open in (R, d).
4. In a discrete metric space (X, ρ) every subset of X is an open set. For if A ⊆ X, and x ∈ A, then
B (x, 1) = {x} ⊆ A.
Before continuing, we recall a few definitions from set theory.
Defintion 3.6. Let X be a set and let { Ai | i ∈ I } be a family (collection) of subsets of X, where the index
set I can be finite or infinite.
We define
[
Ai = { x | x ∈ Ai for some i ∈ I }
i∈I
(the union of the Ai s), and
\
Ai = { x | x ∈ Ai for all i ∈ I }
i∈I
(the intersection of the Ai s).
Theorem 3.7. Let (X, ρ) be a metric space. Then:
1. The empty set ∅ and X are open.
2. The union of any family (finite or infinite) of open sets if X is open.
3. The intersection of a finite family of open sets if X is open.
Proof.
1. The empty set contains no point, and therefore every point in ∅ satisfies the condition of an open set.
For every x ∈ X we have B (x, 1) ⊆ X, so that X is open.
MATH343
Topological Properties of Metric Spaces
2. Let { Ui | i ∈ I } be a family of open sets in X and consider
[
U=
Ui .
i∈I
Let x ∈ U . Then there exists i ∈ I such that x ∈ Ui .
Since Ui is open there exists r > 0 such that B (x, r) ⊆ Ui .
Since Ui ⊆ U we have that for every x ∈ U we can find r > 0 such that B (x, r) ⊆ U . Thus U is open.
3. Let {U1 , U2 , . . . , Un } be a finite family of open sets in X and let
U=
n
\
Ui .
i=1
For every x ∈ U and i = 1 . . . n there exists ri such that B (x, ri ) ⊆ Ui .
Put r = min{r1 , r2 , . . . , rn }. We have r > 0.
Notice that B (x, r) ⊆ B (x, ri ) for all i = 1 . . . n. Therefore B (x, r) ⊆ Ui for all i.
So B (x, r) ⊆ U . (Notice that this argument relies on the fact that we have only finitely many ri , since
otherwise we could not guarantee that r > 0.)
Remark 3.8. The intersection of an infinite family of open sets may not be open. For example in (R, d),
∞ \
1 1
− ,
= {0} ,
n n
n=1
which is not open.
Remark 3.9. All the topological concepts (such as openness, closedness, ...) depend on both X and ρ. For
example a set which is open in ([0, 1], d) may not be open in ((0, 2), d).
Defintion 3.10. Let E be a subset of a metric space (X, ρ). A point x ∈ E is called an interior point of E if
there exists r > 0 such that B (x, r) ⊆ E.
The set of all interior points of E is called the interior of E denoted by E̊.
Thus E̊ = { x ∈ E | B (x, r) ⊆ E, for some r > 0 }.
Example 3.11. In (R, d) we have
˚b] = [a,˚b) = (a, b)
[a,˚b] = (a,˚b) = (a,
and
˚ =∅.
{a}
Theorem 3.12. Let E be a subset of a metric space (X, ρ). Then
1. E̊ is the largest open subset of X, which is contained in E.
2. E is open iff E̊ = E.
MATH343
Topological Properties of Metric Spaces
Proof.
1. We first show that E̊ is open. Let x ∈ E̊. Then there exists r > 0 such that B (x, r) ⊆ E. But B (x, r)
is open itself (see Example ??).
So each point in B (x, r) is the centre of an open ball contained in B (x, r) ⊆ E.
Therefore every point of B (x, r) is in E̊, and hence E̊ is open.
Next suppose that U is an open set in X and U ⊆ E.
Let x ∈ U . By the openness of U there exists r > 0 such that B (x, r) ⊆ U . So B (x, r) ⊆ E, which implies
that x ∈ E̊.
2. Suppose E is open. Then, since E ⊆ E, E ⊆ E̊, by the first part.
But also, by definition, E̊ ⊆ E. Hence E = E̊.
Conversely, assume that E = E̊. Then E is open by the first part.
Theorem 3.13. Let E and F be subsets of a metric space (X, ρ).
1. If E ⊆ F then E̊ ⊆ F̊ .
2. (E ˚
∩ F ) = E̊ ∩ F̊ .
3. E̊ ∪ F̊ ⊆ (E ˚
∪ F ).
Proof.
1. See Tutorial.
2. Notice that E ∩ F ⊆ E and E ∩ F ⊆ F . So by the first part E ˚
∩ F ⊆ E̊ and E ˚
∩ F ⊆ F̊ .
˚
Hence E ∩ F ⊆ E̊ ∩ F̊ .
Conversely let x ∈ E̊ ∩ F̊ . Then x ∈ E̊ and x ∈ F̊ .
So, by definition, there exists r0 > 0 and r1 > 0 such that B (x, r0 ) ⊆ E and B (x, r1 ) ⊆ F . Put r = min{r0 , r1 }.
Then B (x, r) ⊆ E and B (x, r) ⊆ F , which implies that B (x, r) ⊆ E ∩ F .
Hence x ∈ E ˚
∩ F.
3. See Tutorial.
Remark 3.14. The inclusion E̊ ∪ F̊ ⊆ (E ˚
∪ F ), above, can be strict.
For example, in (R, d), we have that Q̊ = ∅ and also I̊ = ∅, where I is the set of all irrational numbers
(notice that every ball of around a rational contains irrational numbers and vice versa). Thus
˚Q .
Q̊ ∪ I̊ = ∅ =
6 R = R̊ = I ∪
3.2
closed sets
Defintion 3.15. Let (X, ρ) be a metric space. A subset F ⊆ X is called closed set if its complement X \ F
is open.
Example 3.16.
1. In (R, d), the interval [a, b] is closed. This is because
R \ [a, b] = (−∞, a) ∪ (b, ∞) ,
being the union of two open sets, is open. In particular a singleton {a} is closed. On the other hand, if
a < b, neither (a, b] not [a, b) is closed.
MATH343
Topological Properties of Metric Spaces
2. Any subset E of a discrete metric space (X, δ) is closed. This is because its complement X \ E is open,
by Example ??.3.
Before stating the next theorem, we recall de Morgan’s laws: Let {Ai }i∈I be a collection of subsets of X. Then
[
\
X\
Ai =
(X \ Ai ) ,
i∈I
X\
\
i∈I
Ai =
i∈I
[
(X \ Ai ) .
i∈I
Theorem 3.17. Let (X, ρ) be a metric space.
1. The empty set ∅ and X are closed.
2. The intersection of any family of closed sets is closed.
3. The union of a finite family of closed sets is closed.
Proof.
1. By Theorem ?? ∅ and X are open. Thus are their complements X and ∅.
2. Suppose {Fi }i∈I is a family of closed sets in (X, ρ). Let
\
F =
Fi .
i∈I
Then X \ F = X \
T
i∈I
Fi =
S
i∈I (X
\ Fi ) .
Since each Fi is closed, (X \ Fi ) is open. Hence
S
i∈I (X
\ Fi ) is open, again by Theorem ??. So F is closed.
3. Suppose F1 , . . . , Fn are closed in (X, ρ). Then
X \ (F1 ∪ · · · ∪ Fn ) = (X \ F1 ) ∩ · · · ∩ (X \ Fn ) .
Since each Fi is closed, (X \ Fi ) is open. Hence (X \ F1 ) ∩ · · · ∩ (X \ Fn ) is open, again by Theorem ??. So F
is closed.
Remark 3.18. The union of an infinite family of closed sets may not be closed. For example for each n ∈ N
the interval
1
0, 1 −
n
is closed in (R, d). But
[
1
0, 1 −
= [0, 1)
n
n>1
is not closed.
So far we have only looked at closed sets as complements of open sets. The next theorem shows, that there is a
more interesting characterisation.
Theorem 3.19. A subset F of a metric space (X, ρ) is closed iff for every convergent sequence (xn )n>1 of
points of F
lim xn ∈ F .
n→∞
MATH343
Topological Properties of Metric Spaces
Proof. ⇒. Suppose F is closed. Let (xn )n>1 be a sequence of points of F that converges to x ∈ X.
Assume that x ∈
/ F . Then x ∈ X \ F , an open set, so there exists r > 0 such that
B (x, r) ⊆ X \ F ,
or equivalently
B (x, r) ∩ F = ∅ .
Since xn ∈ F that means that xn ∈
/ B (x, r), which implies that ρ(x, xn ) > r for all n ∈ N.
But this contradicts (xn )n>1 converging to x. Thus x ∈ F .
⇐. Suppose that F contains limits of all convergent sequences of points of F .
We want to prove that X \ F is open. Let x ∈ X \ F .
Assume there is no n ∈ N such that B x, n1 ⊆ X \ F . Then for every n ∈ N we have
1
∩ F 6= ∅ ,
B x,
n
so we can choose xn ∈ B x, n1 ∩ F for all n ∈ N.
It follows that
1
ρ(x, xn ) < → 0 .
n
Thus (xn )n>1 is a sequence of points of F that
converges to x ∈
/ F ; a contradiction.
Hence there exists n ∈ N such that B x, n1 ⊆ X \ F .
Since this is true for every x ∈ X \ F , X \ F is open, so that F is closed.
Corollary 3.20. If (X, ρ) is a complete metric space and F ⊆ X is closed, then (F, ρ) is complete.