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1. Elementary Topology of Complex Plane
1.1. What is topology. Topology is a branch of geometry that studies the geometric objects, called topological spaces, under continuous
maps. Two topological spaces are considered the same if there is a continuous bijection between them. For example, two circles of different
radius are regarded as the same topological space.
1.2. Basic notions of topology. We mainly concern ourselves with
the complex plane C ∼
= R2 .
Definition 1.1. A set G ⊂ C is open if for every point p ∈ G, there is
r > 0 such that
Bp (r) = {|z − z0 | < r} ⊂ G.
A related concept is the definition of interior points.
Definition 1.2. A point p of a set G ⊂ C is an interior point of G if
there is r > 0 such that
Bp (r) = {|z − z0 | < r} ⊂ G.
In other words, G is open if and only if every point of G is an interior
point.
Example 1.3. Show that D = {|z| < 1} is open.
Proof. For z0 ∈ D, let r = 1 − |z0 |. For every z satisfying |z − z0 | < r,
(1.1)
|z| ≤ |z − z0 | + |z0 | < r + |z0 | = 1.
Therefore, {|z − z0 | < r} ⊂ D and hence D is open.
Example 1.4. Show that D = {|z| ≤ 1} is not open.
Proof. For z0 = 1 ∈ D, 1 + r/2 6∈ D for all r > 0. Therefore,
(1.2)
{|z − 1| < r} 6⊂ D
for all r > 0 and hence D is not open.
Here are a couple of facts about open sets:
Theorem 1.5. In C,
• C and ∅ are open.
• The union of open sets is open.
• The intersection of finitely many open sets is open.
• Let F : G → R be a continuous function on an open set G ⊂ C.
Then {F (z) < 0} is open.
• Let F : G → C be a continuous function on an open set G ⊂ C.
Then F −1 (U ) is open for every open set U ⊂ C.
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Remark 1.6. A subtle point in the above theorem is that the intersection of an infinitely many open sets is not necessarily open. Indeed, for
every point p ∈ C,
\
\
(1.3)
{p} =
Bp (r) =
{|z − p| < r}
r>0
r>0
That is, every set {p} of a single point is the intersection of open sets;
yet {p} is obviously not open. More generally, every closed set G (see
below) is the intersection of open sets given by
!
\ [
(1.4)
G=
Bp (r) .
r>0
p∈G
The only sets that are both open and closed are C and ∅ (see below).
Example 1.7. Show that A = {1 < |z| < 2} is open.
Solution. We can write
A = {1 − |z| < 0} ∩ {|z| − 2 < 0} = A1 ∩ A2 .
(1.5)
Since F1 (z) = 1 − |z| and F2 (z) = |z| − 2 are continuous functions on
C,
A1 = {F1 (z) < 0} and A2 = {F2 (z) < 0}
(1.6)
are open by Theorem 1.5. And since the intersection of two open sets
is open, A is open.
Example 1.8. Find T −1 (D) for D = |z| < 1/2 and
1+z
(1.7)
T (z) =
.
1−z
Show that T −1 (D) is open.
Solution. We have
T
(1.8)
−1
1
z : |T (z)| <
2
(D) = {z : T (z) ∈ D} =
1 + z 1
= z:
<
= {z : 2|1 + z| < |1 − z|}
1 − z 2
= z = x + yi : 4(x + 1)2 + 4y 2 < (x − 1)2 + y 2
5 2
16
2
= x + yi : (x + ) + y <
3
9
5 4
= z : z + <
.
3
3
Obviously, T −1 (D) is open.
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Definition 1.9. A set G ⊂ C is closed if its complement Gc = C\G is
open.
A related concept is the definition of limit points.
Definition 1.10. A point p is a limit point of a set G ⊂ C if there
exists a sequence {pn } ⊂ G such that
(1.9)
lim pn = p.
n→∞
In fact, G is closed if and only if G contains all its limit points.
Here are some basic facts about closed sets:
Theorem 1.11. In C,
• C and ∅ are closed.
• The intersection of closed sets is closed.
• The union of finitely many closed sets is closed.
• Let F : G → R be a continuous function on a closed set G ⊂ C.
Then {F (z) ≤ 0} is closed.
• Let F : G → C be a continuous function on a closed set G ⊂ C.
Then F −1 (V ) is closed for every closed set V ⊂ C.
Definition 1.12. A set G ⊂ C is bounded if there is R > 0 such that
G ⊂ {|z| ≤ R}; namely, there is R > 0 such that |z| ≤ R for all z ∈ G.
Otherwise, G is unbounded.
Example 1.13. Show that G = {|z − 10| < 10} is bounded.
Proof. For every z ∈ G,
(1.10)
|z| ≤ |z − 10| + 10 < 10 + 10 = 20.
Therefore, G is bounded.
Example 1.14. Show that G = {|z − 10| > 10} is unbounded.
Proof. There is a sequence an = n+20 for n = 1, 2, ... such that an ∈ G
and
(1.11)
lim |an | = ∞.
n→∞
Consequently, G is unbounded.
Definition 1.15. A set G ⊂ C is (path-)connected if for every two
points p, q ∈ G, there is a continuous function γ : [0, 1] → G such that
γ(0) = p and γ(1) = q.
Example 1.16. Show that D = {|z| < 1} is connected.
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Proof. For every pair of points p, q ∈ D, the line segment
(1.12)
pq = {(1 − t)p + tq : 0 ≤ t ≤ 1}
is contained in D. Let γ : [0, 1] → D be the function
(1.13)
γ(t) = (1 − t)p + tq.
Then γ is continuous, γ(0) = p and γ(1) = q.
Indeed, a set G is called convex if pq ⊂ G for all p, q ∈ G; every
convex set is connected.
Theorem 1.17. Every convex set in C is connected.
Proof. Use the same argument as in the above example.
Theorem 1.18. Let U and V be two connected sets. If U ∩ V 6= ∅,
then U ∪ V is connected.
Proof. Let q be a point of U ∩ V . For every pair of points p1 ∈ U and
p2 ∈ V , since U and V are connected, there exist continuous functions
γ1 : [0, 1] → U and γ2 : [0, 1] → V satisfying γ1 (0) = p1 , γ1 (1) = q,
γ2 (0) = q and γ2 (1) = p2 . Then we define
(
γ1 (2t)
if 0 ≤ t ≤ 21
(1.14)
γ(t) =
γ2 (2t − 1) if 12 < t ≤ 1
Clearly, γ : [0, 1] → U ∪ V is continuous, γ(0) = p1 and γ(1) = p2 . So
U ∪ V is connected.
Example 1.19. Show that C\[0, ∞) is connected.
Proof. We can write
(1.15)
C\[0, ∞) = U1 ∪ U2 ∪ U3
where U1 = {x < 0}, U2 = {y > 0} and U3 = {y < 0}. All of U1 , U2 , U3
are connected since they are convex.
Since U1 ∩U2 6= ∅, U1 ∪U2 is connected. And since (U1 ∪U2 )∩U3 6= ∅,
U1 ∪ U2 ∪ U3 is connected.
Theorem 1.20. Let G be a set in C satisfying G ⊂ U ∪ V , where U
and V are open, U ∩ V = ∅, U ∩ G 6= ∅ and V ∩ G 6= ∅. Then G is not
connected.
Example 1.21. Show that the complement of {1 < |z| < 2} in C is
not connected.
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Proof. Let G = C\{1 < |z| < 2}. Then
(1.16)
G = {|z| ≤ 1} ∪ {|z| ≥ 2} ⊂ U ∪ V
where U = {|z| < 3/2} and V = {|z| > 3/2}. Clearly, U and V are
open, U ∩ V = ∅, U ∩ G 6= ∅ and V ∩ G 6= ∅. Therefore, G is not
connected.
Corollary 1.22. The only sets that are both open and closed in C are
C and ∅.
Proof. Clearly, C is connected since it is convex. Suppose that there is
a set U 6= ∅, C that is both open and closed in C. Let V = U c = C\U .
Since U is closed, V is open. And since U 6= C, V 6= ∅. Therefore,
C ⊂ U ∪ V , U and V are open, U ∩ V = ∅, U 6= ∅ and V 6= ∅. It follows
from Theorem 1.20 that C is not connected. Contradiction.
In complex analysis, a connected open set G is called a region or
domain. Usually, we study complex functions defined on a region.
Every open set G ⊂ C is a disjoint union of regions:
[
(1.17)
G=
Gi
i
where each Gi is open and connected and Gi ∩ Gj = ∅ for i 6= j. Each
Gi is a connected component of G.