Lecture 2: Elementary Topology on the Complex Plane

Open and closed sets

From what we have seen in the previous lecture, $\mathbb C$ is equivalent to the $xy$ -plane. We can therefore simply translate the usual topology we know for the $xy$ -plane to $\mathbb C$ . From the definition of modulus, it is clear that, for any two points $z\in\mathbb C$ and $\widetilde z\in\mathbb C$ , the number $|z-\widetilde z|$ is the “distance” between the two points.

Definition 1 (Open and Closed Disks). The open and closed disk centered at $z_0\in\mathbb C$ with radius $r$ are respectively the sets $\mathcal B_r(z_0)\equiv \{ z \in \mathbb C\mid | z-z_0| < r\} \qquad \mbox{and}\qquad \overline{\mathcal B}_r(z_0) \equiv \{ z \in\mathbb C\mid |z-z_0| \le r\}.$ The open disk $\mathcal B_r(z_0)$ is usually called the $r$ -neighborhood of $z_0$ .

Geometrically speaking, $\mathcal B_r(z_0)$ is simply the collection of the points strictly inside (that is, not on the boundary of) the disk of radius $r$ centered at $z_0$ , while $\overline{\mathcal B}_r(x)$ is simply the collection of the points inside or on the boundary of that disk.

Example 2. Let $z_0=(0, 1)$ be a point on on $\mathbb C$ , and $r=0.5$ . Then $\mathcal B_r(z_0)$ is the disk $\mathcal B_r(z_0) = \{ z \mid |z-(0, 1)|<0.5\}.$

Definition 3 (Open and Closed Sets). A set $\Omega \subseteq \mathbb C$ is open in $\mathbb C$ if, $\forall z \in\Omega$ , $\exists\varepsilon>0$ s.t. $\mathcal B_\varepsilon(z) \subseteq\Omega$ . A set $\Omega$ is closed if its complement $\Omega^c$ (relative to $\mathbb C$ ) is open.

Example 4. The set $\Omega=\{z\ |\ |z|<1\}$ is open.

Proof. Take any $z_0\in\Omega$ . Let $\varepsilon=(1-|z_0|) /2$ . Then $\mathcal B_\varepsilon(z_0)\subset\Omega$ . In fact, take any $z'\in \mathcal B_\varepsilon(z_0)$ , we have that $|z'|=|z'-z_0+z_0|\le |z'-z_0|+|z_0|<(1-|z_0|)/2+|z_0|=(1+|z_0|)/2<1$ . Therefore $z'\in\Omega$ . ◻

Example 5. The set $\Omega=\{z=x+iy\in\mathbb C\mid 0<y<1\}$ is open.

Proof. Take any $z=x+iy\in\Omega$ . Let $\varepsilon=\min\{y, 1-y\}/2$ . Then $\forall z'\in \mathcal B_\varepsilon(z)$ , $y/2<y'<3y/2$ if $y<1/2$ and $(3y-1)/2<y'<(y+1)/2$ if $y\ge 1/2$ . Therefore $\mathcal B_\varepsilon(z)\in\Omega$ . Therefore $\Omega$ is open. ◻

Exercise 6. Prove that the set $\Omega=\{z\ |\ |z|\le 1\}$ is closed by proving that $\Omega^c:=\{z\ |\ |z|>1\}$ is an open set.

The following result explains why $\mathcal B_r(z)$ is called an “open” disk while $\overline{\mathcal B}_r(z)$ is called a “closed” disk.

Lemma 7. For any $z\in\mathbb C$ and $r>0$ , the set $\mathcal B_r(z)$ is open, and the set $\overline{\mathcal B}_r(z)$ is closed.

Proof. (i) Let $z'\in \mathcal B_r(z)$ be an arbitrary point and define $\varepsilon=(r-|z-z'|)/2>0$ . Then $\forall z''\in \mathcal B_\varepsilon(z')$ , $|z''-z|\le |z''-z'|+|z'-z|<|z-z'|+\varepsilon=(r+|z-z'|)/2<r$ . This means that $z''\in \mathcal B_r(z)$ . Thus $\mathcal B_\varepsilon(z') \subseteq \mathcal B_r(z)$ ; (ii) Take any $z'\in \overline{\mathcal B}_r^c(z)$ , then $|z'-z|>r$ . Let $\varepsilon=(|z'-z|-r)/2>0$ . Then $\mathcal B_\varepsilon(z')\cap \overline{\mathcal B}_r(z)=\emptyset$ . Therefore $\mathcal B_\varepsilon(z')\subseteq\overline{\mathcal B}_r^c(z)$ . This means that $\overline{\mathcal B}_r^c(z)$ is open. Thus $\overline{\mathcal B}_r(z)$ is closed. Figure 2 shows the sets involved in the proof.

Open disk centered at z containing z-prime and a smaller disk centered at z-prime. Complement of a closed disk centered at z, with a small disk around z-prime entirely outside. ◻

Lemma 8 (Axiom of Open Sets). Then the sets $\emptyset$ and $\mathbb C$ are both open and closed.

Proof. (i) A set $\Omega\subseteq\mathbb C$ is open if every point in $\Omega$ has a $\varepsilon$ -neighborhood that is a subset of $\Omega$ . The empty set $\emptyset$ has no element in it, therefore, the statement “ $\emptyset$ is open” is vacuously true. Therefore $\mathbb C=\emptyset^c$ is closed. (ii) Since $\mathbb C$ contains every $z$ in itself, we can simply select $\varepsilon=1$ . Then the statement “ $z'\in \mathcal B_\varepsilon(z,z')$ implies $z'\in \mathbb C$ " holds for all $z\in\mathbb C$ . Therefore $\mathbb C$ is open. Therefore $\emptyset=\mathbb C^c$ is closed. ◻

Remark 9. Sets that are both open and closed are often called clopen set.

Lemma 10 (Union and Intersection of Open Sets). Let $\{O_k\}_{k=1}^\infty$ be a countable collection of open sets in $\mathbb C$ . Then (a) $\cap_{k=1}^N O_k$ is open for any finite $N<\infty$ ; (b) $\cup_{k=1}^\infty O_k$ is open.

Proof. (a) For any $z\in\cap_{k=1}^N O_k$ , it is known that $\exists \varepsilon_k>0$ such that $\mathcal B_{\varepsilon_k}(z)\subseteq O_k (1\le k\le N)$ . Take $\varepsilon=\min\{\varepsilon_1,\cdots,\varepsilon_N\}$ . Then $\mathcal B_\varepsilon(x)\subseteq \cap_{k=1}^N O_k$ . (b) follows directly from the definition. ◻

Lemma 11 (Union and Intersection of Closed Sets). Let $\{\widetilde{O}_k\}_{k=1}^\infty$ be a countable collection of closed sets in $\mathbb C$ . Then (a) $\cup_{k=1}^N \widetilde{O}_k$ is closed for any finite $N<\infty$ ; (b) $\cap_{k=1}^\infty \widetilde{O}_k$ is closed.

Proof. The results follow directly from Lemma Lemma 10 and De Morgan’s law. ◻

Example 12. The set $I_n=\{z\in\mathbb C\mid -\frac{1}{n}<\operatorname{Re}(z)<1,\ \ 0<\operatorname{Im}(z)<1\}$ is open for all $n$ . However, $\cap_{n=1}^\infty I_n=\{z\in\mathbb C\mid 0\le \operatorname{Re}(z)<1,\ \ 0<\operatorname{Im}(z)<1\}$ is not open. Thus an infinite intersection of open sets is not necessarily open.

Example 13. The set $D_n=\{z\in\mathbb C\mid |z|<\dfrac{1}{n}\}$ is open for all $n$ . However, $\cap_{n=1}^\infty D_n=\{0\}$ is not open (why?).

Point Sets on the Complex Plane

Definition 14 (Interior Point and Boundary Point). Let $\Omega$ be a subset of $\mathbb C$ . A point $z_0\in\Omega$ is said to be an interior point of $\Omega$ if we can find an $\varepsilon$ such that $\mathcal B_\varepsilon(z_0)$ is entirely inside $\Omega$ , that is, $\exists \varepsilon>0 \ \mbox{s.t.}\ \mathcal B_\varepsilon(z_0)\subseteq \Omega.$ A point $z_0$ on $\mathbb C$ is called a boundary point of $\Omega$ if every neighborhood of $z_0$ contains at least one point inside $\Omega$ and one point outside of $\Omega$ , that is $\forall \varepsilon>0,\ \exists z',\ z'' \in \mathcal B_\varepsilon(z_0)\ \mbox{s.t.}\ z'\in \Omega\ \mbox{and}\ z''\notin\Omega.$

This definition immediately leads to the following characterization of open sets in $\mathbb C$ .

Lemma 15. A set $\Omega\subseteq\mathbb C$ is open if and only if every point inside it is an interior point.

Example 16. Let $\Omega=\{z\ |\ |z|<1\}$ . Then $z_0=0$ is clearly an interior point. This is because we can find a $\varepsilon$ , for instance $\varepsilon=0.5$ such that $B_\varepsilon(z_0)\equiv \{z\ |\ |z|<\varepsilon\}$ is a subset of $\Omega$ . We can verify that $z_0=0.5+0.5i$ is also an interior point of $\Omega$ .

Example 17. Let $\Omega=\{z\ |\ |z|<1\}$ . Then the point $z_0=(0, i)$ is a boundary point of $\Omega$ .

Remark 18. In the above example, it is clear that $z_0=(0, i)$ is not an element of $\Omega$ since $|z_0-0|=1$ . Therefore, we see that a boundary point of a set $\Omega$ is not necessarily an element of $\Omega$ .

Example 19. Let $\Omega=\{z\ |\ |z|=1\}$ . Then every point of $\Omega$ is a boundary point. Let us prove this statement.

Proof. As illustrated in Figure 3, take any $z\in\Omega$ , then $|z|=1$ . For any $\varepsilon>0$ , take $z'=z+\dfrac{\varepsilon}{2|z|} z$ . Then $|z'-z|=\dfrac{\varepsilon}{2}$ . Therefore $z'\in B_\varepsilon(z)$ . However, $|z'-0|=1+\dfrac{\varepsilon}{2}>1$ . Therefore $z'\notin\Omega$ . Therefore $z$ is a boundary point. ◻

A point z on the unit circle and a nearby point z-prime outside it within a small disk. — The large circle represents $\{z\in\mathbb C\ |\ |z|=1\}$ . The point $z$ is on the large circle and $z'=z+\dfrac{\varepsilon}{2|z|} z$ .

Exercise 20. Verify that it also works if we take $z'=z-\dfrac{\varepsilon}{2|z|} z$ in the proof of the above example. In fact, any point of the form $z'=z\pm\dfrac{\varepsilon}{k|z|} z$ with $1<k<\infty$ would work.

Definition 21 (Limit Point, Point of Closure and Isolated Point). Let $\Omega$ be a subset of $\mathbb C$ . A point $z\in\mathbb C$ is called a limit point (also called an accumulation point) of $\Omega$ if $\forall \varepsilon>0$ , $\exists \widetilde z \neq z \in \mathcal B_\varepsilon(z)$ s.t. $\widetilde z\in \Omega$ . We denote by $\Omega'$ the set of limit points of $\Omega$ . A point $z\in\mathbb C$ is called a point of closure of $\Omega$ if $\forall\varepsilon>0$ , $\exists \widetilde z\in \mathcal B_\varepsilon(z)$ s.t. $\widetilde z\in\Omega$ . We denote by $\overline{\Omega}$ the closure of $\Omega$ , i.e. the set of points of closure of $\Omega$ . A point of closure of $\Omega$ that is not a limit point is called an isolated point.

Example 22. Let $\Omega=(0,\ 1)$ . It is easy to see that $0$ and $1$ are two limit points of $\Omega$ . However $0\notin \Omega$ and $1\notin \Omega$ . We know that in this case $\Omega'=[0,\ 1]$ . We know also that $0$ and $1$ are two points of closure of $\Omega$ and $\overline{\Omega}=[0,\ 1]$ .

Example 23. Let $\Omega=\{z\in\mathbb C\mid |z|<1\}$ . Then every point in $\Omega$ is a limit point. Moreover, every point in $\{z\in\mathbb C\mid |z|=1\}$ is a limit point. Therefore $\Omega'=\{z\in\mathbb C\mid |z|\le 1\}$ . It is also true that $\overline{\Omega}=\{ z\in\mathbb C\mid |z|\le 1\}$ . There is no isolated point for $\Omega$ .

Example 24. Let $\Omega=\{z\in\mathbb C\mid 0< \operatorname{Re}(z)<1\ \ \mbox{or}\ \ z=2+k\pi i,\ k\in\mathbb N\}$ . Then every point in $\{z\in\mathbb C\mid 0\le \operatorname{Re}(z) \le 1\}$ is a limit point of $\Omega$ . Every point in $\{z\in\mathbb C\mid z=2+k\pi i,\ k\in\mathbb N\}$ is an isolated point. Therefore, $\Omega'=\{z\in\mathbb C\mid 0\le \operatorname{Re}(z) \le 1\}$ , while $\overline\Omega=\{z\in\mathbb C\mid 0\le \operatorname{Re}(z) \le 1\ \ \mbox{or}\ \ z=2+k\pi i,\ k\in\mathbb N\}$ . The interior points of $\Omega$ form the set $\{ z\in\mathbb C\mid 0<\operatorname{Re}(z) <1\}$ . The boundary points of $\Omega$ form the set $\{ z\in\mathbb C\mid \operatorname{Re}(z)=0, \ \mbox{or}\ \operatorname{Re}(z)=1,\ \mbox{or}\ z=2+k\pi i,\ k\in\mathbb N\}$ .

Remark 25. The main difference between a limit point and a point of closure is that the open disk centered at a limit point needs to contain at least one point of $\Omega$ that is OTHER THAN the limit point itself. By definition, every limit point is a point of closure but not vice versa. A point $z$ is an isolated point of $\Omega$ if it is an element of $\Omega$ and any open disk centered at it contains no element of $\Omega$ OTHER THAN itself.

It is easy to observe from the definition of limit points that a closet set must contain all of its limit points.

Theorem 26 (Closed Sets Contains All Limit Points). A set $\Omega\subseteq\mathbb C$ is closed iff $\Omega' \subseteq \Omega$ .

Proof. (i) If $\Omega$ is closed, then $\Omega^c$ is open. Take any $z\in \Omega^c$ , $\exists \varepsilon>0$ s.t. $B_\varepsilon(z)\subseteq \Omega^c$ and thus $B_\varepsilon(z)\cap \Omega=\emptyset$ . Thus $z$ is not a limit point of $\Omega$ . Therefore $\Omega^c$ contains no limit points of $\Omega$ . Therefore $\Omega$ contains all its limit points, i.e. $\Omega'\subseteq \Omega$ ; (ii) If $\Omega$ contains all its limit points, then $\forall z\in \Omega^c$ , $z$ is NOT a limit point of $\Omega$ . Therefore, $\exists \varepsilon>0$ s.t. $\forall z'\in B_\varepsilon(z)$ , $z'\in \Omega^c$ , i.e. $B_\varepsilon(z)\subseteq \Omega^c$ . Therefore, $\Omega^c$ is open. This means that $\Omega$ is closed. ◻

Compactness and connectedness

A set $\Omega\subseteq\mathbb C$ is bounded if it can be enclosed in a circle of finite radius.

Definition 27 (Boundedness). A set $\Omega\subseteq\mathbb C$ is said to be bounded if there exists $M>0$ such that for all $z\in\Omega$ , we have that $|z|\le M$ .

Example 28. The set $\Omega=\{ z\ |\ 1<|z|<2\}$ is bounded since for any $z\in\Omega$ , we have $|z|\le 2$ .

Example 29. The set $\Omega=\{ z\ |\ |z-(0.5, 0.5)|\le 3\}$ is bounded since for any $z\in\Omega$ , we have $|z-(0.5, 0.5)|\le 3$ . By the Triangle Inequality, $|z|-|(0.5, 0.5)|\le |z-(0.5, 0.5)|$ . Therefore $|z|\le \widetilde M\equiv |(0.5, 0.5)|+3=3+\sqrt{2}/2$ .

Example 30. The set $\Omega=\{z\in\mathbb C\mid 1<\operatorname{Re}(z)<2\}$ is not bounded.

Proof. (By Contradiction) Assume that $\Omega$ is bounded. Then $\exists\ M>0$ such that $\forall z\in\Omega$ , $|z|\le M$ . Take $z=1.5+M\, i$ . It is clear that $z \in\Omega$ . However, $|z|=\sqrt{1.5^2+M^2} >M$ . We therefore have a contradiction. ◻

Example 31. The set $\Omega=\mathbb C$ is not bounded.

Definition 32 (Compactness). A set $\Omega\subseteq\mathbb C$ is said to be compact if it is closed and bounded.

Example 33. The set $\Omega=\{ z\in\mathbb C\mid 1\le |z|\le 2\}$ is compact.

Example 34. The set $\Omega=\{z\in\mathbb C\mid 1\le \operatorname{Re}(z)\le 2\ \mbox{and}\ 1\le \operatorname{Im}(z)\le 2\}$ is compact.

Example 35. The set $\Omega=\{z\in\mathbb C\mid 1< \operatorname{Re}(z) < 2\ \mbox{and}\ 1\le \operatorname{Im}(z)\le 2\}$ is not compact because it is not closed.

Example 36. The set $\Omega=\{z\in\mathbb C\mid 1\le \operatorname{Re}(z)\le 2\}$ is not compact because it is unbounded.

Remark 37. Compactness is an important property for a set to have. First, for any convergent sequence in the set, it converges to an element in the set because the set is closed (therefore it contains limits of any convergent sequence in it). Second, because the set is bounded, by the Bolzano-Weierstrass theorem (which says every bounded sequence in $\mathbb R^d$ ( $d\ge 1$ ) has a convergent subsequence), every sequence in the set has a convergent subsequence. This property has important consequence when we study functions defined on compact sets.

Definition 38 (Connectedness). An open region is said to be connected if for any two points $z$ and $z'$ in the region, there exists $\{z_1, z_2, \cdots, z_n\}$ such that the line segments $\overrightarrow{zz_1}$ , $\overrightarrow{z_1z_2}$ , $\cdots$ , $\overrightarrow{z_{n-1}z_n}$ , $\overrightarrow{z_nz'}$ are all inside the region.

Example 39. The region $\Omega=\{ z\ |\ 1\le |z|\le 2\}$ is connected.

Example 40. The region $\Omega=\{z\ |\ 1\le \operatorname{Re}(z)\le 2, \ 2<\operatorname{Im}(z)\ \mbox{or}\ \operatorname{Im}(z)<1\}$ is not connected.

In complex analysis, we often call a connected nonempty open set a domain.

Guided review and additional examples

Learning goals

Translate neighborhood language into inequalities involving $|z-z_0|$ .
Distinguish open, closed, bounded, connected, and compact sets.
Identify interior, boundary, isolated, and limit points.
Use sequences to test whether a set contains all of its limit points.

Concept connection: local tests determine global labels

Openness is tested from points inside a set: every point must have a sufficiently small disk that stays in the set. Closedness is tested by limits: every convergent sequence from the set must converge to a point in the set. A set can be both open and closed, or neither; “not open” does not mean “closed.”

Worked example 1: a half-closed annulus

Let $A=\{z\in\mathbb C:1<|z|\le2\}$ . This set is not open because a point with $|z|=2$ has no open disk contained in $A$ . It is not closed because points with $|z|=1$ are limit points of $A$ but do not belong to $A$ . Its interior is $\{1<|z|<2\}$ , and its boundary is the union of the circles $|z|=1$ and $|z|=2$ .

Worked example 2: a sequence viewed as a set

Let $S=\{1/n:n\in\mathbb N\}\subset\mathbb C$ . Every point $1/n$ is isolated: a small enough disk around it contains no other point of $S$ . The only limit point is $0$ , since $1/n\to0$ . Therefore

$S'=\{0\},\qquad \overline S=S\cup\{0\}.$

The set $S$ is bounded but not closed, so it is not compact. Adding $0$ makes it closed and bounded, hence compact in $\mathbb C$ .

Check your understanding

Is $\mathbb C\setminus\overline{\mathcal B}_1(0)$ open, closed, or both?

Show the answer

It is open: if $|z|>1$ , choose a radius smaller than $|z|-1$ to obtain a disk that remains outside the closed unit disk. It is not closed because points on $|z|=1$ are limit points that are missing from the set.