Closure and Limit Points

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Introduction

Definition. Let D ⊆ ℂ be a set of complex numbers. A complex-valued function f of a complex variable defined on D is a rule that assigns to each complex number z belonging to the set D a unique complex number w, $f: D \to \mathbb{C}$.

• Domain D: A subset of $\mathbb{C}$ on which the function is defined. Every $z \in D$ is called a point of the domain.
• Function $f: D \to \mathbb{C}$: A rule that assigns to each $z \in D$ a unique complex number w = f(z).
• Range (or image) f(D): The set of all actual outputs {f(z): z ∈ D}.
• Decomposition into real and imaginary parts: If z = x+ iy ∈ D, then f(z) is called the image of the point z under f. We often write f(z) = u(x ,y) + iv(x, y), where u, v: ℝ² → ℝ are the real and imaginary parts (real‑valued functions of two real variables).

Closure

• Proposition. For any $S \subseteq \mathbb{C}$, its closure $\bar{S}$ is a closed set.

Recall: The closure is defined as $\bar{S} = S \cup \partial S$, where $\partial S$ is the boundary of $S$. Equivalently, $\bar{S}$ is $S$ together with all its limit points. A set S is closed iff its complement $\Complex - \bar{S}$ is open.

Proof

Idea (strategy): Show that $\mathbb{C} \setminus \bar{S}$ is open.

1. Let $z \in \mathbb{C} \setminus \bar{S}$ be arbitrary. By definition of closure, $\bar{S} = S \cup \partial S$. Since $z \notin \bar{S}$, then $z \notin S$ and $z \notin \partial S$. In particular, z is not a limit point of S (boundary points are either isolated (if in S) or limit points).

   A point $z_0$ is a limit point of S if every neighborhood of $z_0$ intersects S in at least one point other than $z_0$. A point $w \in \partial S$ if every neighborhood of w intersects both S and $ \mathbb{C} \setminus \bar{S}$.

2. Since z is not a limit point of S, $\exists \varepsilon > 0 : B(z; \varepsilon) \cap S = \emptyset \quad (\star)$.
3. We claim $B(z; \varepsilon) \cap \bar{S} = \emptyset$.
   First, observe: $B(z; \varepsilon) \cap \bar{S} = B(z; \varepsilon) \cap (S \cup \partial S)$
   $= (B(z; \varepsilon) \cap S) \cup (B(z; \varepsilon) \cap \partial S)$
   $= \emptyset \cup (B(z; \varepsilon) \cap \partial S) \quad \text{(by } (\star))$
   $= B(z; \varepsilon) \cap \partial S$
4. We want to show that $B(z; \varepsilon) \cap \partial S = \emptyset$ by contradiction.
   Suppose $w \in B(z; \varepsilon) \cap \partial S$.
   Since $w \in \partial S$, every neighborhood of $w$ intersects $S$.
   Choose $\delta = \varepsilon - |w - z| > 0$ (since $w \in B(z; \varepsilon)$ means $|w - z| < \varepsilon$).
   Then $B(w; \delta) \subseteq B(z; \varepsilon)$ by the triangle inequality: $w' \in B(w; \delta), | w' - z | \le | w' - w | + |w - z| \lt \delta + |w - z| = \varepsilon - |w - z| + |w - z| = \varepsilon $
   Since $w \in \partial S$, every neighborhood of w intersects S, $B(w; \delta) \cap S \neq \emptyset$.
   But $B(w; \delta) \subseteq B(z; \varepsilon)$, so $B(z; \varepsilon) \cap S \neq \emptyset$. This contradicts $(\star)$! $\Rightarrow\Leftarrow$
5. Therefore $B(z; \varepsilon) \cap \bar{S} = \emptyset$, which means: $B(z; \varepsilon) \subseteq \mathbb{C} \setminus \bar{S}$
6. Since $z$ was arbitrary: $\forall z \in \mathbb{C} \setminus \bar{S}, \exists \varepsilon > 0 : B(z; \varepsilon) \subseteq \mathbb{C} \setminus \bar{S}$. This is the definition of $\mathbb{C} \setminus \bar{S}$ being open. Therefore, $\bar{S}$ is closed. $\blacksquare$

Characterizations of Closed Sets. Let $S \subseteq \mathbb{C}$. The following are equivalent:

1. $S$ is closed in $\mathbb{C}$.
2. S contains all its limit points.
3. $\bar{S} = S$.

Proof

(1) $\Rightarrow$ (2): Closed implies contains limit points.

Suppose $S$ is closed. Then, $\mathbb{C} \setminus S$ is open.

Claim: Every limit point of $S$ is in $S$.

Suppose for contradiction that $z$ is a limit point of $S$ but $z \notin S$. Then, $z \in \mathbb{C} \setminus S$.

Since $\mathbb{C} \setminus S$ is open: $\exists \varepsilon > 0 : B(z; \varepsilon) \subseteq \mathbb{C} \setminus S \implies B(z; \varepsilon) \cap S = \emptyset$. However, $z$ being a limit point requires every neighborhood to meet $S$. Contradiction! $\blacksquare$

(2) $\Rightarrow$ (3): Contains limit points implies set equals closure

By definition: $\bar{S} = S \cup \partial S = S \cup \{\text{limit points of } S\}$ (boundary points are either isolated (if in S) or limit points).

The closure of S is the union of S and its boundary points $\bar{S} = S \cup \partial S$. Alternatively, it can be defined as the union of S and its limit points: $\bar{S} = S \cup \{\text{limit points of } S\}$

If $S$ contains all its limit points, then: $\{\text{limit points of } S\} \subseteq S$. Therefore, $\bar{S} = S \cup \{\text{limit points of } S\} = S$ $\blacksquare$

(3) $\Rightarrow$ (1): Set equals closure implies closed

If $\bar{S} = S$, and we proved that $\bar{S}$ is always closed (Previous proposition, for any $S \subseteq \mathbb{C}$, its closure $\bar{S}$ is a closed set), then the set itself $S$ is closed. $\blacksquare$

Limit Points — Examples and Exercises

• Unit Circle as Limit Points. $S = \{z \in \mathbb{C} : |z| < 1\}$ (open unit disk). Every point with $|z| = 1$ is a limit point of S.

Let $z_0$ with $|z_0| = 1$. For any $\varepsilon > 0$, consider the point $w = \left(1 - \frac{\varepsilon}{2}\right)z_0$.

Then, $|w| = |1 - \frac{\varepsilon}{2}|\cdot |z_0| = 1 - \frac{\varepsilon}{2} < 1$, so $w \in S$ and $|w - z_0| = \left|\left(1 - \frac{\varepsilon}{2}\right)z_0 - z_0\right| = \frac{\varepsilon}{2}|z_0| = \frac{\varepsilon}{2} < \varepsilon$

So $w \in B(z_0; \varepsilon) \cap S$ and $w \neq z_0$. Every neighborhood of $z_0$ contains points of $S$ w ($w \ne z_0$), so $z_0$ is a limit point. $\blacksquare$

• $S = \{z \in \mathbb{C} : |z| < 1\} \cup \{2 + 7i\}$. The point 2 + 7i is an isolated point of S.

Proof:

The distance from $2 + 7i$ to the unit disk is: $|2 + 7i| - 1 = \sqrt{4 + 49} - 1 = \sqrt{53} - 1 \approx 6.28$

Choose $\varepsilon = \sqrt{53} - 1 > 0$. Then, $B(2 + 7i; \varepsilon) \cap S = \{2 + 7i\}$.

The only point of S in this neighborhood is $2 + 7i$ itself, making it isolated. $\blacksquare$

• Rationals are Dense in $\mathbb{C}$. Let $S = \{x + iy \in \mathbb{C} : x, y \in \mathbb{Q}\}$. Every complex number is a limit point of S. In other words, the complex numbers with rational coordinates are dense in $\mathbb{C}$.

Proof:

Let $z = x + iy \in \mathbb{C}$ and let $\varepsilon > 0$.

Goal: Find $w \in S$ with $|w - z| < \varepsilon$ and $w \neq z$.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exist rationals: $x_0 \in \mathbb{Q} : x < x_0 < x + \frac{\varepsilon}{\sqrt{2}}, y_0 \in \mathbb{Q} : y < y_0 < y + \frac{\varepsilon}{\sqrt{2}}$ (Figure B)

Define $w = x_0 + iy_0, w \in S$.

Then, $|z - w| = \sqrt{(x - x_0)^2 + (y - y_0)^2} < \sqrt{\left(\frac{\varepsilon}{\sqrt{2}}\right)^2 + \left(\frac{\varepsilon}{\sqrt{2}}\right)^2} = \sqrt{\frac{\varepsilon^2}{2} + \frac{\varepsilon^2}{2}} = \varepsilon$. Hence, $w \in B(z; \varepsilon) \cap S$.

Since this works for arbitrary $z$ and $\varepsilon$, every complex number is a limit point of $S$, $\bar{S} = \mathbb{C}$ $\blacksquare$

• A set with exactly one limit point. $S = \left\{\frac{1}{k} + i\frac{2}{k} : k \in \mathbb{Z} \setminus \{0\}\right\}$ has exactly one limit point, namely 0.

Part 1. $0$ is a limit point of $S$

Proof:

For any $\varepsilon > 0$: $\left|\frac{1}{k} + i\frac{2}{k}\right| = \sqrt{\frac{1}{k^2} + \frac{4}{k^2}} = \frac{\sqrt{5}}{|k|}$. As $|k| \to \infty$, this approaches $0$.

Choose $N > \frac{\sqrt{5}}{\varepsilon}$. Then. $\forall |k| > N$: $\left|\frac{1}{k} + i\frac{2}{k}\right| = \frac{\sqrt{5}}{|k|} \lt \frac{\sqrt{5}}{N} \lt \sqrt{5}\frac{\varepsilon}{\sqrt{5}} = \varepsilon$

By the Archimedean property, there always exists an integer $N$ such that $N > \frac{\sqrt{5}}{\varepsilon}$.

So for any $\varepsilon > 0$, infinitely many points of $S$ lie in $B(0; \varepsilon)$. Therefore, $0$ is a limit point $\blacksquare$.

Part 2: No other point is a limit point

We must show that for any $z \in \mathbb{C}$ where $z \neq 0$, there exists a neighborhood around $z$ that contains no points of $S$ (or no other points, if $z \in S$ itself).

Observe that all points in $S$ lie on the line $L$ passing through the origin with slope $2$ (i.e., $y = 2x$, since $\frac{2/k}{1/k} = 2$).

Case 1: z is not on the line $L = \{w : \text{Im}(w) = 2\text{Re}(w)\}$, $z = x + iy$

If $z$ is not on the line $y = 2x$, there is a positive distance between $z$ and the line $L$. Let this perpendicular distance be $d = \text{dist}(z, L) > 0$ (Figure C).

Choose $\varepsilon_z = d/2$. Then, $B(z; \varepsilon_z) \cap S = \emptyset$ (the disk $B(z; \varepsilon)$ does not touch the line L. Since all points of $S$ live on $L$, this disk contains no points of S). Hence, $z$ is not a limit point.

Case 2: $z$ is on the line $L$, but outside the range of S. The real parts of points in S are contained in the interval $[-1, 1]$. Suppose $z = x + 2ix$ with $|x| > 1$.

The closest points of $S$ on $L$ are at $x = \pm 1$, i.e., k = 1 (1 + 2i) and k = -1 (-1 - 2i).

Choose radius $\varepsilon_z = |x| - 1 > 0$. Then, $B(z; \varepsilon_z) \cap S = \emptyset$.

Case 3: Suppose $z$ lies on the line segment between two consecutive points of $S$. $z = x + 2ix$ on line $L$, with $|x| \leq 1$, and $z \notin S$

There exists $k \in \mathbb{Z}$ such that $\frac{1}{k+1} < x < \frac{1}{k}$ (or similar for negative $x$).

Choose radius. Let $\varepsilon$ be the distance to the nearest neighbor: $\varepsilon = \min \left( \left|x - \frac{1}{k}\right|, \left|x - \frac{1}{k+1}\right| \right) \cdot \sqrt{5}$ (Note: The $\sqrt{5}$ factor accounts for the distance along the slanted line $L$ vs just the x-axis ($\star$). If we choose $\varepsilon = d$, the neighborhood is the open ball $B(z; d) = \{w : |w - z| < d\}$, the nearest neighbor is at distance equal to $d$. Since $d$ is not strictly less than $d$, the neighbor falls on the boundary of the ball, not inside it). We can still leave a “buffer zone”, $\varepsilon = \min \left( \left|x - \frac{1}{k}\right|, \left|x - \frac{1}{k+1}\right| \right) \cdot \frac{sqrt{5}}{2}$.

Then, this radius fits exactly in the gap and $B(z; \varepsilon_z) \cap S = \emptyset$ (the disk $B(z; \varepsilon)$ contains no points of $S$).

Case 4: $z \in S, z = z_k$ (isolated points), say $z = \frac{1}{k} + i\frac{2}{k}$

The neighbors of $z_k$ are $z_{k+1}$ and $z_{k-1}, \frac{1}{k\pm 1}$.

The distance between consecutive points on $L$ is: $|z_k - z_{k+1}| = \left| \left(\frac{1}{k} - \frac{1}{k+1}\right) + 2i\left(\frac{1}{k} - \frac{1}{k+1}\right) \right| = \frac{\sqrt{5}}{k(k+1)} (\star)$

Choose radius, let $\varepsilon_z = \frac{\sqrt{5}}{2k(k+1)}$.

Then, the punctured neighborhood $B'(z_k; \varepsilon)$ contains no points of S, $B'(z; \varepsilon_z) \cap S = \emptyset$. So $z$ is isolated, not a limit point (of $S \setminus \{z\}$).

Conclusion: $0$ is the only limit point, and the closure of S is: $\bar{S} = S \cup \{0\}$ $\blacksquare$