📂Complex Anaylsis

Non-Openness of the Real Axis in the Complex Plane

Theorem

$$\mathbb{R}^{\circ} \ne \mathbb{R} \subset \mathbb{C}$$

The real axis $\mathbb{R} \subset \mathbb{C}$ is not open in the complex plane $\mathbb{C}$.

Explanation

Therefore, the real axis is not a complex domain in the complex space. Considering that from $\mathbb{R}$ to $\mathbb{R}$ is open, this might seem somewhat contrary to intuition. This is essentially because the concept of being open or closed in a set is a relative concept that changes depending on the given topological space.

This fact is important because it can explain most reasons why the properties of complex functions, which can actually be defined even for real functions, change. Many theorems in complex analysis require a complex domain, i.e., an opened complex subset $\mathscr{R} \subset \mathbb{C}$, as a condition. If there is no restriction of being open, it can be assumed as $\mathscr{R} = \mathbb{R}$, which can cause many problems if left unchecked.

The following proof suggests that “simply being a subset of the complex space does not make it a true complex domain,” and “naturally, $\mathbb{R}$ cannot be considered as a complex space,” rigorously justifying this intuition.

Proof

Definition of open in complex analysis: Let’s say $\alpha \in \mathbb{C}$ and $\delta > 0$ and $S \subset \mathbb{C}$.

• A set like the following is called an Open Neighborhood or Open Ball of $\alpha$. $$B \left( \alpha ; \delta \right) := \left\{ z \in \mathbb{C} : \left| z - \alpha \right| < \delta \right\}$$
• If any open ball of $\alpha$ is included in $S$, then $\alpha$ is called an Interior Point of $S$. $$\exist \delta : B \left( \alpha , \delta \right) \subset S$$
• If all points of $S$ are interior points of $S$, then $S$ is said to be Open, and if $S$ includes all the limit points of $S$, it is said to be Closed.

It is sufficient to show that even a single point in $\mathbb{R}$ is not an interior point. Without loss of generality, if we set that point as $\alpha = 0$, then its open ball $B (0, \delta)$, regardless of the radius $\delta > 0$ chosen, will necessarily include a non-real complex number $z \in \mathbb{C} \setminus \mathbb{R}$, specifically to the extent of $$z_{0} = 0 + i {{ \delta } \over { 2 }}$$ Therefore, $0 \in \mathbb{R}$ is not an interior point of $\mathbb{R}$, and since all points in $\mathbb{R}$ are not interior points in $\mathbb{C}$, $\mathbb{R}$ is not open in $\mathbb{C}$.

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Generalization

Curves $\gamma \subset \mathbb{C}^{n}$ are not open in multidimensional complex space $\mathbb{C}^{n}$.

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From the discussion during the proof, it can be understood that unless it is $\mathbb{R}^{1}$, essentially any point on a thin curve will inevitably include surrounding points, indicating that the theorem holds even outside the real axis and complex plane.