# Closed Set/Complex Analysis/Examples/Closed Unit Circle

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## Example of Closed Set in the context of Complex Analysis

Let $S$ be the subset of the complex plane defined as:

- $\cmod z \le 1$

where $\cmod z$ denotes the complex modulus of $z$.

Then $S$ is closed.

## Proof

By definition, $S$ is closed if and only if $S$ contains all its limit points.

Aiming for a contradiction, suppose $S$ is not closed.

Then $\exists z_1 \in \C: z_1 \notin S$ such that $z_1$ is a limit point of $S$.

We have that $z_1 \notin S$, so $\cmod {z_1} > 1$.

Let $\epsilon \in \R: \epsilon < \cmod {z_1} - 1$.

Then:

- $\map {\N_\epsilon} {z_1} \cap S = \O$

and so $z_1$ is not a limit point of $S$ after all.

Thus there is no such $z_1$.

So by Proof by Contradiction, $S$ is closed.

$\blacksquare$

## Sources

- 1981: Murray R. Spiegel:
*Theory and Problems of Complex Variables*(SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Point Sets: $3.$