# boundary of nonempty set in complex plane

Let us take complex plane to be the space.

In complex plane, the boundary of a set is defined as $\partial D = \overline{D} \cap (\overline{\mathbb{C}\setminus D})$.

I want to prove such assertion holds:

if $D\neq \emptyset$ and $D\neq \mathbb{C}$, then $\partial D \neq \emptyset$.

Intuitively it is straightforward, however I failed to find a formal proof.

I suspect it should be proved by contradiction and using the property of complex plane.

Anyone help me to formulate the proof?

Cheers!

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Recall that $\mathbb{C}$ is connected. If some nonempty subset $D\neq \mathbb{C}$ had empty boundary, then by your definition $\emptyset = \overline{D}\cap\overline{\mathbb{C}-D}$. Now argue that $\mathbb{C}=\overline{D}\cup\overline{\mathbb{C}-D}$. This is a union of disjoint sets. Both sets are clopen (to see, recall that the closure of a set is closed and complements of closed sets are open). Which property of $\mathbb{C}$ does this contradict?