### Author Topic: why complex plane closed  (Read 1143 times)

#### caowenqi

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##### why complex plane closed
« on: October 01, 2020, 05:26:28 PM »
why the complex plane is both open and closed? why did it close? Because by definition, a set is called closed if it contains its boundary. And I don't see the complex plane contains its boundary.

#### Kuba Wernerowski

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##### Re: why complex plane closed
« Reply #1 on: October 01, 2020, 09:05:13 PM »
I used to be really confused by this until I really looked at the definitions of closed and open sets.

By definition, a set $S$ is closed if $S^c$ is open, and vice-versa.

The empty set $\emptyset$ trivially has nothing but interior points, and also trivially includes its boundary. This took me a while to internalize, but I think it's easiest to just ¯\_(ツ)_/¯  and accept it.

Since $\emptyset$ is open, then $\emptyset^c = \mathbb{C}$ is closed. And since $\emptyset$ is closed, then $\emptyset^c = \mathbb{C}$ is open.

Therefore, $\mathbb{C}$ is both open and closed.

#### Lubna Burki

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##### Re: why complex plane closed
« Reply #2 on: October 01, 2020, 09:37:19 PM »
You can also use the different properties to argue that both the empty set and complex plane are open and by complementarity both are closed. Specifically referring to those listed in JB's lecture as attatched,

(1) says that the set S is open iff the intersection between S and its boundary is the empty set. For the empty set itself, the intersection between the empty set and its boundary (also the empty set) is the empty set. So the empty set is open.

By (2), the complex plane is a neighborhood of every element in the complex plane by definition. so (2) is satisfied.

By (4), since the complement of the empty set is the entire plane, then the entire complex plane must also be closed. Vice versa to conclude that both the empty set and complex plane are open and closed at the same time.

#### Victor Ivrii

And what is the boundary of $\mathbb{C}$?