# Complex Analysis: Proof that a straight line in Complex Plane is a connected set

• Sep 12th 2011, 06:46 PM
tarheelborn
Complex Analysis: Proof that a straight line in Complex Plane is a connected set
Prove that a straight line in the complex plane is a connected set.

I have come up with the following:

By the definition of connected sets, the line would have to contain two disjoint open sets, and by the definition of open set, a line would have to contain a neighborhood centered at z with radius r. But by definition of a line, it does not contain a neighborhood and, therefore, cannot be open and must be closed. So it must also be connected.

Does that make sense?
• Sep 13th 2011, 04:25 AM
Plato
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
Quote:

Originally Posted by tarheelborn
Prove that a straight line in the complex plane is a connected set. By the definition of connected sets, the line would have to contain two disjoint open sets, and by the definition of open set, a line would have to contain a neighborhood centered at z with radius r. But by definition of a line, it does not contain a neighborhood and, therefore, cannot be open and must be closed. So it must also be connected.

This question depends a great deal on how much topology your course assumes that you know. It seems to me that you a confused on some basic points. For example, in the complex plane a line cannot contain a open set.
If $a+bi~\&~c+di$ are to points on the line then the parametric equation $[a+t(c-a)]+i[b+(d-b)],~t\in\mathbb{R}$ is that line. So the line is the continuous image of the real numbers. Therefore any line is connected. But that proof depends on knowing a good bit of basic topology.
• Sep 13th 2011, 07:02 AM
tarheelborn
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
Topology is not required for my course. We have hit on a few basic concepts, but that is all. We have definitions of open and closed sets, as you would expect, and connected sets. Is there another way to prove it?
• Sep 13th 2011, 07:10 AM
Plato
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
Quote:

Originally Posted by tarheelborn
Topology is not required for my course. We have hit on a few basic concepts, but that is all. We have definitions of open and closed sets, as you would expect, and connected sets. Is there another way to prove it?

What is the exact definition of connect set from your text/notes?
• Sep 13th 2011, 07:24 AM
tarheelborn
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
The definition is that a set is connected if it is not disconnected. A set $E$ is called disconnected if there exist two disjoint open sets, $U$ and $V$, such that $E\subseteq U\cup V$, $E \cap U \neq\emptyset$, and $E \cap V \neq\emptyset$.
• Sep 13th 2011, 07:44 AM
Plato
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
Quote:

Originally Posted by tarheelborn
The definition is that a set is connected if it is not disconnected. A set $E$ is called disconnected if there exist two disjoint open sets, $U$ and $V$, such that $E\subseteq U\cup V$, $E \cap U \neq\emptyset$, and $E \cap V \neq\emptyset$.

Here is the way you would have to use that definition.
Suppose that a line, $\ell$, is disconnected.
Then use the two open sets $U~\&~V$ from the definition.
$\left( {\exists z_1 \in \ell \cap U} \right)\;\& \;\left( {\exists z_2 \in \ell \cap V} \right)$.
Every point between $z_1~\&~z_2$ is on the line $\ell$ so must belong to $U\text{ or }V$.
Is that possible? They are disjoint.
• Sep 13th 2011, 08:16 AM
tarheelborn
Re: Complex Analysis: Proof that a straight line in Complex Plane is a connected set
OK, I do see that. Thank you very much!