autre said:Homework Statement
Show by example that an infinite union of closed sets in [itex]\mathbb{C}[/itex] need not be closed.
The Attempt at a Solution
In [itex]\mathbb{R}[/itex] I know that an infinite union of the closed sets [itex]A_{n}=[1/n,1-1/n][/itex] is open. Not sure if it works in [itex]\mathbb{C}[/itex] as well.
A closed set in complex numbers is a set that contains all of its boundary points. This means that every sequence of points in the set converges to a point within the set itself.
To show that a set in complex numbers is not closed, you can provide an example of a sequence of points in the set that converges to a point outside of the set. This demonstrates that the set does not contain all of its boundary points.
No, a set in complex numbers cannot be both open and closed. A set is open if it does not contain any of its boundary points, while a set is closed if it contains all of its boundary points. Therefore, a set cannot have both of these properties at the same time.
Yes, all subsets of a closed set in complex numbers are also closed. This is because a subset of a closed set will also contain all of its boundary points, making it a closed set itself.
Understanding closed sets in complex numbers is important in various areas of mathematics, such as analysis and topology. It helps in proving the convergence and continuity of functions, as well as in understanding the behavior of complex sequences and series. Additionally, closed sets play a crucial role in the definition of compactness and connectedness in complex spaces.