- #1
xman
- 93
- 0
Hello,
I am currently working on proving the following theorem
The boundary [tex] \partial A [/tex] and the closure [tex] \overline{A} [/tex] of a subset A of [tex] \mathbb C[/tex] are closed sets.
Proof: Let [tex] A \subset \mathbb C.[/tex] We want to show the set [tex] \partial A \cap \overline{A} [/tex] is closed. To show that [tex] \partial A \cap \overline{A}[/tex] is closed, we will show that the complement of [tex] \partial A \cap \overline{A}[/tex] is open. So the complement of [tex]\partial A \cap \overline{A}[/tex] is the set of all points not in [tex] \partial A \cap \overline{A},[/tex] i.e. [tex] \mathbb C \sim (\partial A \cap \overline{A} ).[/tex] Since [tex] \overline{A} = A\cup \partial S[/tex] and the set [tex]\partial A \cap \overline{A}[/tex] contains all of its boundary points by definition of the intersection. So the complement of [tex] \partial A \cap \overline{A}[/tex] cannot contain any of its boundary points by the definition of the complement. Since the complement does not contain any points on its boundary, then the complement is open. Therefore, since the complement of [tex]\partial A \cap \overline{A}[/tex] is open then the set [tex]\partial A \cap \overline{A}[/tex] is closed.
I am really poor at doing proofs, any help, insight, or questions would be greatly appreciated. Please let me know. Thanks.
I am currently working on proving the following theorem
The boundary [tex] \partial A [/tex] and the closure [tex] \overline{A} [/tex] of a subset A of [tex] \mathbb C[/tex] are closed sets.
Proof: Let [tex] A \subset \mathbb C.[/tex] We want to show the set [tex] \partial A \cap \overline{A} [/tex] is closed. To show that [tex] \partial A \cap \overline{A}[/tex] is closed, we will show that the complement of [tex] \partial A \cap \overline{A}[/tex] is open. So the complement of [tex]\partial A \cap \overline{A}[/tex] is the set of all points not in [tex] \partial A \cap \overline{A},[/tex] i.e. [tex] \mathbb C \sim (\partial A \cap \overline{A} ).[/tex] Since [tex] \overline{A} = A\cup \partial S[/tex] and the set [tex]\partial A \cap \overline{A}[/tex] contains all of its boundary points by definition of the intersection. So the complement of [tex] \partial A \cap \overline{A}[/tex] cannot contain any of its boundary points by the definition of the complement. Since the complement does not contain any points on its boundary, then the complement is open. Therefore, since the complement of [tex]\partial A \cap \overline{A}[/tex] is open then the set [tex]\partial A \cap \overline{A}[/tex] is closed.
I am really poor at doing proofs, any help, insight, or questions would be greatly appreciated. Please let me know. Thanks.