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mahler1
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Homework Statement .
Prove that a closed subset in a metric space ##(X,d)## is the boundary of an open subset if and only if it has empty interior.
The attempt at a solution.
I got stuck in both implications:
##\implies## Suppose ##F## is a closed subspace with ##F=\partial S## for some open subset ##S## of ##(X,d)##. I want to show that ##F^°=\emptyset##. So, let ##x \in F^°##. Then, there is ##\delta## : ##B(x,\delta) \subset F##. But then, ##B(x,\delta) \subset \partial S##, which means there is ##y \in B(x,\delta) \cap S^c## and there is ##z \in B(x,\delta) \cap S##. I couldn't go farther than this.
For the other implication, I need some help too. If ##F^°## is empty, then, for every ##x \in F## and every ##\delta>0##, ##B(x,\delta) \cap F^c \neq \emptyset##. As ##F## is closed, ##F^c## is open. I want to show that ##F=\partial F^c##, we've just seen that ##F \subset \partial F^c##. Now, let ##x \in \partial F^c##, then, for every ##n \in \mathbb N##, there is ##y_n \in B(x, \dfrac{1}{n}) \cap F##. But this means ##x## is a limit point of ##F##, as ##F## is closed ## \implies## ##x \in F##. I've proved ##\partial F^c \subset F##. This proves
##F=\partial F^c##, which means ##F## is the boundary of an open subset.
Could anyone suggest me how to prove the forward implication? Is the other one correct?
Prove that a closed subset in a metric space ##(X,d)## is the boundary of an open subset if and only if it has empty interior.
The attempt at a solution.
I got stuck in both implications:
##\implies## Suppose ##F## is a closed subspace with ##F=\partial S## for some open subset ##S## of ##(X,d)##. I want to show that ##F^°=\emptyset##. So, let ##x \in F^°##. Then, there is ##\delta## : ##B(x,\delta) \subset F##. But then, ##B(x,\delta) \subset \partial S##, which means there is ##y \in B(x,\delta) \cap S^c## and there is ##z \in B(x,\delta) \cap S##. I couldn't go farther than this.
For the other implication, I need some help too. If ##F^°## is empty, then, for every ##x \in F## and every ##\delta>0##, ##B(x,\delta) \cap F^c \neq \emptyset##. As ##F## is closed, ##F^c## is open. I want to show that ##F=\partial F^c##, we've just seen that ##F \subset \partial F^c##. Now, let ##x \in \partial F^c##, then, for every ##n \in \mathbb N##, there is ##y_n \in B(x, \dfrac{1}{n}) \cap F##. But this means ##x## is a limit point of ##F##, as ##F## is closed ## \implies## ##x \in F##. I've proved ##\partial F^c \subset F##. This proves
##F=\partial F^c##, which means ##F## is the boundary of an open subset.
Could anyone suggest me how to prove the forward implication? Is the other one correct?