Definition: Let $(X,d)$ be a metric space and $S \subseteq X$. A point $s \in S$ is said to be a boundary point if $$ \forall r > 0, B_r(s) \cap S \neq \emptyset \text{ and } B_r(s) \cap S^c \neq \emptyset$$ where $B_r(s) = \{y \in X: d(s,y) < r\}$ is the open ball of radius $r$ centered at $s$. The collection of boundary points of $S$ is denoted $\partial S$ and the closure of $S$ is $\overline{S} = S \cup \partial S$. If $A \subseteq B \subseteq X$, prove that $\overline{A} \subseteq \overline{B}$
My Attempt:
Let $x \in \overline{A}$, then $x \in A$ or $x \in \partial A$.
If $x \in A$, then $x \in B$ and $x \in \overline{B}.$ If $x \in \partial A$, then $\forall r >0$
$$ B_r(x) \cap A \neq \emptyset \text{ and } B_r(x) \cap A^c \neq \emptyset$$
Since $A \subseteq B$ we have that $B_r(x) \cap B \neq \emptyset$. I'm having trouble showing that $B_r(x) \cap B^c \neq \emptyset$. Any help would be greatly appreciated.
Best Answer
It suffices to show that $\partial A \subseteq \overline{B}$, as you noted. Suppose we have $x \in \partial A$. This tells us two things.
We will split the proof into 2 cases.
Case 1: there exists some $r > 0$ s.t. $B_r(x) \subseteq B$. In this case, we have in particular that $x \in B_r(x) \subseteq B \subseteq \overline{B}$.
Case 2: For every $r > 0$, $B_r(x) \nsubseteq B$. That is, for every $r > 0$, we have some $y \in B_r(x)$ s.t. $y \notin B$. That is, for every $r > 0$, we have $B_r(x) \cap B^c \neq \emptyset$. And we also have for all $r > 0$, $B_r(x) \cap B \supseteq B_r(x) \cap A \neq \emptyset$. Then $x \in \partial B \subseteq \overline{B}$.