Closures and Interiors.

Question

Let A ⊆ X be a subset of a metric space. Prove the following:

(a) $\overline{A^c}= (\mathring{A})^c$;

(b) $\mathring{A^c}=(\overline{A})^c$.

Answer 1

$\overline{A}$ is the set of elements of $X$ such that any open ball around them intersects $A$, that is, $$\overline{A} = \{ x \in X \mid \forall \epsilon > 0, B(x,\epsilon) \cap A \neq \emptyset \}.$$ Similarly, one definition of interior is the set of elements of $X$ such that you can find an $\epsilon$-ball around them contained in $A$, that is,$$\mathring{A} = \{ x \in X \mid \exists \epsilon > 0 \text{ s.t. } B(x,\epsilon) \subseteq A \}.$$

Now, to prove the first claim, first we will show $\overline{A^{c}} \subseteq (\mathring{A})^{c}:$

Let $x \in \overline{A^{c}}$. Then for every $\epsilon > 0$, $B(x, \epsilon) \cap A^{c} \neq \emptyset$. But this means any ball around $x$ will intersect with $A^{c}$, which means you can never find a ball around $x$ that is contained in $A$. That means $x \not \in \mathring{A}$. Thus, $x \not \in \mathring{A} \implies x \in (\mathring{A})^{c}$.

Now to show $(\mathring{A})^{c} \subseteq \overline{A^{c}}:$

Let $x \in (\mathring{A})^{c}$. Then $x \not \in \mathring{A}$. But this means there does not exist $\epsilon > 0$ such that $B(x,\epsilon) \subseteq A$. But this means for every $\epsilon > 0$, $B(x,\epsilon) \cap A^{c} \neq \emptyset$, and this is precisely what we need for $x \in \overline{A^{c}}$. So we are done.

Can you do the second claim now?