$\newcommand{\Int}{\operatorname{Int}}\newcommand{\Bdy}{\operatorname{Bdy}}$ If $A$ and $B$ are sets in a metric space, show that: (note that $\Int$ stands for interior of the set)
- $\Int (A) \cup \Int (B) \subset \Int (A \cup B)$.
- $(\overline{ A \cup B}) = (\overline A \cup \overline B )$. (note that $\overline A = \Int (A) \cup \Bdy(A)$ )
Now for the first (1) I see why its true for instance in $R$ we can have the intervals set $A=[a,b]$ and $B=[b,c]$ we have $A \cup B=[a,c]$ so $\Int(A \cup B)=(a,c)$ now $\Int(A)=(a,b)$ and $\Int(B)=(b,c)$ so we lose $b$ when we take union to form $\Int(A) \cup \Int(B)=(a,b) \cup (b,c)$.