1. ## metric spaces

Hello

2. The first one is straightforward. If $x\in\mathrm{Int}(A)\cup\mathrm{Int}(B)$, then x is in an open ball S contained in A or one contained in B; in either case, $x\in S\subseteq A\cup B$ and so $x\in\mathrm{Int}(A\cup B)$.

For the second part, proving $\mathrm{Int}(A\cap B)\subseteq \mathrm{Int}(A)\cap\mathrm{Int}(B)$ is equally straightforward. $x\in\mathrm{Int}(A\cap B)$ $\Rightarrow$ x is in an open ball S contained in $A\cap B$ $\Rightarrow$ $\Rightarrow$ $x\in S\subseteq A$ and $x\in S\subseteq B$ $\Rightarrow$ $x\in\mathrm{Int}(A)$ and $x\in\mathrm{Int}(B)$ $\Rightarrow$ $x\in\mathrm{Int}(A)\cap\mathrm{Int}(B)$.

For the reverse inclusion, let $x\in\mathrm{Int}(A)\cap\mathrm{Int}(B)$.

$x\in\mathrm{Int}(A)$ $\Rightarrow$ $\exists\,\delta_1>0$ such that $x\in S_1=\{y:\mathrm{d}(x,y)<\delta_1\}\subseteq A$. Similarly $x\in\mathrm{Int}(B)$. $\Rightarrow$ $\exists\,\delta_2>0$ such that $x\in S_2=\{y:\mathrm{d}(x,y)<\delta_2\}\subseteq B$.

Now take $\delta=\min\{\delta_1,\delta_2\}$.

Then $x\in S=\{y:\mathrm{d}(x,y)<\delta\}\subseteq S_1\cap S_2\subseteq A\cap B$ proving that $x\in \mathrm{Int}(A\cap B)$.

For the example, take the closed intervals $A=[0,1]$ and $B=[1,2]$ in $\mathbb{R}$. Then $\mathrm{Int}(A\cup B)=(0,2)$ but $\mathrm{Int}(A)=(0,1)$ and $\mathrm{Int}(B)=(1,2)$.

3. Thank you so much for your time and effort in replying to my query. I really am grateful.