# metric spaces

• Jan 27th 2008, 09:04 AM
rolylane
metric spaces
Hello
• Jan 27th 2008, 12:41 PM
JaneBennet
The first one is straightforward. If $\displaystyle x\in\mathrm{Int}(A)\cup\mathrm{Int}(B)$, then x is in an open ball S contained in A or one contained in B; in either case, $\displaystyle x\in S\subseteq A\cup B$ and so $\displaystyle x\in\mathrm{Int}(A\cup B)$.

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

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

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

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

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

For the example, take the closed intervals $\displaystyle A=[0,1]$ and $\displaystyle B=[1,2]$ in $\displaystyle \mathbb{R}$. Then $\displaystyle \mathrm{Int}(A\cup B)=(0,2)$ but $\displaystyle \mathrm{Int}(A)=(0,1)$ and $\displaystyle \mathrm{Int}(B)=(1,2)$.
• Jan 28th 2008, 08:29 AM
rolylane
Thank you so much for your time and effort in replying to my query. I really am grateful.