1. ## interior points question

Let A and B be subsets of $\displaystyle R^n$ with $\displaystyle A^0$, $\displaystyle B^0$ denoting the sets of interior points for A and B respectively. Prove that $\displaystyle A^0UB^0$ is a subset of the interior of AUB. Give an example where the inclusion is strict.

2. One reason we ask people to show there work is that there are a number of different ways of defining "interior points" and without seeing what you are trying we don't know which is appropriate for you. I suspect you are using "p is an interior point of p if and only if there exist a neighborhood of p that is a subset of A".

So if p is in $\displaystyle A^O\cup B^O$ then it is in both $\displaystyle A^O$ and $\displaystyle B^O$. Since it is in $\displaystyle A^O$ there exist some neighborhood of p, $\displaystyle N_{\delta_1}(p)$ ($\displaystyle \delta_1$ is the radius), such that $\displaystyle N_{\delta_1}(p)\subset A$. Since p is in $\displaystyle B^O$, there exist some neighborhood of p, $\displaystyle N_{\delta_2}(p)$ such that $\displaystyle N_{\delta_2}(p)\subset B$.

Can you finish now?

3. @HallsofIvy
If $\displaystyle p\in A^0 \cup B^0 \Rightarrow p\in A^0 \text{ or } p\in B^0$. It is not necessary for $\displaystyle p\in A^0 \text{ and } p\in B^0$ that case would be necessary for the intersection not union.

4. Originally Posted by wopashui
Let A and B be subsets of $\displaystyle R^n$ with $\displaystyle A^0$, $\displaystyle B^0$ denoting the sets of interior points for A and B respectively. Prove that $\displaystyle A^0UB^0$ is a subset of the interior of AUB. Give an example where the inclusion is strict.
$\displaystyle \forall p\in A^o\cup B^o\Rightarrow p\in A^o\, or \, p\in B^o \Rightarrow \exists \delta>0 \,such\, that\, N_\delta(p)\subset A \,or\, N_\delta(p)\subset B\Rightarrow$$\displaystyle \Rightarrow\exists \delta>0 \,such\, that\, N_\delta(p)\subset A\cup B \Rightarrow p \in (A\cup B)^o.$

Short version $\displaystyle \forall p\in A^o\cup B^o \Rightarrow p\in (A\cup B)^o,$ hence $\displaystyle A^o\cup B^o\subseteq (A\cup B)^o.$

5. Originally Posted by lvleph
@HallsofIvy
If $\displaystyle p\in A^0 \cup B^0 \Rightarrow p\in A^0 \text{ or } p\in B^0$. It is not necessary for $\displaystyle p\in A^0 \text{ and } p\in B^0$ that case would be necessary for the intersection not union.
Absolutely! For some reason I was thinking "intersection". If $\displaystyle p\in A^O\cup B^O$ then p is in either $\displaystyle A^O$ or $\displaystyle B^O$ so either there exist neighborhood $\displaystyle N_{\delta}(p)\in A^O$ or there exist $\displaystyle N_{\delta}(p)\in B^O$. Fortunately, either of those is enough to conclude that $\displaystyle N_{\delta}(p)\in A^O\cup B^O$, as mathoman shows.