1. ## interior points question

Let A and B be subsets of $R^n$ with $A^0$, $B^0$ denoting the sets of interior points for A and B respectively. Prove that $A^0UB^0$ 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 $A^O\cup B^O$ then it is in both $A^O$ and $B^O$. Since it is in $A^O$ there exist some neighborhood of p, $N_{\delta_1}(p)$ ( $\delta_1$ is the radius), such that $N_{\delta_1}(p)\subset A$. Since p is in $B^O$, there exist some neighborhood of p, $N_{\delta_2}(p)$ such that $N_{\delta_2}(p)\subset B$.

Can you finish now?

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

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

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

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