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Thread: interior points question

  1. #1
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    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.
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  2. #2
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    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?
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    @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.
    Last edited by lvleph; Sep 15th 2010 at 05:14 AM.
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  4. #4
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    Quote Originally Posted by wopashui View Post
    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 $\displaystyle A^o\cup B^o\subseteq (A\cup B)^o.$
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  5. #5
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    Quote Originally Posted by lvleph View Post
    @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.
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