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Thread: interior proofs

  1. #1
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    interior proofs

    For sets A,B subsets of the Reals

    a) int(A) intersect int(B) = int(A intersect B)
    b) bd(A U B) is subset of bd(A) U bd(B)
    c) Give examples of sets A and B where bd(A U B) = empty set and bd(A) = R = bd(B)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Jdg6057 View Post
    For sets A,B subsets of the Reals

    a) int(A) intersect int(B) = int(A intersect B)
    b) bd(A U B) is subset of bd(A) U bd(B)
    c) Give examples of sets A and B where bd(A U B) = empty set and bd(A) = R = bd(B)
    a) Let $\displaystyle x\in int(A)\cap int(B)$. Then $\displaystyle x\in int(A)$ and $\displaystyle x\in int(B)$. So $\displaystyle \exists~\epsilon_1,\epsilon_2$ such that $\displaystyle N(x,\epsilon_1)\subset A$ and $\displaystyle N(x,\epsilon_2)\subset B$. Taking $\displaystyle \epsilon=\min\{\epsilon_1,\epsilon_2\}$ guarantees that $\displaystyle N(x,\epsilon)\subset A\cap B$. So $\displaystyle x\in int(A\cap B)$ and $\displaystyle (int(A)\cap int(B))\subseteq int(A\cap B)$. See if you can prove the other direction.

    b) Let $\displaystyle x\in bd(A\cup B)$. Then $\displaystyle \forall~\epsilon>0$, $\displaystyle N(x,\epsilon)\cap(A\cup B)^c\neq\emptyset$. Note that $\displaystyle (A\cup B)^c=A^c\cap B^c$.
    Spoiler:
    So $\displaystyle N(x,\epsilon)\cap A^c\cap B^c\neq\emptyset$. In particular, $\displaystyle N(x,\epsilon)\cap A^c\neq\emptyset$ and $\displaystyle N(x,\epsilon)\cap B^c\neq\emptyset$, so $\displaystyle x\in bd(A)\cup bd(B)$.


    c) Let $\displaystyle X=\{z\in\mathbb{C}:\Im(z)\geq 0\}$ and $\displaystyle Y=\{z\in\mathbb{C}:\Im(z)\leq0\}$.

    $\displaystyle bd(X)$ and $\displaystyle bd(Y)$ consist of all $\displaystyle z$ with $\displaystyle \Im(z)=0$; in other words, the real line. However, $\displaystyle X\cup Y=\mathbb{C}$, and $\displaystyle bd(\mathbb{C})=\emptyset$.
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