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Thread: Prove the following

  1. #1
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    Prove the following

    If S is a subset of the Reals Prove:

    a) bd(S) = bd(R\S)
    b) int(S) intersect bd(S) = empty set
    c) S is open IFF bd(S) is a subset of R\S
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Jdg6057 View Post
    If S is a subset of the Reals Prove:

    a) bd(S) = bd(R\S)
    b) int(S) intersect bd(S) = empty set
    c) S is open IFF bd(S) is a subset of R\S
    a) A point $\displaystyle x\in\mathbb{R}$ is a boundary point of $\displaystyle S$ iff every neighborhood around $\displaystyle x$ contains points in $\displaystyle S$ and $\displaystyle R\backslash S$. A point $\displaystyle x\in\mathbb{R}$ is a boundary point of $\displaystyle \mathbb{R}\backslash S$ iff every neighborhood around $\displaystyle x$ contains points in $\displaystyle \mathbb{R}\backslash S$ and $\displaystyle S$. What can you conclude from this?

    b) A point $\displaystyle x\in S$ is an interior point of $\displaystyle S$ iff there exists a neighborhood around $\displaystyle x$ that is a proper subset of $\displaystyle S$. Compare this definition with the above definition and deduce that $\displaystyle int(S)\cap bd(S)=\emptyset$.

    c) $\displaystyle \Rightarrow$ If $\displaystyle S$ is open, then every point in $\displaystyle S$ is an interior point, so $\displaystyle bd(S)=\emptyset$. $\displaystyle \emptyset\subset\mathbb{R}\backslash S$ is trivially true.

    $\displaystyle \Leftarrow$ If $\displaystyle bd(S)\subset\mathbb{R}\backslash S$, then $\displaystyle bd(S)\cap S=\emptyset$, meaning that every point in $\displaystyle S$ is an interior point, so $\displaystyle S$ is open.
    Last edited by redsoxfan325; Oct 16th 2009 at 08:27 AM.
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