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Math Help - Proving a set is closed

  1. #1
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    Proving a set is closed

    Let S = [0,4]. Show S is closed.

    How can I show that the bd S is a subset of S, or that R/S is open?
    I know that simply stating the bd should be enough to say that S is closed, but my professor wants us to be able to also show bd{0 , 4}.
    I know that if it were (0,4) I could say that (0 - E, 0+ E) intersected with S and R/S is non empty, but is it sufficient to say that if S is closed then the bd S = bd (R/S) so R/S is empty?

    Also at what point when x is in bd does (x - E, x + E) intersected with S and R/S non empty? This would only be true for an open set correct?
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  2. #2
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    Quote Originally Posted by p00ndawg View Post
    Let S = [0,4]. Show S is closed.
    This is the complement: \left( { - \infty ,0} \right) \cup \left( {4,\infty } \right).
    Can you prove that the union is open?
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  3. #3
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    Quote Originally Posted by Plato View Post
    This is the complement: \left( { - \infty ,0} \right) \cup \left( {4,\infty } \right).
    Can you prove that the union is open?
    yes i think i can, especially since I know that the union of any collection of open sets is open.


    so by showing that the complement is open, we can say that the set is closed.

    nice, thank you.
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  4. #4
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    by bd(s) do you mean the boundary of S ??

    then bd(S) = S closure intersection (X-S) closure. and since S is closed S=S closure so you are done.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by srinivas.sista89 View Post
    by bd(s) do you mean the boundary of S ??

    then bd(S) = S closure intersection (X-S) closure. and since S is closed S=S closure so you are done.
    The point was to prove that S is closed.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    An alternative way to do this (which is just implicitly noting that \mathbb{R}-S is open) is to suppose that \ell\notin[0,4]\implies \ell\in\mathbb{R}-S. We have two cases, either \ell<0 in which case note that \ell+\frac{0-\ell}{2}<0\quad\color{red}\star; or \ell>4 in which case \ell-\frac{\ell-4}{2}>4\quad\color{red}\star\star. And clearly \ell-\frac{0-\ell}{2}<0\quad\color{red}\star and \ell+\frac{\ell-4}{2}>4\quad\color{red}\star\star. If the first case is two then choosing \varepsilon=\color{red}\star shows that \mathcal{U}_{\varepsilon}\subset (-\infty,0) and if the second is true then choosing \varepsilon=\color{red}\star\star shows that \mathcal{U}_{\varepsilon}\subset(4,\infty). In either case \ell is an interior point of \mathbb{R}-S. Thus \mathbb{R}-S is open which implies that S is closed.
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