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Math Help - closure containing smallest closed set

  1. #1
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    closure containing smallest closed set

    Prove: If  S \subset \bold{R} , then  \bar{S} is the smallest closed set containing  S .

    So  \bar{S} = S \cup S' . We want to show that if  T is any closed set and  S \subset T , then  \bar{S} \subset T .

    So do proof by contradiction: Suppose that  \bar{S} is not the smallest closed set containing  S . Is this the correct way to proceed?
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  2. #2
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    I suggest that you start by supposing a close set T such that S \subseteq T\quad \& \quad T \varsubsetneq \overline S .
    Then you know that \left( {\exists x_0  \in \overline S \backslash T} \right). Again the complement of T is open. Get a contradiction
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  3. #3
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    So  T^{c} = \overline{S} \backslash T is open. Suppose that  x_0 \in T^{c} . Then  (\exists \delta > 0)[(x_{0}- \delta, x_{0} + \delta) \subseteq T^{c} = \overline{S} \backslash T] .

    But this is a contradiction because  T \not \subseteq \overline{S} ? So  T \subseteq \overline{S} .
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  4. #4
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    Quote Originally Posted by particlejohn View Post
    So  T^{c} = \overline{S} \backslash T is open. Suppose that  x_0 \in T^{c} . Then  (\exists \delta > 0)[(x_{0}- \delta, x_{0} + \delta) \subseteq T^{c} = \overline{S} \backslash T] .
    But this is a contradiction because  T \not \subseteq \overline{S} ? So  T \subseteq \overline{S} .
    No that does not work. See my suggestion.
    \begin{gathered}<br />
  \left( {\exists x_0  \in \overline S \backslash T} \right) \hfill \\<br />
  x_0  \notin {\rm T}\quad but\quad x_0  \in \overline S  \hfill \\<br />
  S \subseteq T\quad  \Rightarrow \quad x_0  \notin S \hfill \\ <br />
\end{gathered}

    BTW:  T^{c} = \overline{S} \backslash T is not true.
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  5. #5
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    shouldn't we suppose a closed set  T such that  S \subset T and  T \subset \overline{S} ?

    Then we know that  \exists x_{0} \in \overline{S} \backslash T because  \overline{S} is "greater" than  T ? So  x_{0} \not \in T \ \text{but}\ x \in \overline{S} . Then  x \in S \cup S' \implies x \in T , which is a contradiction. Hence  T \supset \overline{S} which implies that  \overline{S} is the smallest set containing  S ?
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  6. #6
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    or is this too simple?
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  7. #7
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    That will work. But you need to suppose that T is proper subset of the closure.
    That insures that \left( {\exists x_0  \in \overline S \backslash T} \right).
    Remind your reader that S \subseteq T.
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