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Thread: closure containing smallest closed set

  1. #1
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    closure containing smallest closed set

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

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

    So do proof by contradiction: Suppose that $\displaystyle \bar{S} $ is not the smallest closed set containing $\displaystyle 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 $\displaystyle S \subseteq T\quad \& \quad T \varsubsetneq \overline S $.
    Then you know that $\displaystyle \left( {\exists x_0 \in \overline S \backslash T} \right)$. Again the complement of T is open. Get a contradiction
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    So $\displaystyle T^{c} = \overline{S} \backslash T $ is open. Suppose that $\displaystyle x_0 \in T^{c} $. Then $\displaystyle (\exists \delta > 0)[(x_{0}- \delta, x_{0} + \delta) \subseteq T^{c} = \overline{S} \backslash T] $.

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

    BTW: $\displaystyle 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 $\displaystyle T $ such that $\displaystyle S \subset T $ and $\displaystyle T \subset \overline{S} $?

    Then we know that $\displaystyle \exists x_{0} \in \overline{S} \backslash T $ because $\displaystyle \overline{S} $ is "greater" than $\displaystyle T $? So $\displaystyle x_{0} \not \in T \ \text{but}\ x \in \overline{S} $. Then $\displaystyle x \in S \cup S' \implies x \in T $, which is a contradiction. Hence $\displaystyle T \supset \overline{S} $ which implies that $\displaystyle \overline{S} $ is the smallest set containing $\displaystyle 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 $\displaystyle \left( {\exists x_0 \in \overline S \backslash T} \right)$.
    Remind your reader that $\displaystyle S \subseteq T$.
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