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Thread: limit points

  1. #1
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    limit points

    Prove that a set $\displaystyle S \subset \bold{R} $ is closed if and only if it contains all of its limit points.

    Suppose that $\displaystyle S $ is closed, and $\displaystyle x_0 $ is a limit point of $\displaystyle S $. Then $\displaystyle S^{C} $ is open so that there is an $\displaystyle \varepsilon > 0 $ such that $\displaystyle (x_{0} - \varepsilon, x_{0} + \varepsilon) \cap S = \emptyset $ which is a contradiction.

    If $\displaystyle S $ contains all its limit points, then there exists an $\displaystyle \varepsilon > 0 $ such that $\displaystyle (x_{0}- \varepsilon, x_{0} + \varepsilon) \cap S \neq \emptyset $. So $\displaystyle S^{C}$ is open $\displaystyle \implies S $ is closed.

    Is this correct?
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  2. #2
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    You have the correct ideas and concepts but an awful form.
    The first part should go something like this. Say that $\displaystyle x_0$ is limit point of of the closed set $\displaystyle S$. Suppose $\displaystyle x_0 \notin S$ then $\displaystyle x_0 \in S^c $ which is an open set. So $\displaystyle \left( {\exists \delta > 0} \right)\left[ {\left( {x_0 - \delta ,x_0 + \delta } \right) \subseteq S^c } \right]$. But that is a contradiction to $\displaystyle x_0$ being a limit point.

    No you try to rewrite the second part.
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  3. #3
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    $\displaystyle S $ contains all of its limit points. Suppose that $\displaystyle x_0 \not \in S $. Then $\displaystyle x_{0} \in S^{c} $, then $\displaystyle (\exists \delta >0)[(x_0- \delta, x_0+\delta) \subseteq S^{c}] $. Thus $\displaystyle S^{c} $ is open which implies that $\displaystyle S $ is closed.
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  4. #4
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    That looks good.
    You might add that $\displaystyle x_0 \in S^c$ means that $\displaystyle x_0 $ is not a limit point of $\displaystyle S$.
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