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Math Help - Basic Topology

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    Question Basic Topology

    1)Show that a set E is open iff E intersection boundary of E = empty set

    2)Show that a set E is closed iff boundary is contained in E
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    1)Show that a set E is open iff E intersection boundary of E = empty set

    2)Show that a set E is closed iff boundary is contained in E
    Don't you dare take a peek until you've tried them yourself.
    Spoiler:

    1) If E is open then every point of E is an interior point of E. Clearly then no point of E may be a boundary point of E so that it follows that E\cap\partial E=\varnothing. Conversely, suppose that that E\cap\partial E=\varnothing, then no point of E is a boundary point of E. Thus, for every \xi\in E we must have that there exists some neighborhood of E, which we'll call N_{\delta}(\xi) such that either N_{\delta}(\xi)\subseteq E or N_{\delta}(\xi)\subseteq E'. But, since \xi\in E the latter is impossible and the conclusion follows.

    2) If a set E is closed then it contains all it's limit points. So let \xi\in\partial E. If \xi is in E we are done so assume that \xi\notin E. Since \xi\in\partial E we must have that every neighborhood N_{\delta}(\xi) contains points of both E and  E', and since \xi\notin E we must conclude that the points of E are distinct from \xi. But, this implies that \xi is a limit point of E and the conclusion follows. Conversely, suppose that \partial E\subseteq E. Let \xi be a limit point of E. If \xi\in E we are done, so assume that \xi\notin E. Then every neighborhood of \xi contains points of E but since every neighborhood of \xi also contains \xi it follows that \xi\in \partial E\implies \xi\in E. We may therefore conclude that E contains all it's limit points. The conclusion follows.
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