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Thread: metric spaces

  1. #1
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    metric spaces

    Let (M,d) be a metric space.
    1. Choose x element of M. Show that a set U subset of M is a neighbourhood of x if and only if x is an element of the Interior of U
    2. Prove that the intersection U intersection V of any two neighbourhoods U and V of a point x element of M is also a neighbourhood of x.
    Last edited by Godisgood; Nov 1st 2010 at 06:13 AM.
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    Quote Originally Posted by Godisgood View Post
    Let (M,d) be a metric space.
    1. Choose x element of M. Show that a set U subset of M is a neighbourhood of x if and only if x is an element of the Interior of U
    2. Prove that the intersection U intersection V of any two neighbourhoods U and V of a point x element of M is also a neighbourhood of x.

    Thanks soo much. I am really struggling with this since it is the first time I am taking a course of this nature, so please go in steps
    Thanks for the help
    What is a neighborhood? Do you take the Bourbaki "set containing an open set containing the point" or just "open set containing point"
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    Quote Originally Posted by Drexel28 View Post
    What is a neighborhood? Do you take the Bourbaki "set containing an open set containing the point" or just "open set containing point"
    @Godisgood,
    Please answer the above question.
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    set containing an open set containing the point
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    O.K. then if $\displaystyle O$ is an open set and $\displaystyle a\in\O\subseteq V$ the by definition $\displaystyle a\in\mathscr{I}(V)$, interior of $\displaystyle V$.
    On the other,hand, if $\displaystyle b\in\mathscr{I}(V)$ the by definition of interior, an open set $\displaystyle \left( {\exists Q} \right)$ such that $\displaystyle b\in Q\subseteq V$ so $\displaystyle V$ is a neighborhood of $\displaystyle b$.

    If each of $\displaystyle V~\&~U$ is a neighborhood of $\displaystyle c$ there are open sets $\displaystyle c \in O \subseteq V\;\& \;c \in Q \subseteq U$.
    But $\displaystyle O\cap Q$ is open and $\displaystyle c\in O\cap Q \subseteq V\cap U$.
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