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Math Help - Open sets in a metric space

  1. #1
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    Open sets in a metric space

    The question is:

    Let (X,d) be a metric space, and Y be a subset of X.

    If G \subset X is open, then G \cap Y is open in the metric space (Y,d).

    Pf:

    Y \subset X implies (Y,d) is a metric space \rightarrow Y is open.
    Therefore Y^{C} is closed

    G is open, thus G^{C} is closed.

    So B = (G^{C} \cup Y^{C}) is closed, since it is the finite union of closed sets.

    Therefore B^{C} = G \cap Y is open.

    I seem to follow all the definitions and theorems correct. Is this ok?

    There is also a second part to this question.

    Conversely, if E_1 \subset Y is open in (Y,d), \exists an open set E \subset X s.t E_1 = E \cap Y

    I don't get this question. Don't we already have an open set that works in the hypothesis? E = E_1?

    Any help is greatly appreciated. Thank you.

    -Jame
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  2. #2
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    Quote Originally Posted by Jame View Post
    Let (X,d) be a metric space, and Y be a subset of X.
    If G \subset X is open, then G \cap Y is open in the metric space (Y,d).
    I do not understand your confusion!
    How are open sets in a sub-space related to open sets in the super-space?
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    Oh right. The are not necessarily related. Something could be open in X but not in a subset of X. Guess I have to start over... and use the definition of open and not try to be tricky with complements.
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  4. #4
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    Actuallly I have a question.

    This is how my book defines open

    For a metric space (X,d) a set G \subset X is open if \forall x in G  \exists an  \epsilon > 0 s.t B(x; \epsilon) \subset G.<br />

    This defiinition doesnt seem to depend on X. Does it matter the space we are in?
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  5. #5
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    Quote Originally Posted by Jame View Post
    This is how my book defines open
    For a metric space (X,d) a set G \subset X is open if \forall x in G  \exists an  \epsilon > 0 s.t B(x; \epsilon) \subset G.
    This defiinition doesnt seem to depend on X. Does it matter the space we are in?
    Actually it does depend upon X~\&~d.
    That is why we use the notation (D,d).
    A metric space is determined by a pair, a set of points and a metric on the set.
    Recall that a ball is defined as \mathcal{B}(x;\delta)=\{y\in X: d(x,y)<\delta\}.
    Balls are the basic open sets of a metric space.
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  6. #6
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    Oh yes!. Thanks so much. I was honestly didn't catch that. I was looking at this definition and was wondering "whats the difference between being open in X and in Y?" Thank you for the clarification.
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