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Thread: Open sets in a metric space

  1. #1
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    Open sets in a metric space

    The question is:

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

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

    Pf:

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

    $\displaystyle G$ is open, thus $\displaystyle G^{C}$ is closed.

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

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

    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 $\displaystyle (X,d)$ be a metric space, and $\displaystyle Y$ be a subset of $\displaystyle X$.
    If $\displaystyle G \subset X$ is open, then $\displaystyle G \cap Y$ is open in the metric space $\displaystyle (Y,d)$.
    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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  3. #3
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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 $\displaystyle (X,d)$ a set $\displaystyle G \subset X$ is open if $\displaystyle \forall x $ in $\displaystyle G$ $\displaystyle \exists $ an $\displaystyle \epsilon > 0$ s.t $\displaystyle B(x; \epsilon) \subset G.
    $

    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 $\displaystyle (X,d)$ a set $\displaystyle G \subset X$ is open if $\displaystyle \forall x $ in $\displaystyle G$ $\displaystyle \exists $ an $\displaystyle \epsilon > 0$ s.t $\displaystyle B(x; \epsilon) \subset G.$
    This defiinition doesnt seem to depend on X. Does it matter the space we are in?
    Actually it does depend upon $\displaystyle X~\&~d$.
    That is why we use the notation $\displaystyle (D,d)$.
    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 $\displaystyle \mathcal{B}(x;\delta)=\{y\in X: d(x,y)<\delta\}.$
    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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