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Thread: prove that D is dense

  1. #1
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    prove that D is dense

    Prove that a subset D of a metric space M is dense in M iff $\displaystyle D \bigcap U$ nonempty for every nonempty open set $\displaystyle U \subset M$

    so given$\displaystyle D \bigcap U$ nonempty we need to show that cl(D)=M, how does this work?
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    Re: prove that D is dense

    Quote Originally Posted by wopashui View Post
    Prove that a subset D of a metric space M is dense in M iff $\displaystyle D \bigcap U$ nonempty for every nonempty open set $\displaystyle U \subset M$

    so given$\displaystyle D \bigcap U$ nonempty we need to show that cl(D)=M, how does this work?
    Here's the basic duality that needs to be noticed $\displaystyle D\cap U=\varnothing$ implies $\displaystyle D\subseteq (M-U)$ and $\displaystyle M-U$ is closed and not the full space (since $\displaystyle U\ne\varnothing$). Get it?
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    Re: prove that D is dense

    Quote Originally Posted by wopashui View Post
    Prove that a subset D of a metric space M is dense in M iff $\displaystyle D \bigcap U$ nonempty for every nonempty open set $\displaystyle U \subset M$ so given$\displaystyle D \bigcap U$ nonempty we need to show that cl(D)=M, how does this work?
    Here is a slightly different discussion of this.
    The statement that $\displaystyle x$ is a contact point of $\displaystyle D$ means that $\displaystyle x\in D$ or $\displaystyle x$ is a limit point of $\displaystyle D$ [i.e. $\displaystyle x\in cl(D)]$.

    So you are asked to show that each point of $\displaystyle M$ is a contact point of $\displaystyle D$.

    Suppose that $\displaystyle t\in M$ but $\displaystyle t\notin cl(D)$. Then there is a ball $\displaystyle \mathcal{B}(t;\delta)$ that contains no point of $\displaystyle D$.
    WHY? And why is that a contradiction?
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