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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- Nov 25th 2011, 05:50 PMwopashuiprove 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? - Nov 25th 2011, 07:54 PMDrexel28Re: prove that D is dense
- Nov 26th 2011, 07:02 AMPlatoRe: prove that D is dense
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?