prove that D is dense

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November 25th 2011, 05:50 PM
wopashui

prove that D is dense

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

so given $D \bigcap U$ nonempty we need to show that cl(D)=M, how does this work?
November 25th 2011, 07:54 PM
Drexel28

Re: prove that D is dense

Quote:

Originally Posted by wopashui

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

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

Here's the basic duality that needs to be noticed $D\cap U=\varnothing$ implies $D\subseteq (M-U)$ and $M-U$ is closed and not the full space (since $U\ne\varnothing$ ). Get it?
November 26th 2011, 07:02 AM
Plato

Re: prove that D is dense

Quote:

Originally Posted by wopashui

Prove that a subset D of a metric space M is dense in M iff $D \bigcap U$ nonempty for every nonempty open set $U \subset M$ so given $D \bigcap U$ nonempty we need to show that cl(D)=M, how does this work?

Here is a slightly different discussion of this.
The statement that $x$ is a contact point of $D$ means that $x\in D$ or $x$ is a limit point of $D$ [i.e. $x\in cl(D)]$ .

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

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

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