# prove that D is dense

• Nov 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?
• Nov 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?
• Nov 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?