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 Post subject: Metric topology Posted: Wed Nov 25, 2015 7:18 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{\left(X,d\right)}$ be a metric space and $\displaystyle{D\subseteq X}$ .

Prove that $\displaystyle{D}$ is dense on $\displaystyle{X}$ if, and only if, for each continuous function

$\displaystyle{f:X\longrightarrow \mathbb{R}}$ holds :

$\displaystyle{f(x)=0\,,\forall\,x\in D\implies f=\mathbb{O}}$ .

Top   Post subject: Re: Metric topology Posted: Thu Nov 26, 2015 12:17 am

Joined: Thu Nov 19, 2015 7:27 pm
Posts: 6
One direction is easy. If $f=0$ on $D$ , then by continuity $f = 0$ on $\bar{D} = X$

Now assume the converse. A different definition of density in a metric space is the following :

"$D$ is dense iff every open set in $X$ intersects $D$ non-trivially".

So assume $D$ is not dense and pick an open set not intersecting $D$. Since we're working in a metric space, there exists an $x$ and an $\epsilon>0$ : $B_{\epsilon} (x) \cap D = \emptyset$.

How can we use this information? Things like Urysohn's lemma come to mind... Indeed, Urysohn gives a continuous function where $f(X -B_{\epsilon} (x)) = 0$ and $f (\bar{B_{\epsilon/2}}(x) ) = 1$ and we are done. (Every metric space is normal)

However, things here are much easier! Just define $A= \bar{B}_{\epsilon/2}(x)$ and $B= B_{\epsilon}(x) ^{\mathsf{c}}$. Notice $D \subset B$ and let $$f(x)= \frac{dist(x,B)} {dist(x,A) +dist(x,B)}$$.

This $f$ does the job , so we get home without any heavy machinery.

Nikos

Top   Post subject: Re: Metric topology Posted: Thu Nov 26, 2015 3:18 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Thank you Nikos.

Here is another proof :

Suppose that $\displaystyle{D}$ is not dense on $\displaystyle{\left(X,d\right)}$, that is $\displaystyle{\overline{D}\neq X}$ .

Then, there exists $\displaystyle{y\in X}$ such that $\displaystyle{d(y,D)>0}$ . The function

$\displaystyle{f:X\longrightarrow \mathbb{R}\,,f(x)=d(x,D)}$ is continuous and $\displaystyle{f(x)=0\,,\forall\,x\in D\subseteq \overline{D}}$ .

Accoding to the hypothesis, $\displaystyle{f=\mathbb{O}}$, a contradiction, since $\displaystyle{f(y)>0}$ .

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