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 Author: Papapetros Vaggelis [ Wed Nov 25, 2015 7:18 pm ] Post subject: Metric topology 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 :$$\displaystyle{f(x)=0\,,\forall\,x\in D\implies f=\mathbb{O}}$$ .

 Author: Nikos Athanasiou [ Thu Nov 26, 2015 12:17 am ] Post subject: Re: Metric topology 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

 Author: Papapetros Vaggelis [ Thu Nov 26, 2015 3:18 pm ] Post subject: Re: Metric topology 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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