# Doubt in open dense sets of a metric space.

Let $$B_d(x,r) \cap G_1$$ where $$G_1$$ is an open dense set. Now this intersection is non empty. Let $$x_1 \in B_d(x,r) \cap G_1$$.Then, $$B_d(x_1,r_1) \subset G_1$$ .Now my doubt is that will there exists a $$r_2 > 0$$ such that $$B_d(x_1,r_2) \subset G_1$$ and $$B_d(x_1,r_2) \subset B_d(x,r)$$.

I think it will be true. However I was wondering if there is a contradicting example.

Whether $$B_d(x_1,r_1)\subseteq G_1$$ depends on $$r_1$$. What is true is that there is an $$r_1>0$$ such that $$B_d(x_1,r_1)\subseteq G_1$$. As for your question, $$B_d(x,r)\cap G_1$$ is an open set containing $$x_1$$, and the open balls are a base for the topology, so necessarily there is an $$r_2>0$$ such that $$B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$$, which of course implies that $$B_d(x_1,r_2)\subseteq B_d(x,r)$$ and $$B_d(x_1,r_2)\subseteq G_1$$.
• What about the closure of $B_d(x_1,r_2)$ ?Will it be contained in $G_1$? Dec 31 '20 at 2:13
• @GuriaSona: Not automatically, but a metric space is regular, so you can always choose $r_2$ small enough so that $\operatorname{cl}B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$. Dec 31 '20 at 2:28
• I am sorry if it sounds stupid but I'm unable to prove the statement that such a $r_2$ will exist such that closure is contained in the ball?Is it because of the fact that a closed ball of radius $r_2<r$ will be contained in $B_d(x,r)$.(If this is the reason then I can probably see it in $\mathbb{R}^k$) .how do I conclude it in case of any general metric space (say discrete metric space)? Dec 31 '20 at 7:28
• @GuriaSona: In any metric space $\langle X,d\rangle$ it’s true that $\operatorname{cl}B_d(x,r)\subseteq\{y\in X:d(x,y)\le r\}$, but since $x_1$ and $x$ are in general different points, that fact doesn’t help here. As I said, use regularity: there is an open set $U$ such that $$x_1\in U\subseteq\operatorname{cl}U\subseteq B_d(x,r)\cap G_1\,.$$ And $U$ is an open nbhd of $x_1$, so there is an $r_2>0$ such that $B_d(x_1,r_2)\subseteq U$, and hence $$\operatorname{cl}B_d(x_1,r_2)\subseteq\operatorname{cl}U\,.$$ Dec 31 '20 at 7:52