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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.

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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 $B_d(x_1,r_2)\subseteq G_1$.

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  • $\begingroup$ What about the closure of $B_d(x_1,r_2)$ ?Will it be contained in $G_1$? $\endgroup$
    – Guria Sona
    Dec 31 '20 at 2:13
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    $\begingroup$ @GuriaSona: Not automatically, but a metric space is regular, so you can always choose $r_2$ small enough so that $\operatorname{cl}B_d(x_1,r_2)\subseteq B_d(x,r)\cap G_1$. $\endgroup$ Dec 31 '20 at 2:28
  • $\begingroup$ 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)? $\endgroup$
    – Guria Sona
    Dec 31 '20 at 7:28
  • $\begingroup$ @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 $$\operatorname{cl}B_d(x_1,r_2)\subseteq\operatorname{cl}U\,.$$ $\endgroup$ Dec 31 '20 at 7:52

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