Prove/disprove the distance between a closed set and a bounded closed set which are disjoint has a positive lower bound (only consider the metric space $\mathbf{R}^n$)
$A$ is a closed set in $\mathbf{R}^{n}$, $B$ is a bounded closed set in $\mathbf{R}^{n}$, and $A\cap B=\emptyset$. How to prove/disprove there exists $\epsilon>0$ such that for all $\mathbf{x}\in A$, $B(\mathbf{x},\epsilon)$ is disjoint from $B$?
My question is related to “Zero distance between a closed set and a bounded closed set which are disjoint in a metric space”, but they are two different questions. My question only considers the metric space $\mathbf{R}^{n}$, but the question in the link considers an arbitrary metric space. Actually, I only have learnt topology of $\mathbf{R}^{n}$.
It seems the question in the link tells us that we can disprove this result in arbitrary metric space.However, only considering the metric space $\mathbf{R}^{n}$, we can prove this result, which means we can prove there exists $\epsilon>0$ such that for all $\mathbf{x}\in A, B(\mathbf{x},\epsilon)$ is disjoint from $B$. Here is my try:
$B$ is compact in $\mathbf{R}^{n}$. We define a function $f\colon B\rightarrow\mathbf{R}$ by $f(\mathbf{x})=d(\mathbf{x},A)$. Here $d(\mathbf{x},A)$ is the distance between the point $\mathbf{x}$ and the set $A$. For all $\mathbf{x}\in B$, $f(\mathbf{x})>0$. $f$ is a continuous function on the compact set $B$. As a result, $f$ has a minimum value $\epsilon>0$ on $B$. Finally, there exists an $\epsilon>0$ such that for all $\mathbf{x}\in A$, $B(\mathbf{x},\epsilon)$ is disjoint from $B$.
Here are my questions:
(i) We can prove there exists $\epsilon>0$ such that for all $\mathbf{x}\in A$, $B(\mathbf{x},\epsilon)$ is disjoint from $B$ in the metric $\mathbf{R}^{n}$ and my “try” above is correct. Am I right?
(ii) We can disprove there exists $\epsilon>0$ such that for all $\mathbf{x}\in A$, $B(\mathbf{x},\epsilon)$ is disjoint from $B$ in arbitrary metric space (the question in the link tells me). Am I right?