Let $A\subset X$ be a nowhere dense set, where $X$ is a complete metric space. My book says there is an open set $S$ of radius less than $1$ such that $S$ is disjoint with $A$. I'm confused as to where that came from. I have formed an argument to support this. Would be great if someone could comment on whether it is sound.
$A$ is nowhere dense implies $\text{Int}(\overline{A})=\emptyset$, where $\overline{A}$ is the closure of $A$ in $X$. We shall prove that there is at least one disjoint open set in $X$ with respect to $A$. Let us assume the contrary. Take a point $x\in X$. Then $B(x,r)$ contains points in $A$ for all $r\in \Bbb{R}$. This is true for every $x$ in $X$. Hence, $\overline{A}=X$. $X$ contains an open set- $X$ itself! This is a contradiction, as $\overline{A}$ shouldnt have been $\emptyset$. If there is a disjoint open set in $X$, then there is definitely a disjoint open set with radius less than $1$.
Thanks in advance!
Best Answer
An easier proof is as follows:
If $A\subset X$ is not dense, then $\text{ext}\left(\overline{A}\right)=\text{ext}(A)=X-\overline{A}\neq\emptyset$. Consider any point $x\in\text{ext}(A)$. Since $\text{ext}(A)$ is an open set, the open ball $B_\epsilon(x)\subset\text{ext}(A)$ for some $\epsilon>0$. Let $\epsilon_0=\min\left(\epsilon,\frac12\right)$.
The open ball $B_{\epsilon_0}(x)$ is disjoint with $\overline{A}$ and thus is disjoint with $A$, and has a radius less than $1$.