# Proving that if $X$ is a complete metric space and $A\subset X$ is nowhere dense in $X$, then there is an open set in $X$ disjoint with $A$.

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 $1$.

• The proof is correct. But I would write cl$(A)$ or $\overline A$ since $A^c$ can easily be confused with the complement. Commented Aug 11, 2013 at 18:08
• I think you can also show that if every open set in (X\A) intersects A, then A is dense in (X\A). Then Cl(A)=(X\A); if Cl(X\A) has empty interior, then X is the union of subspaces with empty interior, and so it is the countable union of nowhere-dense sets, contradicitng Baire.
– FBD
Commented Aug 11, 2013 at 18:55
• Your question has a very easy answer, because $\emptyset$ is an open set in $X$ and is disjoint from $A$. Did you want a nonempty open set disjoint from $A$? That is still easy, if "nowhere dense set" is defined as "a set $A$ such that every nonempty open set contains a nonempty open set which is disjoint from $A$". (That's how I would define "nowhere dense"; what definition are you using?) You don't need a complete metric space for that, but you do need to assume that the space itself is nonempty.
– bof
Commented Jan 1, 2014 at 2:01

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 $1$.
This is because $\bar{A} \neq X$ and $(\bar{A})^c$ is actually an open set where you can find whatever you need.
Assume there is no such open set, so that every open set in $X\setminus A$ intersects $A.$ Then $A$ is dense in $X\setminus A,$ so that $\operatorname{Cl}(A)= X\setminus A.$ Then, either $X\setminus A$ contains an open set, contradicting that $A$ is nowhere-dense in $X,$ or $X\setminus A$ has empty interior, so that $X=A\cup(X\setminus A)$ is a complete metric space, and the countable union of nowhere-dense sets, which cannot happen by Baire Category Theorem.