I want to prove that: a subset $A$ of a metric space $X$ is nowhere dense in $X$ IFF each non-empty open set in $X$ contains an open ball whose closure is disjoint from $A$.

This is a lemma that is used in Baire's category theory.


  • $A \subset X$ is nowhere dense if $\text{int}\left(\bar{A}\right) = \emptyset$.
  • A topologial space is called $T_1$ if for each pair $a$, $b$ of distinct points of $X$, there are open set $U$ and $V$ in $X$ such that $a \in U, a \not \in V, b \not \in U$ and $b \in V$.
  • A topologial space is called regular if it is a $T_1$ space and for each closed subset $C$ and each point $a$ not in $C$, there exist disjoint open sets $U$ and $V$ in $X$ such that $a \in U$ and $C \subset V$.

My proof so far:

$\Rightarrow: $

Assume that $A \subset X$ is nowhere dense. If $X$ is empty, then the statement trivially holds. So assume that $X$ is not empty. If $\bar{A} = X$ then $\text{int}\left( \bar{A} \right) = X \not = \emptyset$, which contradicts the assumption that $A$ is nowhere dense. So we have that $\bar{A} \not = X$. So there exists an $a$ not in $\bar{A}$. Since any metric space is regular and $\bar{A}$ is closed we obtain an open ball $B_a(r)$ such that $B_a(r)$ and $\bar{A}$ are disjoint. But then also $B_a(r)$ and $A$ are disjoint. Now we note that $\overline{B_a(r/2)} \subset B_a(r)$. So we find that $\overline{B_a(r/2)}$ is disjoint from $A$.

The other direction I don't fully have. I made a start.

$\Leftarrow: $

Assume that each non-empty open set in $X$ contains an open ball whose closure is disjoint from $A$. And for a contradiction assume that $A$ is not nowhere dense. Then $\text{int}\left( \bar{A} \right) \not = \emptyset$. So by the assumption there exists an open ball $B$ with ${B} \subset \text{int}\left( \bar{A} \right)$ and $\overline{B} \cap A = \emptyset$. (Here I'm stuck)

It seems to me that we can derive a contradiction, but I don't know how.

I would like to have help with that and also receive comments on the rest of my proof.


Write $B=B(x,r)$ since $B\subset \bar A$ there exists a sequence $x_n$ in $A$ which converges towards $x$, this implies by definition there exists $N$ such that $n>N$ implies that $x_n\in B$. Contradiction since $B\cap A\subset \bar B\cap A$ is empty.


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