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Given $X$ a metric space, $A\subset X$ a nowhere dense set. Show that every open ball $B$ contains another open ball $B_1 \subset B$ such that $B_1 \cap A = \emptyset$.

EDIT: I modify my proof with the help of a great hint!

Given $B(x,\epsilon)$ lets suppose that for every open ball $B_1 \subset B(x,\epsilon)$, $B_1 \cap A \neq \emptyset$.

Given $y \in B(x,\epsilon)$ we have that for every $n>N$ with $N$ sufficiently big, $B(y,\frac{1}{n})\cap A \neq \emptyset$ That is, for every $n>N$ there exists $y_{n} \in B(y,\frac{1}{n})\cap A \neq \emptyset$. If we consider the sequence $(y_{n})_{n>N}$ we have that for every $n$, $y_n \in A$ and $y_n \rightarrow y$. Therefore $y \in cl(A)$ which implies $B(x,\epsilon)\subset cl(A)$ with contradicts the fact that $A$ is nowhere dense.

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That step is not legitimate. Consider the set $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$ in $\Bbb R$. For each $\epsilon>0$ we have $B(0,\epsilon)\cap A\ne\varnothing$, but we can’t take the limit as $\epsilon\to 0^+$ to conclude that $\{0\}\cap A\ne\varnothing$: this is clearly false.

Try this instead. Suppose that every open ball contained in $B(x,\epsilon)$ intersects $A$. Show that this implies that $B(x,\epsilon)\subseteq\operatorname{cl}A$, contradicting the hypothesis that $A$ is nowhere dense in $X$.

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Take an open ball $B$. Since $A$ is nowhere dense, $B$ cannot be contained in $\overline{A}$. Pick $b \in B, b \notin \overline{A}$. Since $\overline{A}^c$ is open, there is a ball $B'$ containing $b$ that doesn't intersect $A$. Let $U:=B \cap B'$. Now, pick an open ball $B_1 \subset U$ and we are done.

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