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The statement that the countable intersection of open dense sets is dense is equivalent to the statement that the countable union of nowhere dense sets contains no balls.

When proving the equivalence I came across the following,

Suppose, $\{\overline{E}_n \}_{n=1}^{\infty}$ is countable collection of nowhere dense sets such that the countable union, $$\bigcup_{n=1}^{\infty}\overline{E}_n $$ contains no balls. Then $$ \bigcap_{n=1}^{\infty}\overline{E_n}^c$$ is the countable intersection of open dense sets. Now, since, $$\left (\bigcup_{n=1}^{\infty}\overline{E}_n \right)^c = \bigcap_{n=1}^{\infty}\overline{E_n}^c$$ We have that, $$\bigcup_{n=1}^{\infty}\overline{E}_n \cup \bigcap_{n=1}^{\infty}\overline{E_n}^c = X$$

Let $x \in X$, and let $B_x$ be ball around $x$, then $B_x \not \subset \bigcup_{n=1}^{\infty}\overline{E}_n$, thus $B_x \subset \bigcap_{n=1}^{\infty}\overline{E_n}^c$

Since we can do this for every $x \in X$, it seems that for every $B_x \in \bigcap_{n=1}^{\infty}\overline{E_n}^c$. Since we can do this for every $x \in X$, it seems that $\bigcap_{n=1}^{\infty}\overline{E_n}^c=X$. However, this is not necessarily true, density only implies that $B_x \cap \bigcap_{n=1}^{\infty}\overline{E_n}^c \neq \emptyset$

What is going wrong in my logic?

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Your mistake is in the line "Let $x\in X,$ and let $B_x$ be a ball around $x$..." Just because $B_x$ is not contained in some set $\cup \bar E_n$ by no means implies $B_x$ is contained in the complement of that set! You could have lots of points of $B_x$ in $\cup \bar E_n$, critically, you could have $x$ in there-just not all of them.

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