[proofplan] We prove the dense $G_\delta$ form directly: given dense open sets $G_1, G_2, \ldots$ and a nonempty open set $U \subset X$, we construct a Cauchy sequence whose limit lies in $U \cap \bigcap_{n=1}^\infty G_n$. At stage $n$, we choose a closed ball $\overline{B}(x_n, r_n) \subset G_n$ with radii $r_n \to 0$ and each ball nested inside the previous one. Completeness forces the centres to converge, and the nesting ensures the limit belongs to every $G_n$. The category form follows by contraposition and the relationship between dense open sets and nowhere dense sets. [/proofplan]

[step:Construct nested closed balls inside each $G_n$]Let $(G_n)_{n=1}^\infty$ be a sequence of dense open subsets of $X$, and let $U \subset X$ be a nonempty open set. We must show $U \cap \bigcap_{n=1}^\infty G_n \neq \varnothing$. Since $G_1$ is dense in $X$ and $U$ is nonempty and open, the intersection $U \cap G_1$ is nonempty and open. Choose $x_1 \in U \cap G_1$. Since $U \cap G_1$ is open, there exists $r_1 > 0$ with $r_1 < 1$ such that \begin{align*} \overline{B}(x_1, r_1) \subset U \cap G_1. \end{align*} The open ball $B(x_1, r_1)$ is nonempty and open, and $G_2$ is dense, so $B(x_1, r_1) \cap G_2$ is nonempty and open. Choose $x_2 \in B(x_1, r_1) \cap G_2$ and $r_2 > 0$ with $r_2 < \frac{1}{2}$ such that \begin{align*} \overline{B}(x_2, r_2) \subset B(x_1, r_1) \cap G_2. \end{align*} Proceeding inductively, suppose we have chosen $x_n \in X$ and $r_n > 0$ with $r_n < \frac{1}{n}$ such that $\overline{B}(x_n, r_n) \subset B(x_{n-1}, r_{n-1}) \cap G_n$. Since $B(x_n, r_n)$ is nonempty and open and $G_{n+1}$ is dense, the intersection $B(x_n, r_n) \cap G_{n+1}$ is nonempty and open. Choose $x_{n+1} \in B(x_n, r_n) \cap G_{n+1}$ and $r_{n+1} > 0$ with $r_{n+1} < \frac{1}{n+1}$ such that \begin{align*} \overline{B}(x_{n+1}, r_{n+1}) \subset B(x_n, r_n) \cap G_{n+1}. \end{align*} This produces sequences $(x_n)_{n=1}^\infty$ in $X$ and $(r_n)_{n=1}^\infty$ in $(0, \infty)$ with $r_n < \frac{1}{n}$ and \begin{align*} \overline{B}(x_{n+1}, r_{n+1}) \subset B(x_n, r_n) \subset \overline{B}(x_n, r_n) \subset G_n \quad \text{for all } n \ge 1. \end{align*}[/step]

[guided]The strategy is to build a sequence of nested closed balls, each contained in the corresponding dense open set $G_n$, with radii shrinking to zero. The nesting will force the centres to be Cauchy, and completeness will provide a limit point in the intersection. We start the construction. The set $U$ is nonempty and open, and $G_1$ is dense, meaning $G_1$ intersects every nonempty open set. Therefore $U \cap G_1 \neq \varnothing$, and since both $U$ and $G_1$ are open, $U \cap G_1$ is open. Pick any $x_1 \in U \cap G_1$. Since $U \cap G_1$ is open, there exists some radius $\varepsilon > 0$ with $B(x_1, \varepsilon) \subset U \cap G_1$. We choose $r_1 := \min(\varepsilon / 2, 1)$, which ensures $\overline{B}(x_1, r_1) \subset B(x_1, \varepsilon) \subset U \cap G_1$ and $r_1 < 1$. Why do we work with closed balls $\overline{B}(x_n, r_n)$ contained inside open balls $B(x_{n-1}, r_{n-1})$? Because we need the nesting to be strict — $\overline{B}(x_{n+1}, r_{n+1}) \subset B(x_n, r_n)$ — so that the sequence of closed balls is decreasing. If we only had $\overline{B}(x_{n+1}, r_{n+1}) \subset \overline{B}(x_n, r_n)$, we could not guarantee the radii shrink or that the centres converge. At stage $n + 1$: we have the open ball $B(x_n, r_n)$, which is nonempty (it contains $x_n$) and open. Since $G_{n+1}$ is dense, $B(x_n, r_n) \cap G_{n+1} \neq \varnothing$. This intersection is open (both sets are open), so we can choose $x_{n+1}$ in it and find $r_{n+1} < \frac{1}{n+1}$ with $\overline{B}(x_{n+1}, r_{n+1}) \subset B(x_n, r_n) \cap G_{n+1}$. The bound $r_n < \frac{1}{n}$ ensures $r_n \to 0$, which will be used in the next step to prove the sequence is Cauchy.[/guided]

[step:Show the sequence $(x_n)$ is Cauchy using the nesting and $r_n \to 0$]For $m > n$, the nesting $\overline{B}(x_{m}, r_{m}) \subset B(x_{m-1}, r_{m-1}) \subset \cdots \subset B(x_n, r_n)$ gives $x_m \in B(x_n, r_n)$, so \begin{align*} d(x_m, x_n) < r_n < \frac{1}{n}. \end{align*} Given $\varepsilon > 0$, choose $N \in \mathbb{N}$ with $\frac{1}{N} < \varepsilon$. For all $m, n \ge N$ (say $m \ge n$), we have $d(x_m, x_n) < \frac{1}{n} \le \frac{1}{N} < \varepsilon$. Therefore $(x_n)_{n=1}^\infty$ is Cauchy in $(X, d)$.[/step]

[guided]We need to verify that the centres $(x_n)$ form a Cauchy sequence. The key observation is that the nesting is strict: for $m > n$, \begin{align*} x_m \in \overline{B}(x_m, r_m) \subset B(x_{m-1}, r_{m-1}) \subset \cdots \subset B(x_n, r_n), \end{align*} so $d(x_m, x_n) < r_n$. Since $r_n < \frac{1}{n} \to 0$, for any $\varepsilon > 0$ we can choose $N$ with $\frac{1}{N} < \varepsilon$, and then for all $m, n \ge N$ (assuming $m \ge n$ without loss of generality): \begin{align*} d(x_m, x_n) < r_n \le r_N < \frac{1}{N} < \varepsilon. \end{align*} This is the Cauchy condition. Note that we have not yet used completeness — we have only used the density of the $G_n$ and the fact that open sets in a metric space contain closed balls of smaller radius.[/guided]

[step:Use completeness to obtain a limit and verify it lies in $U \cap \bigcap_{n=1}^\infty G_n$]Since $(X, d)$ is complete, the Cauchy sequence $(x_n)_{n=1}^\infty$ converges to some $x \in X$. For each fixed $n \ge 1$ and every $m > n$, we have $x_m \in \overline{B}(x_n, r_n)$. Since $\overline{B}(x_n, r_n)$ is closed (as a closed ball in a metric space) and $x_m \to x$, the limit satisfies $x \in \overline{B}(x_n, r_n)$. By construction, $\overline{B}(x_n, r_n) \subset G_n$, so $x \in G_n$. Since this holds for every $n \ge 1$: \begin{align*} x \in \bigcap_{n=1}^\infty G_n. \end{align*} Moreover, $\overline{B}(x_1, r_1) \subset U$, so $x \in \overline{B}(x_1, r_1) \subset U$, giving $x \in U \cap \bigcap_{n=1}^\infty G_n$. Since $U$ was an arbitrary nonempty open set, $\bigcap_{n=1}^\infty G_n$ is dense in $X$.[/step]

[guided]This is where completeness enters. The Cauchy sequence $(x_n)$ converges to some limit $x \in X$. We must verify two things: that $x$ lies in every $G_n$, and that $x$ lies in $U$. **Membership in $G_n$.** Fix $n \ge 1$. For every $m > n$, the nesting gives $x_m \in \overline{B}(x_n, r_n)$. The closed ball $\overline{B}(x_n, r_n) = \{y \in X : d(x_n, y) \le r_n\}$ is closed (as the preimage of $[0, r_n]$ under the continuous map $y \mapsto d(x_n, y)$). Since the sequence $(x_m)_{m > n}$ lies in the closed set $\overline{B}(x_n, r_n)$ and converges to $x$, we conclude $x \in \overline{B}(x_n, r_n)$. By construction, $\overline{B}(x_n, r_n) \subset G_n$, so $x \in G_n$. This holds for every $n$, so $x \in \bigcap_{n=1}^\infty G_n$. **Membership in $U$.** The outermost ball satisfies $\overline{B}(x_1, r_1) \subset U \cap G_1 \subset U$. Since $x \in \overline{B}(x_1, r_1)$, we have $x \in U$. Therefore $x \in U \cap \bigcap_{n=1}^\infty G_n$, which shows this intersection is nonempty. Since $U$ was an arbitrary nonempty open set, $\bigcap_{n=1}^\infty G_n$ meets every nonempty open set — that is, $\bigcap_{n=1}^\infty G_n$ is dense in $X$. This completes the proof of the dense $G_\delta$ form.[/guided]

[step:Deduce the category form from the dense $G_\delta$ form]Suppose $(F_n)_{n=1}^\infty$ is a sequence of nowhere dense closed subsets of $X$. For each $n$, define $G_n := X \setminus F_n$. Each $G_n$ is open (as the complement of a closed set). We claim each $G_n$ is dense: by the [Equivalent Characterisations of Nowhere Dense Sets](/theorems/1084), $F_n$ is nowhere dense if and only if $X \setminus \overline{F_n}$ is dense. Since $F_n$ is closed, $\overline{F_n} = F_n$, so $G_n = X \setminus F_n = X \setminus \overline{F_n}$ is dense. By the dense $G_\delta$ form, $\bigcap_{n=1}^\infty G_n$ is dense in $X$. In particular, $\bigcap_{n=1}^\infty G_n \neq \varnothing$ (provided $X \neq \varnothing$), so \begin{align*} X \neq \bigcup_{n=1}^\infty F_n. \end{align*} Moreover, $\bigcap_{n=1}^\infty G_n = X \setminus \bigcup_{n=1}^\infty F_n$ is dense, which means $\bigcup_{n=1}^\infty F_n$ has empty interior: if $V \subset \bigcup_{n=1}^\infty F_n$ were a nonempty open set, then $V \cap \bigcap_{n=1}^\infty G_n = \varnothing$, contradicting the density of $\bigcap_{n=1}^\infty G_n$.[/step]

[guided]The category form states that a countable union of nowhere dense closed sets cannot have nonempty interior. We derive this from the dense $G_\delta$ form by passing to complements. Let $(F_n)_{n=1}^\infty$ be a sequence of nowhere dense closed subsets of $X$. Set $G_n := X \setminus F_n$, which is open since $F_n$ is closed. We need to verify that each $G_n$ is dense. By the [Equivalent Characterisations of Nowhere Dense Sets](/theorems/1084), a set $A$ is nowhere dense if and only if $X \setminus \overline{A}$ is dense in $X$ (characterisation (ii)). Since $F_n$ is already closed, $\overline{F_n} = F_n$, so $X \setminus \overline{F_n} = X \setminus F_n = G_n$ is dense. The dense $G_\delta$ form now applies: $\bigcap_{n=1}^\infty G_n$ is dense in $X$. By De Morgan's law, \begin{align*} \bigcap_{n=1}^\infty G_n = \bigcap_{n=1}^\infty (X \setminus F_n) = X \setminus \bigcup_{n=1}^\infty F_n. \end{align*} Since this set is dense, its complement $\bigcup_{n=1}^\infty F_n$ has empty interior. Indeed, if there were a nonempty open set $V \subset \bigcup_{n=1}^\infty F_n$, then $V$ would be disjoint from $X \setminus \bigcup_{n=1}^\infty F_n = \bigcap_{n=1}^\infty G_n$, contradicting the density of $\bigcap_{n=1}^\infty G_n$ (which must meet every nonempty open set). In particular, when $X$ is nonempty, $X$ itself is a nonempty open set, so $X \not\subset \bigcup_{n=1}^\infty F_n$, giving $X \neq \bigcup_{n=1}^\infty F_n$.[/guided]