[proofplan] We prove each direction separately. The forward direction ($G_\delta \Rightarrow$ Polish) is the substantive one: given $G = \bigcap_{n=1}^\infty U_n$ with each $U_n$ open in the Polish space $(X, d)$, we embed $G$ into the product $X \times \mathbb{R}^\mathbb{N}$ via $x \mapsto (x, f_1(x), f_2(x), \ldots)$ where $f_n(x) = 1/\operatorname{dist}(x, X \setminus U_n)$. These distance-to-complement functions blow up as $x$ approaches the boundary of any $U_n$, which forces any convergent sequence in the image to have its limit inside $G$ — making the image a closed subset of the Polish product, hence Polish. The reverse direction ($G$ Polish $\Rightarrow$ $G_\delta$) uses the interplay between the ambient metric $d$ and a complete metric $d_G$ on $G$: the identity map's [continuity](/page/Continuity) provides, for each $n$, a family of $d$-balls whose traces on $G$ are $d_G$-small, and the completeness of $d_G$ forces any point in the resulting [countable](/page/Countable%20Set) intersection to lie in $G$. [/proofplan]

[step:Show that a Polish subspace of a metrizable space is $G_\delta$]Let $G \subset X$ carry a Polish [topology](/page/Topology) that agrees with the subspace topology inherited from $(X, d)$. Let $d_G$ be a complete metric on $G$ generating this topology. Since both $d|_G$ and $d_G$ generate the same topology on $G$, the identity map $\operatorname{id}: (G, d|_G) \to (G, d_G)$ is [continuous](/page/Continuity). For each $n \in \mathbb{N}$ and each $x \in G$, continuity of $\operatorname{id}$ at $x$ provides a radius $r_n(x) > 0$ satisfying $r_n(x) \le 1/n$ and \begin{align*} B_d(x, r_n(x)) \cap G \subset B_{d_G}(x, 1/n). \end{align*} Define the [open sets](/page/Open%20Set) \begin{align*} V_n := \bigcup_{x \in G} B_d(x, r_n(x)). \end{align*} Each $V_n$ is open in $X$ (as a union of $d$-open balls) and $G \subset V_n$ (since $x \in B_d(x, r_n(x))$ for every $x \in G$). We show $\bigcap_{n=1}^\infty V_n \subset G$. Let $y \in \bigcap_{n=1}^\infty V_n$. For each $n$, choose $x_n \in G$ with $d(y, x_n) < r_n(x_n) \le 1/n$. Then $x_n \to y$ in $(X, d)$. For $m > n$, once $m$ is large enough that $d(x_m, x_n) < r_n(x_n)$ (which holds since $d(x_m, x_n) \le d(x_m, y) + d(y, x_n) < 1/m + 1/n$ and $r_n(x_n) \le 1/n$, so for $m$ large we have $d(x_m, x_n) < r_n(x_n)$), we get $x_m \in B_d(x_n, r_n(x_n)) \cap G \subset B_{d_G}(x_n, 1/n)$, giving $d_G(x_m, x_n) < 1/n$. Therefore $(x_n)$ is Cauchy in the complete metric $d_G$, so $x_n \to z$ for some $z \in G$ in $d_G$. Since $d_G$ and $d|_G$ generate the same topology, $x_n \to z$ in $(X, d)$ as well. By uniqueness of limits in the Hausdorff space $X$, $y = z \in G$. Therefore $G = \bigcap_{n=1}^\infty V_n$ is a $G_\delta$ subset of $X$.[/step]

[guided]The goal is to express $G$ as a [countable](/page/Countable%20Set) intersection of open subsets of $X$. **Why is this non-trivial?** Completeness is not a topological invariant — [Completeness is Not a Topological Invariant](/theorems/293) shows that $(0,1)$ and $\mathbb{R}$ are homeomorphic but only $\mathbb{R}$ is complete. So knowing that $G$ is "complete" does not directly yield a topological conclusion like being $G_\delta$. The argument must exploit the specific interplay between two metrics: the ambient metric $d$ (from $X$) and the complete metric $d_G$ (from the Polish structure on $G$). Let $d_G$ be a complete metric on $G$ generating the subspace topology $\tau_d|_G$. Since both $d|_G$ and $d_G$ generate the same topology on $G$, the identity map $\operatorname{id}: (G, d|_G) \to (G, d_G)$ is continuous. For each $n \in \mathbb{N}$ and each $x \in G$, continuity of $\operatorname{id}$ at $x$ provides a radius $r_n(x) > 0$ such that \begin{align*} B_d(x, r_n(x)) \cap G \subset B_{d_G}(x, 1/n). \end{align*} We may assume $r_n(x) \le 1/n$ (replace $r_n(x)$ by $\min(r_n(x), 1/n)$ if needed — this only shrinks the ball, preserving the inclusion). Define the open sets \begin{align*} V_n := \bigcup_{x \in G} B_d(x, r_n(x)). \end{align*} Each $V_n$ is open in $X$ (a union of $d$-open balls) and $G \subset V_n$ (since $x \in B_d(x, r_n(x))$). We show $\bigcap_{n=1}^\infty V_n \subset G$, establishing $G = \bigcap_{n=1}^\infty V_n$. Take $y \in \bigcap_{n=1}^\infty V_n$. For each $n$, choose $x_n \in G$ with $y \in B_d(x_n, r_n(x_n))$, so $d(y, x_n) < r_n(x_n) \le 1/n$. Therefore $x_n \to y$ in $(X, d)$. **Showing $(x_n)$ is $d_G$-Cauchy.** Fix $\varepsilon > 0$ and choose $N$ with $1/N < \varepsilon$. For $m > n \ge N$, we estimate $d(x_m, x_n) \le d(x_m, y) + d(y, x_n) < 1/m + 1/n$. For $m$ sufficiently large (specifically $m > n/(r_n(x_n) \cdot n - 1)$, but concretely: since $1/m + 1/n \to 1/n$ as $m \to \infty$ and $r_n(x_n) \le 1/n$, for large $m$ we have $d(x_m, x_n) < r_n(x_n)$). Then $x_m \in B_d(x_n, r_n(x_n)) \cap G \subset B_{d_G}(x_n, 1/n)$, so $d_G(x_m, x_n) < 1/n \le 1/N < \varepsilon$. This confirms $(x_n)$ is Cauchy in $(G, d_G)$. Since $d_G$ is complete, $x_n \to z$ for some $z \in G$ in the $d_G$-metric. Since $d_G$ and $d|_G$ generate the same topology, $x_n \to z$ also in $(X, d)$. The ambient space $X$ is [metrizable](/page/Metrizable%20Space) (hence Hausdorff), so limits are unique: $y = z \in G$. This proves $\bigcap_{n=1}^\infty V_n \subset G$, so $G$ is $G_\delta$ in $X$.[/guided]

[step:Define the distance-to-complement functions and the product embedding]For the forward direction, let $(X, d)$ be a Polish space and $G = \bigcap_{n=1}^\infty U_n$ where each $U_n$ is open in $X$. Replacing $U_n$ by $U_1 \cap \cdots \cap U_n$, we may assume $U_1 \supset U_2 \supset \cdots \supset G$. For each $n \in \mathbb{N}$, define \begin{align*} f_n: G &\to \mathbb{R} \\ x &\mapsto \frac{1}{\operatorname{dist}(x, X \setminus U_n)}, \end{align*} where $\operatorname{dist}(x, X \setminus U_n) := \inf\{d(x, z) : z \in X \setminus U_n\}$. Since $x \in G \subset U_n$ and $U_n$ is open, $\operatorname{dist}(x, X \setminus U_n) > 0$, so $f_n$ is well-defined and positive on $G$. Each $f_n$ is [continuous](/page/Continuity) on $G$: the map $x \mapsto \operatorname{dist}(x, X \setminus U_n)$ is $1$-Lipschitz on $(X, d)$ (since $|\operatorname{dist}(x, A) - \operatorname{dist}(y, A)| \le d(x, y)$ for any $A \subset X$), positive on $G$, and $t \mapsto 1/t$ is continuous on $(0, \infty)$. Define the embedding \begin{align*} \Phi: G &\to X \times \mathbb{R}^\mathbb{N} \\ x &\mapsto (x, f_1(x), f_2(x), \ldots). \end{align*} Equip $\mathbb{R}^\mathbb{N}$ with the [product topology](/page/Product%20Topology), which is [metrizable](/page/Metrizable%20Space) by [Metrizability of Countable Products](/theorems/959) (each factor $(\mathbb{R}, |\cdot|)$ is metrizable). Equip $X \times \mathbb{R}^\mathbb{N}$ with the product [topology](/page/Topology).[/step]

[guided]The idea is to encode the $G_\delta$ structure metrically. The functions $f_n$ serve as "sentinels" guarding the boundary: as $x$ approaches $X \setminus U_n$, $\operatorname{dist}(x, X \setminus U_n) \to 0$, so $f_n(x) \to +\infty$. A [Cauchy sequence](/page/Cauchy%20Sequence) in the product cannot have $f_n$ blowing up (Cauchy sequences are bounded in each coordinate), so it is forced to stay away from the boundary of every $U_n$ — hence its limit remains in $\bigcap U_n = G$. For each $n \in \mathbb{N}$, define \begin{align*} f_n: G &\to \mathbb{R} \\ x &\mapsto \frac{1}{\operatorname{dist}(x, X \setminus U_n)}. \end{align*} We verify well-definedness. Since $x \in G \subset U_n$ and $U_n$ is open, there exists $r > 0$ with $B_d(x, r) \subset U_n$, so $\operatorname{dist}(x, X \setminus U_n) \ge r > 0$ and $f_n(x) < \infty$. Continuity of $f_n$ follows from two facts: (i) the map $x \mapsto \operatorname{dist}(x, X \setminus U_n)$ is $1$-Lipschitz on $(X, d)$ (by the triangle inequality: for any $A \subset X$ and $x, y \in X$, $\operatorname{dist}(x, A) \le d(x, y) + \operatorname{dist}(y, A)$, so $|\operatorname{dist}(x, A) - \operatorname{dist}(y, A)| \le d(x, y)$), and (ii) $t \mapsto 1/t$ is continuous on $(0, \infty)$. The composition is continuous on $G$. We embed $G$ into a product: \begin{align*} \Phi: G &\to X \times \mathbb{R}^\mathbb{N} \\ x &\mapsto (x, f_1(x), f_2(x), \ldots). \end{align*} By [Metrizability of Countable Products](/theorems/959), the [countable](/page/Countable%20Set) product $\mathbb{R}^\mathbb{N}$ is metrizable (each factor $(\mathbb{R}, |\cdot|)$ is metrizable). The full product $X \times \mathbb{R}^\mathbb{N}$ is also metrizable (as a countable product of metrizable spaces). Moreover, it is Polish: $X$ is Polish and $\mathbb{R}$ is Polish, and a countable product of Polish spaces is Polish — completeness follows from the bounded product metric $\rho((a_k), (b_k)) = \sum_{k=0}^\infty 2^{-k} \min(d_k(a_k, b_k), 1)$ applied to complete metrics, and separability follows from [Countable Products of Separable Spaces](/theorems/946).[/guided]

[step:Show $\Phi$ is a homeomorphism onto its image]The map $\Phi$ is injective: the first coordinate determines the point. It is [continuous](/page/Continuity): by the universal property of the [product topology](/page/Product%20Topology), it suffices that each component is continuous — the first component is the inclusion $G \hookrightarrow X$, and each $f_n$ is continuous. We show $\Phi$ is open onto $\Phi(G)$. Let $V \subset G$ be open in the subspace [topology](/page/Topology), so $V = G \cap W$ for some open $W \subset X$. Let $\pi_0: X \times \mathbb{R}^\mathbb{N} \to X$ denote the projection onto the first factor. Since $\pi_0 \circ \Phi = \operatorname{id}_G$, we have \begin{align*} \Phi(V) = \Phi(G) \cap \pi_0^{-1}(W). \end{align*} The set $\pi_0^{-1}(W)$ is open in $X \times \mathbb{R}^\mathbb{N}$ (since $\pi_0$ is continuous), so $\Phi(V)$ is open in $\Phi(G)$. Therefore $\Phi: G \to \Phi(G)$ is a homeomorphism.[/step]

[guided]We verify three properties: injectivity, continuity, and openness of $\Phi$ onto its image. **Injectivity.** If $\Phi(x) = \Phi(y)$, then the first coordinates agree: $x = y$. **Continuity.** By the universal property of the product topology, $\Phi$ is continuous if and only if each component is continuous. The first component is the inclusion $G \hookrightarrow X$ (continuous since $G$ carries the subspace topology). Each subsequent component $f_n: G \to \mathbb{R}$ is continuous as established above. **Openness onto $\Phi(G)$.** We show $\Phi$ maps open subsets of $G$ to relatively open subsets of $\Phi(G)$. Let $V \subset G$ be open, so $V = G \cap W$ for an [open set](/page/Open%20Set) $W \subset X$. The projection $\pi_0: X \times \mathbb{R}^\mathbb{N} \to X$ onto the first factor is continuous, so $\pi_0^{-1}(W)$ is open in $X \times \mathbb{R}^\mathbb{N}$. We claim \begin{align*} \Phi(V) = \Phi(G) \cap \pi_0^{-1}(W). \end{align*} Indeed, $\Phi(x) \in \pi_0^{-1}(W)$ if and only if $\pi_0(\Phi(x)) = x \in W$. So $\Phi(x) \in \Phi(G) \cap \pi_0^{-1}(W)$ if and only if $x \in G \cap W = V$, if and only if $\Phi(x) \in \Phi(V)$. The set $\Phi(V)$ is thus the intersection of $\Phi(G)$ with an open set, hence open in $\Phi(G)$. Therefore $\Phi: G \to \Phi(G)$ is a continuous bijection with continuous inverse — a homeomorphism.[/guided]

[step:Show $\Phi(G)$ is closed in $X \times \mathbb{R}^\mathbb{N}$]Let $(x_k)_{k=1}^\infty$ be a sequence in $G$ such that $\Phi(x_k) \to (y, t_1, t_2, \ldots)$ in $X \times \mathbb{R}^\mathbb{N}$. Convergence in the product means $x_k \to y$ in $(X, d)$ and $f_n(x_k) \to t_n$ in $\mathbb{R}$ for each $n$. We show $y \in G$. Fix $n \in \mathbb{N}$. Since $f_n(x_k) \to t_n < \infty$, the sequence $(f_n(x_k))$ is bounded: there exists $M_n > 0$ with $f_n(x_k) \le M_n$ for all $k$. By definition of $f_n$, \begin{align*} \operatorname{dist}(x_k, X \setminus U_n) = \frac{1}{f_n(x_k)} \ge \frac{1}{M_n} > 0 \quad \text{for all } k. \end{align*} The distance function $z \mapsto \operatorname{dist}(z, X \setminus U_n)$ is $1$-Lipschitz, so [continuity](/page/Continuity) and $x_k \to y$ give \begin{align*} \operatorname{dist}(y, X \setminus U_n) = \lim_{k \to \infty} \operatorname{dist}(x_k, X \setminus U_n) \ge \frac{1}{M_n} > 0. \end{align*} Therefore $y \notin X \setminus U_n$, i.e., $y \in U_n$. Since $n$ was arbitrary, $y \in \bigcap_{n=1}^\infty U_n = G$. With $y \in G$ established, continuity of $f_n$ gives $f_n(y) = \lim_{k} f_n(x_k) = t_n$, so $(y, t_1, t_2, \ldots) = \Phi(y) \in \Phi(G)$. Every convergent sequence in $\Phi(G)$ has its limit in $\Phi(G)$; since $X \times \mathbb{R}^\mathbb{N}$ is [metrizable](/page/Metrizable%20Space), sequential closedness implies closedness. Therefore $\Phi(G)$ is closed.[/step]

[guided]This is the heart of the argument — the step where the distance-to-complement functions do their work. Let $(x_k)_{k=1}^\infty$ be a sequence in $G$ with $\Phi(x_k) \to (y, t_1, t_2, \ldots)$ in $X \times \mathbb{R}^\mathbb{N}$. Convergence in the product means $x_k \to y$ in $(X, d)$ and $f_n(x_k) \to t_n$ in $\mathbb{R}$ for each $n$. **Why might $y$ fail to lie in $G$?** The set $G = \bigcap U_n$ could be a "thin" intersection — a sequence in $G$ converging in $X$ might drift toward the boundary of some $U_n$ and escape. The functions $f_n$ prevent this: if $x_k$ were approaching the boundary of $U_n$, then $\operatorname{dist}(x_k, X \setminus U_n) \to 0$ and $f_n(x_k) \to +\infty$, contradicting the convergence $f_n(x_k) \to t_n < \infty$. Fix $n \in \mathbb{N}$. The real sequence $(f_n(x_k))_{k=1}^\infty$ converges to $t_n$, so it is bounded: $f_n(x_k) \le M_n$ for some $M_n > 0$ and all $k$. Unraveling $f_n(x_k) = 1/\operatorname{dist}(x_k, X \setminus U_n)$: \begin{align*} \operatorname{dist}(x_k, X \setminus U_n) = \frac{1}{f_n(x_k)} \ge \frac{1}{M_n} > 0. \end{align*} The function $z \mapsto \operatorname{dist}(z, X \setminus U_n)$ is $1$-Lipschitz (for any $A \subset X$ and $z_1, z_2 \in X$: $\operatorname{dist}(z_1, A) \le d(z_1, z_2) + \operatorname{dist}(z_2, A)$ by the triangle inequality, so $|\operatorname{dist}(z_1, A) - \operatorname{dist}(z_2, A)| \le d(z_1, z_2)$). Since $x_k \to y$ in $(X, d)$, \begin{align*} \operatorname{dist}(y, X \setminus U_n) = \lim_{k \to \infty} \operatorname{dist}(x_k, X \setminus U_n) \ge \frac{1}{M_n} > 0. \end{align*} Therefore $y$ has positive distance from $X \setminus U_n$, forcing $y \in U_n$. Since $n$ was arbitrary, $y \in \bigcap_{n=1}^\infty U_n = G$. With $y \in G$ established, continuity of each $f_n$ on $G$ gives $f_n(y) = \lim_k f_n(x_k) = t_n$, so $(y, t_1, t_2, \ldots) = \Phi(y) \in \Phi(G)$. We have shown that every convergent sequence in $\Phi(G)$ has its limit in $\Phi(G)$. Since $X \times \mathbb{R}^\mathbb{N}$ is metrizable (sequential closedness equals closedness in metrizable spaces), $\Phi(G)$ is closed.[/guided]

[step:Conclude that $G$ is Polish]The space $X \times \mathbb{R}^\mathbb{N}$ is Polish: $X$ is Polish by hypothesis, $\mathbb{R}$ is Polish (complete and [separable](/page/Separable)), and a [countable](/page/Countable%20Set) product of Polish spaces is Polish. For completeness: the bounded product metric $\rho((a_k), (b_k)) = \sum_{k=0}^\infty 2^{-k} \min(d_k(a_k, b_k), 1)$ applied to complete metrics $d_k$ on each factor is complete (a $\rho$-[Cauchy sequence](/page/Cauchy%20Sequence) is Cauchy in each coordinate, hence convergent in each coordinate, and coordinate-wise convergence implies $\rho$-convergence). For separability: by [Countable Products of Separable Spaces](/theorems/946), a countable product of [separable spaces](/page/Separable%20Space) is separable. The set $\Phi(G)$ is a closed subset of the Polish space $X \times \mathbb{R}^\mathbb{N}$. By [Completeness and Closedness of Subspaces](/theorems/287) (hypothesis: $X \times \mathbb{R}^\mathbb{N}$ is complete and $\Phi(G)$ is closed), $\Phi(G)$ is complete in the induced metric. By [Subspaces of Separable Metrizable Spaces](/theorems/942) (hypothesis: $X \times \mathbb{R}^\mathbb{N}$ is separable and [metrizable](/page/Metrizable%20Space)), $\Phi(G)$ is separable. Therefore $\Phi(G)$ is Polish. Since $\Phi: G \to \Phi(G)$ is a homeomorphism, $G$ is homeomorphic to a Polish space, hence Polish in the subspace [topology](/page/Topology). Combined with the reverse direction, this completes the proof: $G \subset X$ is Polish (in the subspace topology) if and only if $G$ is $G_\delta$ in $X$.[/step]

[guided]We assemble the ingredients. The product $X \times \mathbb{R}^\mathbb{N}$ is Polish because: **Completeness.** Choose a complete metric $d_X$ on $X$ (exists since $X$ is Polish) and the standard complete metric $|\cdot|$ on each copy of $\mathbb{R}$. The bounded product metric \begin{align*} \rho\!\left((x, (s_n)), (y, (t_n))\right) := \min(d_X(x, y), 1) + \sum_{n=1}^\infty 2^{-n} \min(|s_n - t_n|, 1) \end{align*} is complete on $X \times \mathbb{R}^\mathbb{N}$. To verify: a $\rho$-Cauchy sequence $(\Phi(x_k))_k$ is Cauchy in each coordinate (since each projection is Lipschitz with respect to $\rho$), hence convergent in each coordinate by completeness of $X$ and $\mathbb{R}$. Coordinate-wise convergence implies $\rho$-convergence (by dominated convergence applied to the series defining $\rho$, with dominating series $\sum 2^{-n}$). **Separability.** $X$ is separable, each $\mathbb{R}$ is separable, and by [Countable Products of Separable Spaces](/theorems/946) (hypothesis: countably many separable factors), the countable product is separable. The set $\Phi(G)$ is closed in this Polish space (shown in the previous step). [Completeness and Closedness of Subspaces](/theorems/287) states that a closed subset of a complete [metric space](/page/Metric%20Space) is complete in the induced metric. We verify: $X \times \mathbb{R}^\mathbb{N}$ is complete (just established) and $\Phi(G)$ is closed (previous step). Therefore $\Phi(G)$ is complete. By [Subspaces of Separable Metrizable Spaces](/theorems/942), every subspace of a separable metrizable space is separable. We verify: $X \times \mathbb{R}^\mathbb{N}$ is separable (just established) and metrizable (by construction). Therefore $\Phi(G)$ is separable. Combining: $\Phi(G)$ is a complete, separable metrizable space — i.e., Polish. Since $\Phi: G \to \Phi(G)$ is a homeomorphism, $G$ inherits the Polish property. Together with the reverse direction, this gives the full equivalence.[/guided]