[proofplan] The strategy is to embed $X$ into the Hilbert cube $[0,1]^{\mathbb{N}}$. Since $X$ is regular and second-[countable](/page/Countable%20Set), it is normal (every regular Lindelöf space is normal, and second-countable implies Lindelöf). We enumerate a countable basis and, for each basis element, apply [Urysohn's Lemma](/theorems/887) to separate its closure from the complement of a larger basis element. The resulting countable family of continuous functions $f_n: X \to [0,1]$ separates points from [closed sets](/page/Closed%20Set). We then define an embedding $F: X \to [0,1]^{\mathbb{N}}$ by $F(x) = (f_1(x), f_2(x), \ldots)$, show it is injective and continuous, and verify that $F$ maps [open sets](/page/Open%20Set) in $X$ to relatively open sets in $F(X)$ — making $F$ a homeomorphism onto its image. Since $[0,1]^{\mathbb{N}}$ is metrizable (via the weighted supremum metric), so is $X$. [/proofplan] [step:Establish normality of $X$ from regularity and second countability] Let $\mathcal{B} = \{B_n\}_{n \in \mathbb{N}}$ be a countable basis for $X$. Since $X$ is second-countable, every open cover of $X$ admits a countable subcover (the Lindelöf property). We use this together with regularity to prove normality. Let $C_0, C_1 \subset X$ be disjoint closed sets. For each $x \in C_0$, since $C_1$ is closed, the set $X \setminus C_1$ is an open neighbourhood of $x$. By regularity, there exists an open set $V_x$ with $x \in V_x \subset \overline{V_x} \subset X \setminus C_1$. Since $\mathcal{B}$ is a basis, there exists $B_{n_x} \in \mathcal{B}$ with $x \in B_{n_x} \subset V_x$. The collection $\{B_{n_x}\}_{x \in C_0}$ covers $C_0$. Since $X$ is Lindelöf, there is a countable subcover $\{U_k\}_{k \in \mathbb{N}}$ of $C_0$ such that $\overline{U_k} \cap C_1 = \varnothing$ for each $k$. By the same argument applied to $C_1$ and $C_0$, there is a countable collection $\{V_k\}_{k \in \mathbb{N}}$ covering $C_1$ with $\overline{V_k} \cap C_0 = \varnothing$ for each $k$. Define \begin{align*} U_k' &:= U_k \setminus \bigcup_{j=1}^{k} \overline{V_j}, \qquad V_k' := V_k \setminus \bigcup_{j=1}^{k} \overline{U_j}. \end{align*} Each $U_k'$ is open (as the difference of an open set and a finite union of closed sets), and similarly for $V_k'$. Set $U := \bigcup_{k=1}^{\infty} U_k'$ and $V := \bigcup_{k=1}^{\infty} V_k'$. Then $C_0 \subset U$, $C_1 \subset V$, and $U \cap V = \varnothing$. To verify disjointness: if $x \in U_k' \cap V_j'$ for some $k, j$, then $x \in U_k$ and $x \in V_j$. If $j \le k$, then $x \in U_k' \subset X \setminus \overline{V_j}$, contradicting $x \in V_j \subset \overline{V_j}$. If $k < j$, then $x \in V_j' \subset X \setminus \overline{U_k}$, contradicting $x \in U_k \subset \overline{U_k}$. Therefore $X$ is normal. [guided] Why do we need normality? The embedding strategy relies on [Urysohn's Lemma](/theorems/887), which requires the space to be normal. The theorem statement gives us regularity and second countability, but not normality directly. We must bridge the gap. The key observation is that second countability implies the Lindelöf property: given any open cover $\{W_\alpha\}_{\alpha \in A}$, each $W_\alpha$ is a union of basis elements from the countable basis $\mathcal{B} = \{B_n\}_{n \in \mathbb{N}}$. The collection of all basis elements appearing in these unions is a subset of $\mathcal{B}$, hence countable, and still covers $X$. Selecting one $W_\alpha$ for each such basis element yields a countable subcover. Now we prove normality. Let $C_0, C_1 \subset X$ be disjoint closed sets. For each $x \in C_0$, the set $X \setminus C_1$ is an open neighbourhood of $x$. By regularity (the $T_3$ axiom), there exists an open set $V_x$ such that $x \in V_x \subset \overline{V_x} \subset X \setminus C_1$. Since $\mathcal{B}$ is a basis, there exists $B_{n_x} \in \mathcal{B}$ with $x \in B_{n_x} \subset V_x$, and consequently $\overline{B_{n_x}} \subset \overline{V_x} \subset X \setminus C_1$. The collection $\{B_{n_x}\}_{x \in C_0}$ is an open cover of $C_0$, and by the Lindelöf property applied to the subspace $C_0$ (which inherits the Lindelöf property), we extract a countable subcover $\{U_k\}_{k \in \mathbb{N}}$ with $\overline{U_k} \cap C_1 = \varnothing$ for each $k$. By the symmetric argument applied to $C_1$ (separating each point of $C_1$ from the closed set $C_0$), we obtain a countable collection $\{V_k\}_{k \in \mathbb{N}}$ covering $C_1$ with $\overline{V_k} \cap C_0 = \varnothing$ for each $k$. The difficulty is that $\bigcup_k U_k$ and $\bigcup_k V_k$ need not be disjoint. We resolve this by a standard "shaving" construction. Define \begin{align*} U_k' &:= U_k \setminus \bigcup_{j=1}^{k} \overline{V_j}, \qquad V_k' := V_k \setminus \bigcup_{j=1}^{k} \overline{U_j}. \end{align*} Each $U_k'$ is open because it is the difference of the open set $U_k$ and the closed set $\bigcup_{j=1}^{k} \overline{V_j}$. The sets $U_k'$ still cover $C_0$: for $x \in C_0$, we have $x \in U_k$ for some $k$, and $x \notin \overline{V_j}$ for any $j$ (since $\overline{V_j} \cap C_0 = \varnothing$), so $x \in U_k'$. Similarly, the sets $V_k'$ cover $C_1$. Set $U := \bigcup_{k=1}^{\infty} U_k'$ and $V := \bigcup_{k=1}^{\infty} V_k'$. These are open, cover $C_0$ and $C_1$ respectively, and are disjoint. To verify disjointness, suppose for contradiction that $x \in U_k' \cap V_j'$ for some $k, j$. Then $x \in U_k$ and $x \in V_j$. If $j \le k$, then $x \in U_k' \subset X \setminus \overline{V_j}$, contradicting $x \in V_j \subset \overline{V_j}$. If $k \le j$, then $x \in V_j' \subset X \setminus \overline{U_k}$, contradicting $x \in U_k \subset \overline{U_k}$. In either case we reach a contradiction, so $U \cap V = \varnothing$. This establishes that $X$ is normal. [/guided] [/step] [step:Construct a countable family of Urysohn functions that separates points from closed sets] Since $X$ is normal and $\mathcal{B} = \{B_n\}_{n \in \mathbb{N}}$ is a countable basis, consider all pairs $(n, m) \in \mathbb{N} \times \mathbb{N}$ such that $\overline{B_n} \subset B_m$. This collection of pairs is countable. For each such pair, by [Urysohn's Lemma](/theorems/887) applied to the disjoint closed sets $\overline{B_n}$ and $X \setminus B_m$ in the normal space $X$, there exists a continuous function \begin{align*} g_{n,m}: X &\to [0,1] \end{align*} satisfying $g_{n,m}|_{\overline{B_n}} \equiv 0$ and $g_{n,m}|_{X \setminus B_m} \equiv 1$. Enumerate these functions as $\{f_k\}_{k \in \mathbb{N}}$. This family separates points from closed sets: if $x \in X$ and $C \subset X$ is closed with $x \notin C$, then $X \setminus C$ is an open neighbourhood of $x$. By regularity, there exists an open set $W$ with $x \in W \subset \overline{W} \subset X \setminus C$. Since $\mathcal{B}$ is a basis, there exist $B_n, B_m \in \mathcal{B}$ with $x \in B_n \subset \overline{B_n} \subset B_m \subset X \setminus C$. The corresponding function $g_{n,m}$ satisfies $g_{n,m}(x) = 0$ (since $x \in \overline{B_n}$) and $g_{n,m}(y) = 1$ for all $y \in C$ (since $C \subset X \setminus B_m$). [guided] The goal of this step is to build enough continuous functions to detect the topology of $X$. A single Urysohn function separates one pair of disjoint closed sets; we need a countable family that collectively separates every point from every closed set not containing it. The idea is to parametrise the applications of [Urysohn's Lemma](/theorems/887) by pairs of basis elements. Since $\mathcal{B} = \{B_n\}_{n \in \mathbb{N}}$ is a countable basis and $X$ is regular, for any point $x$ and open neighbourhood $U$ of $x$, there exist basis elements $B_n, B_m$ with $x \in B_n \subset \overline{B_n} \subset B_m \subset U$. (To find these: regularity gives an open $W$ with $x \in W \subset \overline{W} \subset U$; then pick $B_m$ with $x \in B_m \subset W$, and apply regularity again inside $B_m$ to get $B_n$ with $x \in B_n \subset \overline{B_n} \subset B_m$.) For each pair $(n, m) \in \mathbb{N} \times \mathbb{N}$ with $\overline{B_n} \subset B_m$, the sets $\overline{B_n}$ and $X \setminus B_m$ are disjoint and closed (the latter because $B_m$ is open). Since $X$ is normal (established in the previous step), [Urysohn's Lemma](/theorems/887) applies: there exists a continuous function \begin{align*} g_{n,m}: X &\to [0,1] \end{align*} with $g_{n,m}|_{\overline{B_n}} \equiv 0$ and $g_{n,m}|_{X \setminus B_m} \equiv 1$. The set of pairs $(n,m)$ is a subset of $\mathbb{N} \times \mathbb{N}$, hence countable. Enumerate the corresponding functions as $\{f_k\}_{k \in \mathbb{N}}$. To verify the separation property: let $x \in X$ and let $C \subset X$ be closed with $x \notin C$. Then $X \setminus C$ is open and contains $x$. By the observation above, there exist $B_n, B_m$ with $x \in B_n \subset \overline{B_n} \subset B_m \subset X \setminus C$. The function $g_{n,m}$ satisfies $g_{n,m}(x) = 0$ (since $x \in \overline{B_n}$) and $g_{n,m}(y) = 1$ for every $y \in C$ (since $C \subset X \setminus B_m$). In particular, $f_k = g_{n,m}$ for some index $k$, and $f_k(x) \ne f_k(y)$ for all $y \in C$. This is exactly the condition needed for the embedding to detect the topology. [/guided] [/step] [step:Define the embedding $F: X \to [0,1]^{\mathbb{N}}$ and verify [continuity](/page/Continuity)] Define \begin{align*} F: X &\to [0,1]^{\mathbb{N}} \\ x &\mapsto (f_1(x), f_2(x), f_3(x), \ldots). \end{align*} Here $[0,1]^{\mathbb{N}} = \prod_{n=1}^{\infty} [0,1]$ carries the [product topology](/page/Product%20Topology). **Continuity.** The [product topology](/page/Product%20Topology) on $[0,1]^{\mathbb{N}}$ is the coarsest topology making each coordinate projection $\pi_k: [0,1]^{\mathbb{N}} \to [0,1]$ continuous. Since $\pi_k \circ F = f_k$ is continuous for every $k \in \mathbb{N}$, the universal property of the product topology implies that $F$ is continuous. **Injectivity.** Let $x, y \in X$ with $x \ne y$. Since $X$ is $T_1$ (every regular space is $T_1$), the singleton $\{y\}$ is closed and $x \notin \{y\}$. By the separation property established in the previous step, there exists $k \in \mathbb{N}$ with $f_k(x) \ne f_k(y)$. Therefore $F(x) \ne F(y)$. [guided] We now assemble the Urysohn functions into a single map. Define \begin{align*} F: X &\to [0,1]^{\mathbb{N}} \\ x &\mapsto (f_1(x), f_2(x), f_3(x), \ldots). \end{align*} The target space $[0,1]^{\mathbb{N}} = \prod_{n=1}^{\infty} [0,1]$ is equipped with the [product topology](/page/Product%20Topology), which is generated by the subbasis of sets $\pi_k^{-1}(V)$ where $\pi_k: [0,1]^{\mathbb{N}} \to [0,1]$ is the $k$-th coordinate projection and $V \subset [0,1]$ is open. **Continuity.** To check that $F$ is continuous, it suffices (by the universal property of the product topology) to verify that each composition $\pi_k \circ F: X \to [0,1]$ is continuous. But $\pi_k \circ F = f_k$, which is continuous by construction (it was produced by Urysohn's Lemma). Hence $F$ is continuous. **Injectivity.** Suppose $x, y \in X$ with $x \ne y$. Since $X$ is regular, it is in particular $T_1$: singletons are closed. Therefore $\{y\}$ is a closed set not containing $x$. By the separation property from the previous step, there exists an index $k \in \mathbb{N}$ with $f_k(x) = 0$ and $f_k(y) = 1$. In particular $F(x)$ and $F(y)$ differ in the $k$-th coordinate, so $F(x) \ne F(y)$. [/guided] [/step] [step:Show that $F$ is an open map onto its image, completing the embedding] It remains to show that $F: X \to F(X)$ is an open map (equivalently, that $F^{-1}: F(X) \to X$ is continuous), which will establish that $F$ is a homeomorphism onto its image. Let $U \subset X$ be open and let $x \in U$. We must find an open set $W$ in $[0,1]^{\mathbb{N}}$ with $F(x) \in W \cap F(X) \subset F(U)$. Since $X \setminus U$ is closed and $x \notin X \setminus U$, the separation property gives an index $k \in \mathbb{N}$ with $f_k(x) = 0$ and $f_k(y) = 1$ for all $y \in X \setminus U$. Define \begin{align*} W := \pi_k^{-1}\bigl([0, \tfrac{1}{2})\bigr) = \bigl\{z \in [0,1]^{\mathbb{N}} : z_k < \tfrac{1}{2}\bigr\}. \end{align*} This is open in $[0,1]^{\mathbb{N}}$ (it is a subbasis element). We have $F(x) \in W$ since $f_k(x) = 0 < \tfrac{1}{2}$. If $F(y) \in W \cap F(X)$ for some $y \in X$, then $f_k(y) < \tfrac{1}{2}$, which forces $y \notin X \setminus U$ (since $f_k \equiv 1$ on $X \setminus U$), i.e., $y \in U$. Therefore $W \cap F(X) \subset F(U)$. Since $x \in U$ was arbitrary, $F(U)$ is open in the subspace topology on $F(X)$. [guided] This is the most delicate part of the proof. We have shown that $F$ is a continuous injection; we must upgrade it to a homeomorphism onto its image. A continuous bijection need not be a homeomorphism in general (consider the identity from $\mathbb{R}$ with the discrete topology to $\mathbb{R}$ with the standard topology). The key property that makes it work here is the separation property of the family $\{f_k\}$. Let $U \subset X$ be open and let $x \in U$. We need to show that $F(x)$ is an interior point of $F(U)$ relative to $F(X)$. Consider the closed set $C := X \setminus U$. Since $x \notin C$, the separation property provides an index $k \in \mathbb{N}$ with $f_k(x) = 0$ and $f_k(y) = 1$ for all $y \in C$. Define the open set \begin{align*} W := \pi_k^{-1}\bigl([0, \tfrac{1}{2})\bigr) = \bigl\{z \in [0,1]^{\mathbb{N}} : z_k < \tfrac{1}{2}\bigr\}. \end{align*} This is a subbasis element of the product topology, hence open. We verify: - $F(x) \in W$: since $f_k(x) = 0$, the $k$-th coordinate of $F(x)$ is $0 < \tfrac{1}{2}$. - $W \cap F(X) \subset F(U)$: if $F(y) \in W$ for some $y \in X$, then $f_k(y) < \tfrac{1}{2}$. But $f_k \equiv 1$ on $X \setminus U$, so $y \notin X \setminus U$, hence $y \in U$ and $F(y) \in F(U)$. Therefore $F(x) \in W \cap F(X) \subset F(U)$, and since $x \in U$ was arbitrary, $F(U)$ is open in $F(X)$. This shows $F: X \to F(X)$ is an open map, hence a homeomorphism onto its image. [/guided] [/step] [step:Conclude [metrizability](/page/Metrizable%20Space) from the metrizability of the Hilbert cube] The space $[0,1]^{\mathbb{N}}$ is metrizable: the function \begin{align*} d: [0,1]^{\mathbb{N}} \times [0,1]^{\mathbb{N}} &\to [0, \infty) \\ (a, b) &\mapsto \sum_{n=1}^{\infty} \frac{|a_n - b_n|}{2^n} \end{align*} is a metric that induces the product topology. (The series converges since $|a_n - b_n| \le 1$ and $\sum 2^{-n} < \infty$. The triangle inequality follows from $|a_n - c_n| \le |a_n - b_n| + |b_n - c_n|$ applied term by term. The metric topology agrees with the product topology: each $\pi_k$ satisfies $|\pi_k(a) - \pi_k(b)| \le 2^k d(a,b)$, so the $d$-topology is at least as fine as the product topology, and conversely, convergence in $d$ implies coordinatewise convergence.) Moreover, $[0,1]^{\mathbb{N}}$ is compact by [Tychonoff's Theorem](/theorems/953), since each factor $[0,1]$ is compact. Since $F: X \to F(X) \subset [0,1]^{\mathbb{N}}$ is a homeomorphism, the subspace $F(X)$ is metrizable (the restriction of $d$ to $F(X)$ is a compatible metric), and therefore $X$ is metrizable. This completes the proof. [guided] To conclude, we must verify that the target space $[0,1]^{\mathbb{N}}$ is itself metrizable — otherwise the embedding would not yield metrizability of $X$. Define the metric \begin{align*} d: [0,1]^{\mathbb{N}} \times [0,1]^{\mathbb{N}} &\to [0, \infty) \\ (a, b) &\mapsto \sum_{n=1}^{\infty} \frac{|a_n - b_n|}{2^n}. \end{align*} The series converges absolutely for all $a, b \in [0,1]^{\mathbb{N}}$, since $|a_n - b_n| \le 1$ and $\sum_{n=1}^{\infty} 2^{-n} = 1$. The function $d$ satisfies the axioms of a metric: positivity and symmetry are immediate; definiteness holds because $d(a,b) = 0$ implies $a_n = b_n$ for all $n$; and the triangle inequality follows from $|a_n - c_n| \le |a_n - b_n| + |b_n - c_n|$ applied term by term. We verify that $d$ induces the product topology. For the forward direction, each projection $\pi_k$ satisfies $|\pi_k(a) - \pi_k(b)| = |a_k - b_k| \le 2^k \cdot d(a,b)$, so $\pi_k$ is Lipschitz (hence continuous) with respect to $d$. Since the product topology is the coarsest making all projections continuous, the $d$-topology is at least as fine as the product topology. For the converse, if $a_n \to a$ coordinatewise, then for any $\varepsilon > 0$, choose $N$ with $\sum_{n > N} 2^{-n} < \varepsilon/2$, then choose $n_0$ so that $\sum_{n=1}^{N} |a_n^{(k)} - a_n|/2^n < \varepsilon/2$ for $k \ge n_0$. This gives $d(a^{(k)}, a) < \varepsilon$, so convergence in the product topology implies convergence in $d$. Hence the two topologies coincide. Finally, $[0,1]^{\mathbb{N}}$ is compact by [Tychonoff's Theorem](/theorems/953): each factor $[0,1]$ is compact, and Tychonoff's theorem guarantees that the product of an arbitrary family of [compact spaces](/page/Compact%20Space) is compact in the product topology. Since $F: X \to F(X)$ is a homeomorphism and $F(X) \subset [0,1]^{\mathbb{N}}$ inherits the metric $d|_{F(X) \times F(X)}$, the space $X$ is metrizable. The embedding $F$ simultaneously witnesses that $X$ embeds homeomorphically into the Hilbert cube $[0,1]^{\mathbb{N}}$. [/guided] [/step]