[proofplan] The Sobolev norm $\|u\|_{W^{k,p}(U)} = \bigl(\sum_{|\alpha| \le k} \|D^\alpha u\|_{L^p(U)}^p\bigr)^{1/p}$ controls $u$ and all its [weak derivatives](/page/Weak%20Derivative) simultaneously. This structure means that the map $\Phi: u \mapsto (D^\alpha u)_{|\alpha| \le k}$ embeds $W^{k,p}(U)$ isometrically into the finite product $\prod_{|\alpha| \le k} L^p(U)$, equipped with the $\ell^p$-sum norm. Since each factor $L^p(U)$ is separable for $1 \le p < \infty$ by the [Separability of $L^p$ Spaces](/theorems/548), and a finite product of separable [metric spaces](/page/Metric%20Space) is separable by [Countable Products of Separable Spaces](/theorems/946), the product is separable. The [Separability via Isometric Embedding](/theorems/948) theorem then transfers separability from the product back to $W^{k,p}(U)$. [/proofplan]

[step:Define the isometric embedding $\Phi$ from $W^{k,p}(U)$ into a finite product of $L^p(U)$ spaces]Fix an enumeration $\alpha_1, \alpha_2, \ldots, \alpha_N$ of all multi-indices $\alpha \in \mathbb{N}_0^n$ with $|\alpha| \le k$, where $N = \binom{n+k}{k}$. Define the map \begin{align*} \Phi: W^{k,p}(U) &\to \underbrace{L^p(U) \times \cdots \times L^p(U)}_{N \text{ factors}} \\ u &\mapsto (D^{\alpha_1} u, \, D^{\alpha_2} u, \, \ldots, \, D^{\alpha_N} u). \end{align*} Here $D^{\alpha} u$ denotes the [weak derivative](/page/Weak%20Derivative) of $u$ of order $\alpha$, with the convention that $D^{(0,\ldots,0)} u = u$. Each component $D^{\alpha_j} u$ belongs to $L^p(U)$ by the definition of $W^{k,p}(U)$, so $\Phi$ is well-defined.[/step]

[guided]The definition of $W^{k,p}(U)$ requires that $u$ and all its weak derivatives $D^\alpha u$ with $|\alpha| \le k$ belong to $L^p(U)$. The Sobolev norm aggregates these $L^p$-norms into a single quantity. This suggests a natural strategy: treat $u \in W^{k,p}(U)$ as a "vector" of $L^p$ functions — one component for $u$ itself, one for each weak derivative — and embed the [Sobolev space](/page/Sobolev%20Space) into a product of $L^p$ spaces. The number of multi-indices $\alpha \in \mathbb{N}_0^n$ with $|\alpha| \le k$ is $N = \binom{n+k}{k}$. This is finite for any fixed $n$ and $k$, which is essential: the product of finitely many separable spaces is separable, but a product of uncountably many need not be. We fix an enumeration $\alpha_1, \alpha_2, \ldots, \alpha_N$ of all multi-indices $\alpha \in \mathbb{N}_0^n$ with $|\alpha| \le k$. The choice of enumeration does not affect the argument — any ordering produces the same image set and the same distance function, since the $\ell^p$-sum norm is symmetric in its terms. We define the map \begin{align*} \Phi: W^{k,p}(U) &\to \underbrace{L^p(U) \times \cdots \times L^p(U)}_{N \text{ factors}} \\ u &\mapsto (D^{\alpha_1} u, \, D^{\alpha_2} u, \, \ldots, \, D^{\alpha_N} u), \end{align*} where $D^{\alpha} u$ denotes the weak derivative of $u$ of order $\alpha$, with the convention $D^{(0,\ldots,0)} u = u$. Each component $D^{\alpha_j} u$ belongs to $L^p(U)$ by the definition of $W^{k,p}(U)$: membership in the Sobolev space requires precisely that all weak derivatives of order up to $k$ are $L^p$-integrable. Therefore $\Phi$ is well-defined as a map into the product.[/guided]

[step:Verify that $\Phi$ is an isometry with respect to the $\ell^p$-sum norm on the product] Equip the product $\prod_{j=1}^N L^p(U)$ with the $\ell^p$-sum norm: \begin{align*} \|(f_1, f_2, \ldots, f_N)\|_{\ell^p} := \left(\sum_{j=1}^N \|f_j\|_{L^p(U)}^p\right)^{1/p}. \end{align*} This makes $\prod_{j=1}^N L^p(U)$ a [normed vector space](/page/Normed%20Vector%20Space), and the induced metric is $d_{\ell^p}((f_1, \ldots, f_N), (g_1, \ldots, g_N)) = \|(f_1 - g_1, \ldots, f_N - g_N)\|_{\ell^p}$. For any $u, v \in W^{k,p}(U)$, the distance between their images under $\Phi$ is: \begin{align*} d_{\ell^p}(\Phi(u), \Phi(v)) &= \left(\sum_{j=1}^N \|D^{\alpha_j} u - D^{\alpha_j} v\|_{L^p(U)}^p\right)^{1/p} \\ &= \left(\sum_{|\alpha| \le k} \|D^\alpha(u - v)\|_{L^p(U)}^p\right)^{1/p} \\ &= \|u - v\|_{W^{k,p}(U)}. \end{align*} The first equality substitutes the definition of $\Phi$ and uses linearity of weak differentiation: $D^{\alpha_j} u - D^{\alpha_j} v = D^{\alpha_j}(u - v)$. The second equality re-indexes the sum over the enumeration $\alpha_1, \ldots, \alpha_N$ as a sum over all multi-indices $\alpha$ with $|\alpha| \le k$. The third equality is the definition of the Sobolev norm on $W^{k,p}(U)$. Since $d_{\ell^p}(\Phi(u), \Phi(v)) = \|u - v\|_{W^{k,p}(U)}$ for all $u, v \in W^{k,p}(U)$, the map $\Phi$ is an isometric embedding of $W^{k,p}(U)$ into $\prod_{j=1}^N L^p(U)$. [/step]

[step:Conclude separability by combining separability of $L^p$, finite products, and the isometric embedding theorem]We assemble three results. **Separability of each factor.** Since $U \subset \mathbb{R}^n$ is open, the Borel $\sigma$-algebra $\mathcal{B}(U)$ is countably generated (by rectangles with rational endpoints intersected with $U$), and $\mathcal{L}^n$ is $\sigma$-finite on $U$. By the [Separability of $L^p$ Spaces](/theorems/548), each factor $L^p(U)$ is separable for $1 \le p < \infty$. **Separability of the product.** The product $\prod_{j=1}^N L^p(U)$ is a finite product of separable [topological](/page/Topology) spaces. By the [Countable Products of Separable Spaces](/theorems/946) theorem (applied to the finite case $N = \binom{n+k}{k}$), this product is separable. Since $\prod_{j=1}^N L^p(U)$ is equipped with the $\ell^p$-sum norm — which induces the same topology as the [product topology](/page/Product%20Topology) on a finite product of [normed spaces](/page/Normed%20Vector%20Space) — the product is separable as a [metric space](/page/Metric%20Space). **Transfer via isometric embedding.** The map $\Phi: W^{k,p}(U) \to \prod_{j=1}^N L^p(U)$ is an isometric embedding into a separable metric space. By [Separability via Isometric Embedding](/theorems/948), $W^{k,p}(U)$ is separable.[/step]

[guided]We must verify the hypotheses of each theorem we invoke and assemble the conclusion explicitly. **Separability of each factor.** We apply the [Separability of $L^p$ Spaces](/theorems/548) to each factor $L^p(U)$. Theorem 548 requires: (i) the underlying measure space is $\sigma$-finite, and (ii) the $\sigma$-algebra is countably generated. We verify both. Since $U \subset \mathbb{R}^n$ is open, we can write $U = \bigcup_{m=1}^\infty K_m$ where $K_m = \{x \in U : |x| \le m \text{ and } \operatorname{dist}(x, \partial U) \ge 1/m\}$ are [compact](/page/Compact%20Space) sets with $\mathcal{L}^n(K_m) < \infty$, so $(U, \mathcal{B}(U), \mathcal{L}^n)$ is $\sigma$-finite. The Borel $\sigma$-algebra $\mathcal{B}(U)$ is generated by the [countable](/page/Countable%20Set) family of open rectangles with rational endpoints intersected with $U$, so it is countably generated. Both hypotheses hold, and theorem 548 gives: $L^p(U)$ is separable for each $1 \le p < \infty$. This is where the hypothesis $1 \le p < \infty$ is consumed. For $p = \infty$, the conclusion fails — $L^\infty(U)$ is not separable — and correspondingly $W^{k,\infty}(U)$ cannot be separable either, since the $|\alpha| = 0$ component of $\Phi$ embeds $L^\infty(U)$ isometrically into $W^{k,\infty}(U)$. **Separability of the product.** We now have $N = \binom{n+k}{k}$ separable metric spaces $L^p(U)$. The [Countable Products of Separable Spaces](/theorems/946) theorem states that a countable (in particular, finite) product of separable topological spaces is separable in the product topology. Applying this to the $N$ factors: $\prod_{j=1}^N L^p(U)$ is separable in the product topology. We must verify that the product topology agrees with the metric topology induced by the $\ell^p$-sum norm. On a finite product of normed spaces, the product topology is the topology induced by any norm of the form $\|(f_1, \ldots, f_N)\|_q = (\sum_{j=1}^N \|f_j\|^q)^{1/q}$ for any $1 \le q \le \infty$, since all such norms are equivalent on $\mathbb{R}^N$. In particular, the $\ell^p$-sum norm induces the product topology, so separability in the product topology implies separability in the $\ell^p$-sum metric. **Transfer via isometric embedding.** We established in Step 2 that $\Phi: W^{k,p}(U) \to \prod_{j=1}^N L^p(U)$ is an isometric embedding with respect to the $\ell^p$-sum norm. The [Separability via Isometric Embedding](/theorems/948) theorem states: if $(X, d_X)$ admits an isometric embedding into a separable metric space $(Y, d_Y)$, then $X$ is separable. We apply this with $X = W^{k,p}(U)$ (metrized by the Sobolev norm), $Y = \prod_{j=1}^N L^p(U)$ (metrized by the $\ell^p$-sum norm, which we just showed is separable), and $\Phi$ as the isometric embedding. The hypotheses are satisfied: - $\Phi$ preserves distances: $d_{\ell^p}(\Phi(u), \Phi(v)) = \|u - v\|_{W^{k,p}(U)}$ (Step 2). - $Y$ is separable (established above). Therefore $W^{k,p}(U)$ is separable. Alternatively, one can bypass theorem 948 and argue directly: the image $\Phi(W^{k,p}(U))$ is a subspace of the separable metric space $\prod_{j=1}^N L^p(U)$, hence separable by [Subspaces of Separable Metrizable Spaces](/theorems/942). Let $E = \{e_m\}_{m=1}^\infty$ be a countable [dense subset](/page/Dense%20Subset) of $\Phi(W^{k,p}(U))$. Since $\Phi$ is an isometry, it is injective, so $\Phi^{-1}: \Phi(W^{k,p}(U)) \to W^{k,p}(U)$ is well-defined. For any $u \in W^{k,p}(U)$ and $\varepsilon > 0$, density of $E$ provides $e_m$ with $d_{\ell^p}(e_m, \Phi(u)) < \varepsilon$, and the isometry gives $\|\Phi^{-1}(e_m) - u\|_{W^{k,p}} = d_{\ell^p}(e_m, \Phi(u)) < \varepsilon$. So $D := \{\Phi^{-1}(e_m)\}_{m=1}^\infty$ is a countable dense subset of $W^{k,p}(U)$.[/guided]