[proofplan] The proof has two layers. The outer layer represents an arbitrary $f \in \mathcal{O}(K)$ as a Cauchy-type integral on a contour $\gamma$ surrounding $K$ in the open set $U$ where $f$ is holomorphic, and approximates the integral by a Riemann sum — this expresses $f$ uniformly on $K$ as a limit of finite linear combinations of "Cauchy kernels" $\zeta_j \mapsto (\zeta_j - z)^{-1}$, i.e.\ rational functions whose poles $\zeta_j$ lie on the contour, hence in $\mathbb{C} \setminus K$. The inner layer is **pole-pushing**: a single Cauchy kernel with pole at $\zeta_0 \in \mathbb{C} \setminus K$ can be approximated uniformly on $K$ by rational functions with poles at any prescribed point $\lambda$ in the same connected component of $\mathbb{C} \setminus K$ as $\zeta_0$. By chaining a finite number of pole-pushings we can move all the poles of the Riemann-sum approximant into the prescribed set $\Lambda$. If $\zeta_0$ lies in the unbounded component, pole-pushing to infinity reduces the kernel to a polynomial. Combining the two layers gives the precise statement. [/proofplan] [step:Set up an admissible contour around $K$ in the open set of holomorphy] Let $f \in \mathcal{O}(K)$, so $f$ is holomorphic on some open set $U \subseteq \mathbb{C}$ with $K \subset U$. We construct a contour $\gamma$ — a finite union of oriented piecewise-linear closed curves in $U \setminus K$ — such that the winding numbers satisfy \begin{align*} n(\gamma, z) = \begin{cases} 1, & z \in K, \\ 0, & z \in \mathbb{C} \setminus U. \end{cases} \end{align*} The construction is the standard grid argument: by compactness of $K$ and openness of $U$, $\delta := \frac{1}{4}\operatorname{dist}(K, \mathbb{C} \setminus U) > 0$. Tile the plane by closed squares of side $\delta$ aligned with the axes. By translating the grid by a generic vector if necessary, we may assume no grid line meets $K$ — equivalently, the boundary of every grid square is disjoint from $K$. This is possible because $K$ is compact and the set of bad translations (those for which some grid line meets $K$) has two-dimensional Lebesgue measure zero: the set of horizontal translations placing a grid line through $K$ is a countable union of translates of $\pi_1(K)$ (a compact subset of $\mathbb{R}$), each a $\mathcal{L}^1$-null set when $K$ has empty interior, and analogously for vertical translations; in any case the set of bad two-dimensional translations is contained in a countable union of lines, hence is $\mathcal{L}^2$-null. Pick any good translation. Let $\mathcal{Q}$ be the finite set of closed squares meeting $K$. Each $Q \in \mathcal{Q}$ is contained in $U$: $Q$ meets $K$, the diameter of $Q$ is $\delta\sqrt{2}$, so every point of $Q$ lies within $\delta\sqrt{2}$ of a point of $K$; since $\operatorname{dist}(K, \mathbb{C} \setminus U) \ge 4\delta$, we have $\operatorname{dist}(Q, \mathbb{C} \setminus U) \ge 4\delta - \delta\sqrt{2} > 0$, so $Q \subset U$. Take $\gamma$ to be the formal sum of the boundary edges of squares in $\mathcal{Q}$ that bound exactly one square in $\mathcal{Q}$ (i.e.\ "outer" edges), each oriented counterclockwise relative to the square in $\mathcal{Q}$ it bounds. Because no boundary of any grid square meets $K$, we have $\mathrm{image}(\gamma) \subset U \setminus K$. Then: - For $z \in K$: by the genericity of the translation, $z$ lies in the open interior of a unique $Q_0 \in \mathcal{Q}$ (no grid line meets $K$). The winding number $n(\gamma, z)$ counts $1$ for $Q_0$ and the contributions of internal edges (shared between two adjacent squares in $\mathcal{Q}$) cancel; so $n(\gamma, z) = 1$. - For $z \in \mathbb{C} \setminus U$, $z$ lies outside every $Q \in \mathcal{Q}$ (since each $Q \subset U$), so $n(\gamma, z) = 0$. Hence $\gamma$ has the required winding properties, and $\mathrm{image}(\gamma) \subset U \setminus K$. By the Cauchy Integral Formula on cycles homologous to zero in $U$ applied to the holomorphic $f$: \begin{align*} f(z) = \frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\zeta - z}\, d\zeta \qquad (z \in K). \tag{$\ast$} \end{align*} <!-- illustration-needed: the grid argument — show K as a compact set, the tiling by squares of side delta, the squares of Q meeting K shaded, the outer boundary edges (forming gamma) oriented counterclockwise, and the cancellation of internal shared edges --> [/step] [step:Approximate the contour integral uniformly on $K$ by a Riemann sum] Parametrise $\gamma$ as a finite union of segments. Let $L = \ell(\gamma)$ be the total arc length. Subdivide $\gamma$ into $N$ arcs of equal length $L/N$, with endpoints $\zeta_0, \zeta_1, \dots, \zeta_N = \zeta_0$ (within each component of $\gamma$). Define the Riemann sum \begin{align*} S_N(z) := \frac{1}{2\pi i}\sum_{j=1}^N \frac{f(\zeta_j)}{\zeta_j - z}(\zeta_j - \zeta_{j-1}). \end{align*} This is a rational function of $z$ with poles at $\zeta_1, \dots, \zeta_N$, all lying on $\mathrm{image}(\gamma) \subset U \setminus K \subseteq \mathbb{C} \setminus K$. We show $S_N \to f$ uniformly on $K$ as $N \to \infty$. The integrand \begin{align*} F : \mathrm{image}(\gamma) \times K &\to \mathbb{C} \\ (\zeta, z) &\mapsto \frac{f(\zeta)}{\zeta - z} \end{align*} is continuous on the compact product $\mathrm{image}(\gamma) \times K$ (the denominator is bounded below by $d_0 := \operatorname{dist}(K, \mathrm{image}(\gamma)) > 0$, since $K$ and $\mathrm{image}(\gamma)$ are disjoint compact sets), hence uniformly continuous. For any $\varepsilon > 0$ there exists $N_0$ such that for $N \ge N_0$ and any $\zeta, \zeta'$ on the same arc of length $\le L/N$, \begin{align*} \sup_{z \in K}\left|\frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta')}{\zeta' - z}\right| < \varepsilon. \end{align*} For $z \in K$, applying ($\ast$) [Cauchy Integral Formula]: \begin{align*} |S_N(z) - f(z)| &= \left|\frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{\zeta - z}\, d\zeta - \frac{1}{2\pi i}\sum_{j=1}^N \frac{f(\zeta_j)}{\zeta_j - z}(\zeta_j - \zeta_{j-1})\right|\\ &\le \frac{1}{2\pi}\sum_{j=1}^N \int_{\gamma_j}\left|\frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta_j)}{\zeta_j - z}\right|\, |d\zeta|\\ &\le \frac{1}{2\pi}\sum_{j=1}^N \varepsilon \cdot |\gamma_j| = \frac{\varepsilon L}{2\pi}, \end{align*} where $\gamma_j$ is the $j$-th arc of length $L/N$. The bound is uniform in $z \in K$, so $S_N \to f$ uniformly on $K$. We have shown: every $f \in \mathcal{O}(K)$ is the uniform limit on $K$ of rational functions whose poles lie in $\mathbb{C} \setminus K$. Hence $\mathcal{O}(K) \subseteq \mathcal{R}(K)$. [/step] [step:Pole-push: a Cauchy kernel can be approximated by rational functions with poles in any prescribed point of the same connected component] Let $\Omega \subseteq \mathbb{C} \setminus K$ be a connected component of $\mathbb{C} \setminus K$. We claim: \begin{align*} \text{For all } a, b \in \Omega: \quad \frac{1}{(\cdot) - a} \text{ is the uniform limit on } K \text{ of polynomials in } \frac{1}{(\cdot) - b}. \end{align*} This is the **pole-pushing lemma** for Runge's theorem. Define $\mathcal{A}_b \subseteq C(K)$ to be the closed unital subalgebra of $C(K)$ generated by the single function $(\cdot - b)^{-1}$. Concretely, $\mathcal{A}_b$ is the closure (in $\|\cdot\|_K$) of the algebra of polynomials in $(\cdot - b)^{-1}$, equivalently the closure of $\mathrm{span}\{1, (\cdot - b)^{-k} : k \ge 1\}$. **Claim: $\mathcal{A}_b$ is closed under pointwise multiplication.** Suppose $f_n \to f$ and $g_n \to g$ uniformly on $K$, with $f_n, g_n \in \mathrm{span}\{1, (\cdot - b)^{-k} : k \ge 1\}$. Since uniformly convergent sequences in $C(K)$ are bounded, there exists $M > 0$ with $\|f\|_K, \|g\|_K, \|f_n\|_K, \|g_n\|_K \le M$ for all $n$. Then \begin{align*} \|f_n g_n - fg\|_K \le \|f_n g_n - f_n g\|_K + \|f_n g - fg\|_K \le M\|g_n - g\|_K + M\|f_n - f\|_K \to 0, \end{align*} so $f_n g_n \to fg$ in $C(K)$. Thus $\mathcal{A}_b$ is closed under multiplication: it is a closed unital subalgebra. In particular it contains all powers of any of its elements. Define \begin{align*} \mathcal{P} := \{a \in \Omega : \tfrac{1}{(\cdot) - a} \in \mathcal{A}_b\}. \end{align*} We show $\mathcal{P} = \Omega$ by proving $\mathcal{P}$ is non-empty, open, and closed in $\Omega$ (and using that $\Omega$ is connected). **$\mathcal{P}$ contains $b$.** The function $(\cdot - b)^{-1}$ generates $\mathcal{A}_b$, so it lies in $\mathcal{A}_b$. Hence $b \in \mathcal{P}$. **$\mathcal{P}$ is open in $\Omega$.** Let $a_0 \in \mathcal{P}$ and let $r := \operatorname{dist}(a_0, K) > 0$ (positive since $a_0 \in \mathbb{C} \setminus K$ and $K$ is closed). For $a$ with $|a - a_0| < r/2$, expand \begin{align*} \frac{1}{(\cdot) - a} = \frac{1}{((\cdot) - a_0) - (a - a_0)} = \sum_{k=0}^\infty \frac{(a - a_0)^k}{((\cdot) - a_0)^{k+1}}, \end{align*} which converges in $C(K)$ uniformly on $K$ because for $z \in K$, $|z - a_0| \ge r$ and $|a - a_0| < r/2$, so the geometric series in $(a - a_0)/(z - a_0)$ has ratio of modulus $\le 1/2$. Hence $\frac{1}{(\cdot) - a}$ is the limit (in $\|\cdot\|_K$) of finite linear combinations of $(\cdot - a_0)^{-k}$ for $k \ge 1$. Since $a_0 \in \mathcal{P}$, $(\cdot - a_0)^{-1} \in \mathcal{A}_b$; because $\mathcal{A}_b$ is closed under multiplication, $(\cdot - a_0)^{-k} \in \mathcal{A}_b$ for all $k \ge 1$. Linear combinations of these still lie in $\mathcal{A}_b$, and $\mathcal{A}_b$ is closed, so $\frac{1}{(\cdot) - a} \in \mathcal{A}_b$. Therefore $a \in \mathcal{P}$. So $B(a_0, r/2) \cap \Omega \subseteq \mathcal{P}$, proving $\mathcal{P}$ open. **$\mathcal{P}$ is closed in $\Omega$.** Let $a_n \in \mathcal{P}$ with $a_n \to a \in \Omega$. We need $\frac{1}{(\cdot) - a}$ to be in $\mathcal{A}_b$. For all $z \in K$: \begin{align*} \left|\frac{1}{z - a_n} - \frac{1}{z - a}\right| = \frac{|a - a_n|}{|z - a_n||z - a|} \le \frac{|a - a_n|}{d_n d}, \quad d := \operatorname{dist}(K, a),\ d_n := \operatorname{dist}(K, a_n). \end{align*} Since $a_n \to a \in \Omega$, $a$ is at positive distance $d$ from $K$, and $d_n \to d$. Hence $\sup_{z \in K}|\cdots| \to 0$, so $\frac{1}{(\cdot) - a_n} \to \frac{1}{(\cdot) - a}$ in $C(K)$. Each $\frac{1}{(\cdot) - a_n} \in \mathcal{A}_b$, and $\mathcal{A}_b$ is closed, so $\frac{1}{(\cdot) - a} \in \mathcal{A}_b$. Hence $a \in \mathcal{P}$. By connectedness of $\Omega$, $\mathcal{P} = \Omega$. **Unbounded component case (pole pushing to infinity).** Suppose $\Omega$ is the unbounded component of $\mathbb{C} \setminus K$. Then $\Omega$ contains all $a$ with $|a|$ sufficiently large. For such $a$ with $|a| > \sup_{z \in K}|z|$ (so $|z/a| < 1$ for $z \in K$), \begin{align*} \frac{1}{(\cdot) - a} = -\frac{1}{a}\cdot\frac{1}{1 - (\cdot)/a} = -\frac{1}{a}\sum_{k=0}^\infty \frac{(\cdot)^k}{a^k}, \end{align*} a series of polynomials in $(\cdot)$ converging uniformly on $K$. So $\frac{1}{(\cdot) - a}$ lies in the closed algebra of polynomials in $C(K)$, and by the pole-pushing argument above (applied within $\Omega$ to the closed polynomial algebra rather than $\mathcal{A}_b$), the same holds for every $\frac{1}{(\cdot) - \zeta}$ with $\zeta \in \Omega$. [guided] [Standalone proof of pole-pushing.] We prove: if $f$ is holomorphic on an open set $\Omega' \subseteq \mathbb{C}$ containing $K$ except for a single pole at $a_0 \in \Omega' \setminus K$, with $a_0$ in a connected component $\Omega$ of $\mathbb{C} \setminus K$, and $\Lambda \cap \Omega = \{\lambda\}$ (one prescribed pole per bounded component, or pole at infinity = polynomial for the unbounded component), then $f$ can be approximated uniformly on $K$ by rational functions whose only finite poles lie in $\Lambda$ — concretely, in $\{\lambda\}$ if $\Omega$ is bounded, or no finite poles if $\Omega$ is unbounded. The Laurent expansion at $a_0$ writes $f$ as the sum of a holomorphic part and a finite linear combination of $(\cdot - a_0)^{-k}$ for $k \ge 1$. The holomorphic part is already pole-free on a neighbourhood of $K$. So it suffices to handle each $(\cdot - a_0)^{-k}$ — and in fact, since $\mathcal{A}_b$ (defined below) is closed under multiplication, it suffices to handle the case $k = 1$: $(\cdot - a_0)^{-1}$. **Reduction.** Fix $\lambda \in \Omega$ (the prescribed point in the same component, or any large $\lambda$ for the unbounded component — see end of this block). Let $\mathcal{A}_\lambda$ be the closed unital subalgebra of $C(K)$ generated by $(\cdot - \lambda)^{-1}$. We must show $(\cdot - a_0)^{-1} \in \mathcal{A}_\lambda$. **Step (i): $\mathcal{A}_\lambda$ is closed under pointwise multiplication.** We have already proved this: if $f_n \to f$, $g_n \to g$ uniformly on $K$, with $|f_n|, |g_n|, |f|, |g| \le M$ on $K$, then $\|f_n g_n - fg\|_K \le M\|f_n - f\|_K + M\|g_n - g\|_K \to 0$. Combined with the fact that products of polynomials in $(\cdot - \lambda)^{-1}$ are again polynomials in $(\cdot - \lambda)^{-1}$, this shows $\mathcal{A}_\lambda$ is closed under products. In particular, $(\cdot - \lambda)^{-k} = \big((\cdot - \lambda)^{-1}\big)^k \in \mathcal{A}_\lambda$ for all $k \ge 1$. **Step (ii): the reachable set is non-empty.** Let \begin{align*} \mathcal{P} := \{a \in \Omega : (\cdot - a)^{-1} \in \mathcal{A}_\lambda\}. \end{align*} Then $\lambda \in \mathcal{P}$, since $(\cdot - \lambda)^{-1}$ is the generator of $\mathcal{A}_\lambda$ and lies in $\mathcal{A}_\lambda$. **Step (iii): the reachable set is open in $\Omega$.** Fix $a_1 \in \mathcal{P}$. Let $r := \operatorname{dist}(a_1, K) > 0$ (positive because $a_1 \in \Omega \subseteq \mathbb{C} \setminus K$ and $K$ is closed). For any $a \in \mathbb{C}$ with $|a - a_1| < r/2$, and any $z \in K$, $|z - a_1| \ge r$, and so $|(a - a_1)/(z - a_1)| \le 1/2 < 1$. The geometric series \begin{align*} \frac{1}{z - a} = \frac{1}{(z - a_1) - (a - a_1)} = \frac{1}{z - a_1}\cdot\frac{1}{1 - (a-a_1)/(z-a_1)} = \sum_{k=0}^\infty \frac{(a - a_1)^k}{(z - a_1)^{k+1}} \end{align*} converges uniformly in $z \in K$ (the tail bound is the geometric tail with ratio $\le 1/2$ and is independent of $z$). Hence $(\cdot - a)^{-1}$ is the uniform limit on $K$ of partial sums $\sum_{k=0}^M (a - a_1)^k (\cdot - a_1)^{-(k+1)}$. Each partial sum is a finite linear combination of $(\cdot - a_1)^{-(k+1)}$ for $k \ge 0$, all of which lie in $\mathcal{A}_\lambda$ (by step (i) since $(\cdot - a_1)^{-1} \in \mathcal{A}_\lambda$). Linear combinations are in $\mathcal{A}_\lambda$ (linear subspace), and the limit is in $\mathcal{A}_\lambda$ (closed). Hence $a \in \mathcal{P}$, so $B(a_1, r/2) \cap \Omega \subseteq \mathcal{P}$. **Step (iv): the reachable set is closed in $\Omega$.** Let $a_n \in \mathcal{P}$ with $a_n \to a \in \Omega$. Let $d := \operatorname{dist}(K, a) > 0$ and $d_n := \operatorname{dist}(K, a_n)$; by continuity of $\operatorname{dist}(K, \cdot)$, $d_n \to d$, so $d_n \ge d/2$ for $n$ large. For such $n$ and all $z \in K$: \begin{align*} \left|\frac{1}{z - a_n} - \frac{1}{z - a}\right| = \frac{|a - a_n|}{|z - a_n||z - a|} \le \frac{|a - a_n|}{d_n d} \le \frac{2|a - a_n|}{d^2}. \end{align*} So $\sup_{z \in K}|(z - a_n)^{-1} - (z - a)^{-1}| \le 2|a - a_n|/d^2 \to 0$. Hence $(\cdot - a_n)^{-1} \to (\cdot - a)^{-1}$ in $C(K)$. Each $(\cdot - a_n)^{-1} \in \mathcal{A}_\lambda$ and $\mathcal{A}_\lambda$ is closed, so $(\cdot - a)^{-1} \in \mathcal{A}_\lambda$, i.e.\ $a \in \mathcal{P}$. **Step (v): connectedness conclusion.** $\mathcal{P} \subseteq \Omega$ is non-empty (contains $\lambda$), open in $\Omega$, and closed in $\Omega$. Since $\Omega$ is connected, $\mathcal{P} = \Omega$. In particular, the original $a_0$ in the same component as $\lambda$ satisfies $a_0 \in \mathcal{P}$, so $(\cdot - a_0)^{-1} \in \mathcal{A}_\lambda$ — it is approximable uniformly on $K$ by polynomials in $(\cdot - \lambda)^{-1}$, which are rational functions with all finite poles at $\lambda$. **Step (vi): unbounded component, pole at infinity.** If $\Omega$ is the unbounded component, $\Lambda \cap \Omega = \emptyset$ (the unbounded component contributes no finite pole) and we approximate by polynomials, i.e.\ rational functions with no finite pole at all. Pick any $\lambda^*$ in $\Omega$ with $|\lambda^*| > \sup_{z \in K}|z|$ (which exists since $\Omega$ is unbounded). The geometric series \begin{align*} \frac{1}{z - \lambda^*} = -\frac{1}{\lambda^*}\sum_{k=0}^\infty \frac{z^k}{(\lambda^*)^k} \end{align*} converges uniformly on $K$ (ratio $|z/\lambda^*| < 1$ uniformly in $z \in K$), so $(\cdot - \lambda^*)^{-1}$ is the uniform limit of polynomials in $(\cdot)$. Let $\mathcal{A}_\infty$ be the closed unital subalgebra of $C(K)$ generated by $(\cdot)$, i.e.\ the closure of polynomials in $C(K)$. Then $(\cdot - \lambda^*)^{-1} \in \mathcal{A}_\infty$. By steps (i)–(v) with $\mathcal{A}_b$ replaced by $\mathcal{A}_\infty$ throughout (the multiplication-closure argument is identical, and $\lambda^* \in \mathcal{P}$), we get $\mathcal{P} = \Omega$, so for every $a_0 \in \Omega$, $(\cdot - a_0)^{-1}$ lies in the closed polynomial algebra — i.e.\ is approximable uniformly on $K$ by polynomials. This completes the standalone proof of pole-pushing. [/guided] [/step] [step:Combine pole-pushing with the Riemann-sum approximation to prove the precise statement] Let $\Lambda \subseteq \mathbb{C} \setminus K$ contain exactly one point from each bounded connected component of $\mathbb{C} \setminus K$. We show every $f \in \mathcal{O}(K)$ is a uniform limit on $K$ of rational functions with poles in $\Lambda$. Let $f \in \mathcal{O}(K)$ and $\varepsilon > 0$. By Step 2, there is a rational function $S$ of the form \begin{align*} S(z) = \sum_{j=1}^N \frac{c_j}{\zeta_j - z} \end{align*} with $\zeta_j \in \mathbb{C} \setminus K$ and $\sup_{z \in K}|S(z) - f(z)| < \varepsilon/2$. For each $j$, the pole $\zeta_j$ lies in some connected component $\Omega_j$ of $\mathbb{C} \setminus K$. - **Case 1**: $\Omega_j$ is bounded. Then there is a unique $\lambda_j \in \Lambda \cap \Omega_j$. By Step 3 (pole-pushing within $\Omega_j$), $\frac{c_j}{\zeta_j - (\cdot)} = -\frac{c_j}{(\cdot) - \zeta_j}$ can be approximated uniformly on $K$ by rational functions with poles only at $\lambda_j$. Choose such an approximant $T_j$ within $\varepsilon/(2N)$ in $\|\cdot\|_K$. - **Case 2**: $\Omega_j$ is the unbounded component of $\mathbb{C} \setminus K$. Then $\Lambda \cap \Omega_j = \emptyset$, and by Step 3 (unbounded component case), $\frac{c_j}{\zeta_j - (\cdot)}$ can be approximated uniformly on $K$ by polynomials. Polynomials are rational functions with no finite poles, hence (vacuously) have all finite poles in $\Lambda$. Choose a polynomial approximant $T_j$ within $\varepsilon/(2N)$ in $\|\cdot\|_K$. Set $R := \sum_{j=1}^N T_j$. Each $T_j$ has all poles in $\Lambda \cup \emptyset = \Lambda$, so $R$ has all poles in $\Lambda$. By the triangle inequality, \begin{align*} \|R - S\|_K \le \sum_{j=1}^N \|T_j - \tfrac{c_j}{\zeta_j - (\cdot)}\|_K \le \sum_{j=1}^N \frac{\varepsilon}{2N} = \frac{\varepsilon}{2}. \end{align*} Combining with $\|S - f\|_K < \varepsilon/2$: \begin{align*} \|R - f\|_K \le \|R - S\|_K + \|S - f\|_K < \varepsilon. \end{align*} Hence rational functions with poles in $\Lambda$ approximate $f$ uniformly on $K$ — this is the precise statement. To deduce $\overline{\mathcal{O}(K)} = \mathcal{R}(K)$ in $C(K)$: rational functions with poles in $\Lambda$ form a subset of rational functions with poles in $\mathbb{C} \setminus K$, whose closure in $C(K)$ is exactly $\mathcal{R}(K)$. The argument above shows every $f \in \mathcal{O}(K)$ is a uniform limit of such rational functions, so $\mathcal{O}(K) \subseteq \mathcal{R}(K)$ and hence $\overline{\mathcal{O}(K)} \subseteq \mathcal{R}(K)$ (since $\mathcal{R}(K)$ is closed by definition). Conversely, every rational function with poles in $\mathbb{C} \setminus K$ is itself in $\mathcal{O}(K)$ (it is holomorphic on the open set $\mathbb{C} \setminus \{\text{its poles}\}$, which contains $K$), so $\mathcal{R}(K) \subseteq \overline{\mathcal{O}(K)}$. Therefore $\overline{\mathcal{O}(K)} = \mathcal{R}(K)$. [/step]