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Since you reformulated your question, it deserves a new answer. Now, the theorem in the gray box in the new version is true, but what does it really say? It says that the map $\Psi^{-1}: \Psi(\mathcal{R}_d) \mapsto \mathcal{R}_d$ is continuous! (The topology on $\mathcal{R}_d$ is inherited from $\mathbb{P}^{2d+2}$; on $\Psi(\mathcal{R}_d)$ we have, as you state, the topology of uniform convergence with respect to e.g. chordal metric). So what you are really asking is: how to prove that $\Psi^{-1}$ is continuous in these topologies? This is not so hard if you are used to working with homogeneous coordinates on projective spaces, but there is a subtle point involved. Here is the argument: Let $[r:Z]=[a_0:...:a_d:b_0:...:b_d:Z_0:Z_1] \in \mathbb{P}^{2d+2}\times \mathbb{P}^1$ and define $p([r:Z] \in \mathbb{P}^1$ as $p([r:Z]=[a_0{Z_0}^d+a_1{Z_0}^{d-1}Z_1+...+a_d{Z_1}^d:b_0{Z_0}^d+b_1{Z_0}^{d-1}Z_1+...:b_d{Z_1}^d]$. Then the map $p$ is continuous (being given by homogeneous polynomials in homogeneous coordinates), so as $[r_n:Z^{(n)}] \to [r,Z]$, we have $p[r_n:Z^{(n)}] \to p[r:Z]$. Note that $p[r:Z]$ is the value of the rational function $R$ with the coefficients $r$ at the point $[Z_0:Z_1]\in \mathbb{P}^1$. Note also that the continuity of $p$ means that $R_n$ converge continuously to $R$ on $\mathbb{P}^1$. On a (locally) compact space , continuous convergence of a sequence of continuous functions is equivalent to its uniform convergence.

The subtlety is the following: the domain of the map $p$ is $\mathcal{R}_d \times \mathbb{P}^1$, not the whole product $\mathbb{P}^{2d+2}\times \mathbb{P}^1$. (The ratio of two polynomials is not defined at their common zeros. We saw this in Beardon's counterexample: the limit expression $z/z$ (or $[Z_1:Z_1]$) in the limit is not defined at the point $0=[1:0]$.) So the assumption that all $R_n$ and $R$ have the same degree $d$ guarantees that there are no factors in common for each numerator and denominator and we do not consider fractional expressions that are not rational functions on $\mathbb{P}^1$.

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Since you reformulated your question, it deserves a new answer. Now, the theorem in the gray box in the new version is true, but what does it really say? It says that the map $\Psi^{-1}: \Psi(\mathcal{R}_d) \mapsto \mathcal{R}_d$ is continuous! (The topology on $\mathcal{R}_d$ is inherited from $\mathbb{P}^{2d+2}$; on $\Psi(\mathcal{R}_d)$ we have, as you state, the topology of uniform convergence with respect to e.g. chordal metric). So what you are really asking is: how to prove that $\Psi^{-1}$ is continuous in these topologies? This is not so hard if you are used to working with homogeneous coordinates on projective spaces, but there is a subtle point involved. Here is the argument: Let $[r:Z]=[a_0:...:a_d:b_0:...:b_d:Z_0:Z_1] \in \mathbb{P}^{2d+2}\times \mathbb{P}^1$ and define $p([r:Z] \in \mathbb{P}^1$ as $p([r:Z]=[a_0{Z_0}^d+a_1{Z_0}^{d-1}Z_1+...+a_d{Z_1}^d:b_0{Z_0}^d+b_1{Z_0}^{d-1}Z_1+...:b_d{Z_1}^d]$. Then the map $p$ is continuous (being given by homogeneous polynomials of degree two in homogeneous coordinates), so as $[r_n:Z^{(n)}] \to [r,Z]$, we have $p[r_n:Z^{(n)}] \to p[r:Z]$. Note that $p[r:Z]$ is the value of the rational function $R$ with the coefficients $r$ at the point $[Z_0:Z_1]\in \mathbb{P}^1$. Note also that the continuity of $p$ means that $R_n$ converge continuously to $R$ on $\mathbb{P}^1$. On a (locally) compact space , continuous convergence of a sequence of continuous functions is equivalent to its uniform convergence.

The subtlety is the following: the domain of the map $p$ is $\mathcal{R}_d \times \mathbb{P}^1$, not the whole product $\mathbb{P}^{2d+2}\times \mathbb{P}^1$. (The ratio of two polynomials is not defined at their common zeros. We saw this in Beardon's counterexample: the limit expression $z/z$ (or $[Z_1:Z_1]$) in the limit is not defined at the point $0=[1:0]$.) So the assumption that all $R_n$ and $R$ have the same degree $d$ guarantees that there are no factors in common for each numerator and denominator and we do not consider fractional expressions that are not rational functions on $\mathbb{P}^1$.

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Since you reformulated your question, it deserves a new answer. Now, the theorem in the gray box in the new version is true, but what does it really say? It says that the map $\Psi^{-1}: \Psi(\mathcal{R}_d) \mapsto \mathcal{R}_d$ is continuous! (The topology on $\mathcal{R}_d$ is inherited from $\mathbb{P}^{2d+2}$; on $\Psi(\mathcal{R}_d)$ we have, as you state, the topology of uniform convergence with respect to e.g. chordal metric). So what you are really asking is: how to prove that $\Psi^{-1}$ is continuous in these topologies? This is not so hard if you are used to working with homogeneous coordinates on projective spaces, but there is a subtle point involved. Here is the argument: Let $[r:Z]=[a_0:...:a_d:b_0:...:b_d:Z_0:Z_1] \in \mathbb{P}^{2d+2}\times \mathbb{P}^1$ and define $p([r:Z] \in \mathbb{P}^1$ as $p([r:Z]=[a_0{Z_0}^d+a_1{Z_0}^{d-1}Z_1+...+a_d{Z_1}^d:b_0{Z_0}^d+b_1{Z_0}^{d-1}Z_1+...:b_d{Z_1}^d]$. Then the map $p$ is continuous (being given by homogeneous polynomials of degree two in homogeneous coordinates), so as $[r_n:Z^{(n)}] \to [r,Z]$, we have $p[r_n:Z^{(n)}] \to p[r:Z]$. Note that $p[r:Z]$ is the value of the rational function $R$ with the coefficients $r$ at the point $[Z_0:Z_1]\in \mathbb{P}^1$. Note also that the continuity of $p$ means that $R_n$ converge continuously to $R$ on $\mathbb{P}^1$. On a (locally) compact space , continuous convergence of a sequence of continuous functions is equivalent to its uniform convergence.

The subtlety is the following: the domain of the map $p$ is $\mathcal{R}_d \times \mathbb{P}^1$, not the whole product $\mathbb{P}^{2d+2}\times \mathbb{P}^1$. (The ratio of two polynomials is not defined at their common zeros. We saw this in Beardon's counterexample: the limit expression $z/z$ (or $[Z_1:Z_1]$) in the limit is not defined at the point $0=[1:0]$.) So the assumption that all $R_n$ and $R$ have the same degree $d$ guarantees that there are no factors in common for each numerator and denominator and we do not consider fractional expressions that are not rational functions on $\mathbb{P}^1$.