Let $R$ be a rational function of degree $d$ mapping the Riemann sphere to itself:$$R(z) = \frac{a\_d z^d + a\_{d-1} z^{d-1} + \cdots + a\_0}{b\_d z^d + b\_{d-1} z^{d-1} + \cdots + b\_0}$$
where $a\_d$ and $b\_d$ are not both zero. And suppose that a sequence of coefficients $\{(a\_d, a\_{d-1}, \ldots, a\_0; b\_d, b\_{d-1}, \ldots, b\_0)\_n\}$ converges to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. Let $R\_n$ be the rational function with coefficients $(a\_d, a\_{d-1}, \ldots, a\_0; b\_d, b\_{d-1}, \ldots, b\_0)\_n$.



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> Is it a theorem that: The sequence $\{R\_n\}$ converges uniformly to $R$ on the Riemann sphere if and only if for all $n$ sufficiently large the degree of $R\_n$ matches the degree of $R$?
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I ask because I am reading Beardon's *Iteration of Rational Functions* and in section 2.8 he introduces a mapping $\Psi:\mathcal{R}\to\mathbb{C}\mathbb{P}^{2d+1}$, which takes a rational function to its coefficients. Beardon omits the details, but claims that $\Psi$ is a homeomorphism of $\mathcal{R}\_d$ onto $\Psi(\mathcal{R}\_d)$, where $\mathcal{R}\_d$ is the space of rational functions having degree exactly $d$.


My thought is, if for $n$ sufficiently large $R\_n$ has degree $d$, then $(a\_d, a\_{d-1}, \ldots, a\_0; b\_d, b\_{d-1}, \ldots, b\_0)\_n \in \Psi(\mathcal{R}\_d)$. Now $\Psi^{-1}$ is defined and continuous at $(a\_d, a\_{d-1}, \ldots, a\_0; b\_d, b\_{d-1}, \ldots, b\_0)\_n$, because $\Psi$ is a homeomorphism of $\mathcal{R}\_d$ onto $\Psi(\mathcal{R}\_d)$. And (up to multiplication by a constant) $R\_n = \Psi^{-1}((a\_d, a\_{d-1}, \ldots, a\_0; b\_d, b\_{d-1}, \ldots, b\_0)\_n)$. Since $\Psi^{-1}$ is continuous on the sequence of coefficients and at the limit of the sequence of coefficients, $R\_n$ converges to $R$. But I would like to know:



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> Is the convergence always uniform under these circumstances?
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Conversely, if for all $n$ there exists $R\_{m>n}$ in the sequence with degree not equal to $d$, then $\{R\_n\}$ cannot converge to $R$ because (Beardon also states that) the spaces $\{\mathcal{R}\_0, \mathcal{R}\_1, \ldots, \mathcal{R}\_d\}$ are the connected components of the space of rational functions with degree at most $d$.


The reason that I think this might be a theorem is that Beardon introduces a metric $\rho(R,S) = \sup\_{z\in\mathbb{C}\_\infty} \sigma\_0 (R(z),S(z))$, $\sigma\_0$ the chordal metric on $\mathbb{C}\_{\infty}$, just a couple of pages earlier than the point where he introduces the mapping $\Psi$. It is tempting to read "$\Psi$ is a homeomorphism of $\mathcal{R}\_d$ onto $\Psi(\mathcal{R}\_d)$" as "$\Psi$ is a homeomorphism of the metric space $(\mathcal{R}\_d,\rho)$ onto $\Psi(\mathcal{R}\_d)$." If that is the correct reading then $R\_n$ converges to $R$ under the metric $\rho$ and the convergence is clearly uniform. But I am not sure that this is the correct reading since Beardon is only explicitly treating $\mathcal{R}\_d$ and $\mathbb{C}\mathbb{P}^{2d+1}$ as topological spaces at the point where he introduces $\Psi$.


Note: I have edited the question in response to Margaret Friedland's answer, in an attempt to clarify my assumptions (and hopefully my source of confusion). I am a little worried that the reason my point of confusion was unclear is that it is actually trivial. I apologize in advance if that is the case.



When a sequence of coefficients converges to the coefficients of a rational function $R$, when does the sequence $R_n$ converge uniformly to $R$?
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 
 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 $\mathbb{P}^1$.


