Thank you Thomas for your answer. But I mean class of continuous functions from 2-sphere( = { x \in R^3 | ||x||=1}) to itself, topologically.................. Please feel free in case you need any further clarification..............

Thank you Thomas.  Actually I want to consider interesting subclasses of class  of continuous functions from one point compactification of C to itself. Class of continuous functions from one point compactification of C to itself is a bigger class than meromorphic functions on this space.  If I am not wrong  then class of meromorphic functions from S^2 to itself is the class of quotients of complex polynomials..............

Yes, Thomas. There are many results related to meromorphic functions. But I want to know if this class is dense in the class of continuous functions with respect to compact open topology.

Meromorphic functions at least do have the property that  they give you every homotopy class since [S^2, S^2] = \pi_2(S^2) = H_2(S^2) =  Z, . The integer is given by the total degree (i.e. degree of the nominator - degree of denominator) of the meromorphic function.

EDIT: this is wrong

 This is because the degree of a meromorphic function (i.e a holomorphic map CP^1 to CP^1) is non negative. The meromorphic functions are not dense in the continuous functions C^0(S^2, S^2),  for trivial reasons.

 Any meromorphic function f can be written in homogenous coordinates as a quotient  of two relatively prime homogeneous polynomials P and Q i.e. as

 f(Z_1 : Z_2) = (P(Z_1 : Z_2) :  Q(Z_1 : Z_2))   where   P and Q have no common zero’s.

 the degree of this function is degree (P) = degree (Q). In the non homogeneous picture this means that when f is written as a quotient of relatively prime polynomials the degree is the maximal degree of the nominator and denominator. For example  the meromorphic function 

f(z)  = (z^2 + 1) / (z^3 + z + 2)

becomes in homogeneous coordinates:

f(Z_1 : Z_2) = ((Z_1^2Z_2 + Z_2^3) : (Z_1^3 + Z_1Z_2^2 + 2 Z_2^3))

and so has degree 3 instead of degree -1.

However  we can easily construct algebraic functions with non positive degree by taking anti  meromorphic functions using complex conjugation i.e. functions of the form

 g(z_1 : Z_2) = (P(\bar Z_1, \bar Z_2) : Q(\bar Z_1, \bar Z_2))

 Meromorphic functions  on S^2  are rational functions .

We can ask 

Q1. What are the interesting classes of continuous functions from 2-sphere S^2 to itself?.

Q2.   if rational functions    are  dense in the class of continuous functions with respect to compact open topology.

Q3. Whether  continous  functions on S^2   can be approximated uniformly by  rationals?

It seems that the answer is negative  and    that   the question is related to  Mergelyan's theorem.

It  states the following:

Let K be a compact subset of the complex plane C such that C\K is connected. Then, every continuous function f : K- C, such that the restriction f to int(K) is holomorphic, can be approximated uniformly on K with polynomials. Here, int(K) denotes the interior of K.

Mergelyan's theorem is the ultimate development and generalization of the Weierstrass approximation theorem and Runge's theorem.  In the case that C\K is not connected, in the initial approximation problem the polynomials have to be replaced by rational functions

Mergelyan's theorem is a famous result from complex analysis proved by the Armenian mathematician Sergei Nikitovich Mergelyan in 1951.

 Thank you prof  Miodrag Mateljević  for your answer. I also think that your qn no.2 has negative answer but I want confirmed answer.  Yes, Mergelyan's theorem is a related interesting theorem but I could not find answer to my qn. using this theorem...........

I think you can use theorem:If  the sequence R_n  of rational functions   converge  uniformly  on the complex sphere  to a function R,

then R is  rational.

It seems to  me if  we consider  a functon f in in C(S^2, S^2)   which is not rational then 

f can not be uniformly approximated by rational functions, because   the set of rational functions is closed  in C(S^2, S^2)  in compact open topology .

Thomas: In  a standard  Complex Analysis book    there is   Weisstras theorem:

If  the sequence f_n  o f analytic   functions   converge  uniformly  on the complex sphere  to a function f,

then  f  is  analytic.

Also as far as I remember  there is theorem  concerning  rational functions:

If  the sequence R_n  of rational functions   converge  uniformly  on the complex sphere  to a function R,then R is  rational.  We need to find the source.

Thomas:   For example  let    R_n = 1/(z-n). R_n    is a sequence of rational functions which  converge    to  R=0 pointwise.Bur it does not converge  uniformly on the complex sphere to a function R=0.

@ThomasKorimort if a closed set Z is dense in X,  then Z = X  (because being dense means that the closure of Z is X and being closed means that the closure of Z is Z). However non rational functions definitively exist.

A simple example which is still somehow algebraic is  the extension of complex conjugation on the complex plane

(z1 : z2)  \to  (bar(z_1): \bar(z_2)).

it has degree -1.  

More generally the anti rational functions

(P(\bar z1: \bar z2) : Q(\bar z_1: \bar z_2))) for

homogeneous polynomials P and Q of degree -d.

@ Prof Miodrag Mateljević , I am not able to understand following:

If  the sequence R_n  of rational functions   converge  uniformly  on the complex sphere  to a function R,then R is  rational.  

Because, how do you define uniform convergence of sequences on sphere? 

The chordal distance is $ c(z,\omega) = 2 k\, e$, where $k =

k(z,\omega)= \frac{1}{\sqrt{1 +|z |^2}\,\sqrt{1 +| \omega|^2}}$

and $ e(z,\omega)= |z - \omega |$..

We define  uniform convergence of sequences on sphere  with respect to   the chordal distance  (or the spherical metric ). Then we define the distance between  f , g  in C(S^2, S^2)   by   d(f,g)=   max { c(fz,gz): z \in S^2}.  A  sequence f_n o f  functions converge uniformly on the complex sphere to a function f  if   d(f_n,f)   converges to 0.  As soon as posible I will find the literature and   try to attach   some of my files related to the subject.

@Miodrag Mateljević

If  $S^2= C \cup \{\infty\}, then what is $c(z,\infty$, for $z\ne \infty$?

The chordal distance is $ c(z,\omega) = 2 k\, e$, where $k =

k(z,\omega)= \frac{1}{\sqrt{1 +|z |^2}\,\sqrt{1 +| \omega|^2}}$

and $ e(z,\omega)= |z - \omega |$. Since $ h^2 = 1 -

(\frac{c}{2})^2 $, $ h = k\, |1 + z \overline{\omega}|$, it

follows $ s(z,\omega) = 2 arcsin \, \frac{c(z,\omega)}{2} = 2 \,

arc tg\, \frac{c}{2\, h}= 2\, arc tg\, \frac{|z - \omega |}{ |1 +

z\, \overline{\omega}| }$.

In particular, $c(z, \infty) = \frac{2}{1 +

|z|^2}.$

The interesting classes of continuous functions from 2-sphere S^2 to itself are  also Mobius transformations,  homeomorphisams and diffeomorphisms, quasiconformal mappings (qc)     of  2-sphere S^2  onto  itself.

 The meromorphic functions are not dense in the continuous functions C^0(S^2, S^2),  for trivial reasons. This is because the degree of a meromorphic function (i.e a holomorphic map CP^1 to CP^1) is non negative.  Any meromorphic function f can be written in homogenous coordinates as a quotient  of two relatively prime homogeneous polynomials P and Q i.e. as

f(Z_1 : Z_2) = (P(Z_1 : Z_2) :  Q(Z_1 : Z_2))   where   P and Q have no common zero’s.

the degree of this function is degree (P) = degree (Q).

 In a previous post I claimed that the degree of a meromorphic function written as a quotient of the inhomogeneous variable z is degree of the nominator - degree of the denominator. This is wrong. E.g.

f(z)  = (z^2 + 1) / (z^3 + z + 2)

becomes in homogeneous coordinates:

 f(Z_1 : Z_2) = ((Z_1^2Z_2 + Z_2^3) : (Z_1^3 + Z_1Z_2^2 + 2 Z_2^3))

 and so has degree 3 instead of degree -1.

However  we can easily construct algebraic functions with non positive degree by taking anti  meromorphic functions using complex conjugation i.e. functions of the form

 g(z_1 : Z_2) = (P(\bar Z_1, \bar Z_2) : Q(\bar Z_1, \bar Z_2))

Rogier Brussee  is right:

the meromorphic functions are not dense in the continuous functions C^0(S^2, S^2),. This is because the degree of a meromorphic function (i.e a holomorphic map CP^1 to CP^1) is non negative. But as I said before  the set of rational functions  R(S^2, S^2)  on $S^2$ is closed  with respect to spherical metric.  Of course there is  a continuous function in  C^0(S^2, S^2)  which is not rational and there fThe meromorphic functions are not dense in the continuous functions . Perhaps  an iteresting question is:

whether  every homeomorphisam can be aproximated by quasiconformal mappings on $S^2$.

Prof Miodrag Mateljević,

I will be really gratefull  if you can provide some literature  related to some results about rational functions and metric d(f, g)= max { c(fz,gz): z \in S^2}, in C(S^2, S^2). ......................... 

RB Yadav,   see  for example  M. Fragoulopoulou, V. Nestoridis and I. Papadoperakis, Some results on spherical approximation, Bull. London Math. Soc.

45 (2013) 1171–1180

See  Function Theory of One Complex Variable,

by Robert Everist Greene, Steven George Krantz, for analytic functions.

It seems that we can reduce proof to  analytic functions  case:

If the sequence $R_n$ of rational functions converge uniformly on the complex sphere to a continuous function R, then R is rational. We need to find the source.

Set $V= R^{-1} (C)$ and $F=\overline{C}\setminus V$. Consider $A(w)= 1/w$ and $A\circ R_n$. One can conclude that $R$ is analytic on V and $A\circ R$ on a nbg of F. Hence R is rational function.

See https://www.researchgate.net/profile/Maria_Fragoulopoulou/publication/260244698_Some_results_on_spherical_approximation/links/02e7e5304bacf0ff18000000.pdf

If the sequence $R_n$ of rational functions converge uniformly on the complex sphere to a continuous function R, then R is rational. We need to find the source.

Set $V= R^{-1} (C)$ and $F=\overline{C}\setminus V$. Consider $A(w)= 1/w$ and $1/ R_n$. One can conclude that $R$ is analytic on V and $A\circ R$ on a nbg of F. Hence R is rational function.

Article Some results on spherical approximation