Analytic contraction of the Stone-Cech compactification of $\mathbb C$

Let $S$ be the Stone-Cech compactification of $\mathbb C$. Then any meromorphic function $f:\mathbb C \to \mathbb{CP}^1$ extends to $S$.

Do the meromorphic functions separate the points of $S$?

I doubt there are enough such functions, though I could also believe that this could depend on our set theory.

If not, "what is" the space $S/\!\sim$, where $p\sim q$ if $f(p)=f(q)$ for all meromorphic $f$?

Again, I could easily believe that this question has very different answers depending on one's set theory.

I don't have any especially good reason for asking this question. Feel free to retag.

• If I'm not mistaken about what you mean, there are only $2^{\aleph_0}$ meromorphic functions $\mathbb{C} \to \mathbb{CP}^1$. But $\beta\mathbb{C}$ has size $2^{2^{\aleph_0}}$ so there aren't enough meromorphic functions to separate points of $\beta\mathbb{C}$. – François G. Dorais Jan 7 '14 at 18:01
• I hope I'm not misunderstanding Francois's counting argument: There are $2^{\aleph_0}$ meromorphic functions, so there are $2^{2^{\aleph_0}}$ functions from the set of meromorphic functions to $\mathbb{CP}^1$. This is enough for each element of $\beta\mathbb{C}$ to determine a different function {meromorphic functions}$\to\mathbb{CP}^1$ (i.e., for meromorphic functions to separate points of $\beta\mathbb{C}$). Also, aren't there only $2^{\aleph_0}$ continuous functions $\mathbb{C}\to\mathbb{CP}^1$? – Julian Rosen Jan 7 '14 at 18:39
• I don't understand how the cardinality argument is supposed to work. After all, the identity function $\mathbb{R}\to\mathbb{R}$ separates points in $\mathbb{R}$ all by itself, so it can't just be a matter of simple counting. – Joel David Hamkins Jan 7 '14 at 18:41
• I believe the first question is equivalent to the question of whether the space of continuous combinations $F(f_1,\dots,f_r)$ of finitely many meromorphic functions $f_1,\dots,f_r: {\bf C} \to {\bf CP}^1$ is dense in the space of all bounded continuous functions $G: {\bf C} \to {\bf C}$ on the complex plane (in the sup norm topology). I'm not getting good intuition as to whether the latter question has a positive answer or not though - it depends on how "rigid" the class of meromorphic functions are. – Terry Tao Jan 8 '14 at 0:50
• Unfortunately, it's the non-compactly supported version of the sup norm which is relevant here; I don't see for instance how to approximate $\sin(x)$ uniformly on ${\bf C}$ (as opposed to uniformly on compacta) by a continuous function of meromorphic functions. Actually even $x$ is problematic because $z \mapsto (z+\bar{z})/2$ is not continuous at the north pole $\infty$ of the Riemann sphere ${\bf CP}^1$. (Indeed, one cannot hope to get any unbounded function as a continuous function of ${\bf CP}^1$-valued functions.) – Terry Tao Jan 8 '14 at 2:03