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.

  • 1
    $\begingroup$ 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 $\beta\mathbb{C}$. $\endgroup$ – François G. Dorais Jan 7 '14 at 18:01
  • 1
    $\begingroup$ 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 $\mathbb{C}\to\mathbb{CP}^1$? $\endgroup$ – Julian Rosen Jan 7 '14 at 18:39
  • 4
    $\begingroup$ 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. $\endgroup$ – Joel David Hamkins Jan 7 '14 at 18:41
  • 3
    $\begingroup$ 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. $\endgroup$ – Terry Tao Jan 8 '14 at 0:50
  • 3
    $\begingroup$ 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.) $\endgroup$ – Terry Tao Jan 8 '14 at 2:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.