Let $X$ be a non-compact holomorphic manifold of dimension $1$. Is there a compact Riemann surface $\bar{X}$ suc that $X$ is biholomorphic to an open subset of $\bar{X}$ ?


**Edit:** To rule out the case where $X$ has infinite genus, perhaps one could add the hypothesis that the topological space $X^{\mathrm{end}}$ (is it a topological surface?), obtained by adding the *ends* of $X$, has finitely generated $\pi\_1$ (or $H\_1$ ). Would the new question make sense and/or be of any interest?


**Edit2:** What happens if we require that $X$ has finite genus? (the *genus* of a non-compact surface, as suggested in a comment below, can be defined as the maximal $g$ for which a compact Riemann surface $\Sigma\_g$ minus one point embeds into $X$)



Is a non-compact Riemann surface an open subset of a compact one ?
You should probably check the following article:
Migliorini, Luca, "On the compactiﬁcation of Riemann surfaces". 
Here is the Mathscinet review about it:
"In this paper the author studies some questions concerning the compactiﬁcations of Riemann
surfaces. It is proved that if $X$ is an open connected Riemann surface then X has ﬁnite genus if and only if there exists a holomorphic injection $i: X \hookrightarrow \tilde{X}$ 
(with $\tilde{X}$ a compact Riemann surface), $i(X)$ being dense in $\tilde{X}$..."


