Variant of Riemann mapping theorem Put $D=\{z\in \mathbb C: |z|<1\}$ (open disk) and let $\Omega$ be non empty open simply connected in $\mathbb C$ and $\Omega \neq \mathbb C.$ Then Riemann mapping theorem tells us that
there exists biholomorphic map $f:\Omega \to D.$

My Question is: Suppose that $\Omega$ is a nonempty open connected subset of $\mathbb C$ and $\Omega \subset (\mathbb C \setminus X),$ (where $X$ is closed connected and not singleton).  Can we expect to get a map $f:\Omega \to \mathbb C$ which is holomorphic , one-one and bounded ? If yes, how?

(We note that if $\Omega$ is simply connected then of course this is true by RMT)
 A: In general, it depends. But with $X = \mathbb{C}\setminus \Omega$ connected and not a singleton, things are easy. If $X$ is bounded, then $\widehat{\mathbb{C}} \setminus X$ is by the Riemann mapping theorem biholomorphically equivalent to the unit disk, and thus $\Omega$ is biholomorphically equivalent to the punctured unit disk. If $X$ is unbounded, $X\cup \{\infty\}$ is closed (in the Riemann sphere) and connected, and $\Omega = \mathbb{C}\setminus X = \widehat{\mathbb{C}} \setminus (X\cup \{\infty\})$ is biholomorphically equivalent to the unit disk. So under these assumptions, there are bounded injective holomorphic functions on $\Omega$.

If $\mathbb{C}\setminus \Omega$ is not connected, things are more complicated.
If $\mathbb{C}\setminus \Omega$ consists only of isolated points for example, the Riemann removable singularity theorem asserts that every bounded holomorphic $f\colon \Omega \to \mathbb{C}$ has an extension to a bounded entire function, and is therefore constant. We can iterate the removal of removable singularities, so if any iterated derived set of $\mathbb{C}\setminus \Omega$ is empty, all bounded holomorphic functions on $\Omega$ are constant. We can thus assume that $\mathbb{C}\setminus \Omega$ contains no isolated points.
If $X$ contains a connected set $K$ with more than one point, then by the above $C\setminus K$ is biholomorphically equivalent to either the unit disk or the punctured unit disk, and hence $\Omega$ is biholomorphically equivalent to a subset of the unit disk, so then there is a bounded injective holomorphic function on $\Omega$.
If the complement of $\Omega$ is a perfect totally disconnected set, things become difficult. If the analytic capacity of $(\mathbb{C}\cup\{\infty\}) \setminus \Omega$ is $0$, every bounded holomorphic function on $\Omega$ is constant, so no injective bounded holomorphic function on $\Omega$ exists. If the analytic capacity is positive, I don't know if there is always an injective bounded holomorphic function on $\Omega$, I suspect that to guarantee the existence of that, more than positive analytic capacity is required, but that's just a guess.
