The classification of surfaces Can we completely classify the simply-connected surfaces (with or without boundary) in $\mathbb R^3$ up to homeomorphism?
 A: The classification of surfaces is well-known. Every surface is homeomorphic to either $mT+nD$ or $mP+nD$ where $T$ is a torus, $P$ is a projective plane, $D$ is a disk, $m,n$ are non-negative integers, and $+$ is connected sum. 
Now you want the surfaces to be in ${\bf R}^3$, so that rules out $mP$ for $m\ge1$. 
You also want them simply-connected. That rules out $mT+nD$ for $m\ge1$ and $mP+nD$ for $m\ge1$ and $nD$ for $n\ge2$, leaving only two surfaces: the sphere, and the disk. 
A: For noncompact surfaces --- with additional assumptions that the boundary is empty and the surface is connected --- the answer is the similar to the one given for compact surfaces in the answer of @GerryMyerson, namely the only possibility is the open disc also known as $\mathbb{R}^2$.
To see why one needs a bit of the classification theory for noncompact surfaces. The surface may contain neither a Möbius band nor a punctured torus, for then it would not be simply connected. It follows from the classification theory that the surface embeds in $\mathbb{R}^2$. Nor may it have more than one end (in the sense of Freudenthal), for then there would be an embedded circle separating the two ends from each other, and again it would not be simply connected. It remains to know that every planar, one-ended surface is homeomorphic to $\mathbb{R}^2$.
