A smooth curve can be expressed as a smooth projective curve of genus g minus r points, with g and r unique? I have some familiarity with the notions of modern algebraic geometry, but have little knowledge in the concrete theory of curves, so I am stuck at the following questions. 
Let $Y$ be a smooth geometrically integral curve over a field $K$. I have seen the Euler characteristic $\chi(Y)$ be defined as follows: if $Y_{\overline{K}}$ is expressed as a smooth projective curve of genus $g$ minus $r$ points, then $\chi(Y):=2-2g-r$. This raises the following questions: 

$(1)$ Why can any smooth integral curve over an algebraically closed field be written as a smooth projective curve minus finitely many points?
  $(2)$ How unique is this description, e.g. is the pair of numbers $(g,r)$ unique or is only the sum $2g+r$ unique?

Now here are my ideas on the matter, though it may be that they lead nowhere. Corollary on page $5$ of http://math.stanford.edu/~vakil/725/class25.pdf says that any nonsingular curve is either projective or affine. (Incidentally, I don't see how this corollary follows from the Proposition stated before it in Vakil's class notes, so an explanation for that would also be appreciated :))
Hence $Y_{\overline{K}}$ is either projective (in which case we are done) or affine, so we may assume that $Y_{\overline{K}}$ is affine, say $Y_{\overline{K}} \subset \mathbb{A}^n$. Then we may take the projective closure of $Y_{\overline{K}}$, i.e. embed $\mathbb{A}^n$ into $\mathbb{P}^n$ (e.g. as the complement of $\{[x_1,...,x_n,0]\}$) and then take the closure $V$ of $Y_{\overline{K}}$ in $\mathbb{P}^n$, which is a projective curve. But I am not sure if $V$ is necessarily smooth, in general smoothness is not preserved by taking the projective closure. Moreover, I'm not sure if the complement $V - Y_{\overline{K}}$ is finitely many points and have no idea as to question $(2)$.
 A: As you said, any smooth integral curve $Y$ (over an algebraically closed field) is a Zariski open subset of a projective curve $Z$. Let $\pi: X\to Z$ be the normalization. Then, $X$ is smooth (integrally closed=smooth in this situation, for curves). Also, $\pi^{-1}(Y)\to Y$ is an isomorphism, since $Y$ is smooth. Thus, $Y$ is an open set of $X$, which is smooth.
For any integral curve (say $X$), the closed sets are, 1) $X$; 2) $\emptyset$; 3) finite set of points. It should be clear now why $Y$ is the complement of finite set of points in $X$.
The final point, there are several arguments, may be the one I describe is not the simplest. If $Y\subset X, X'$ both $X, X'$ smooth projective, then using identity map on $Y$, we get a rational map $f:X\to X'$. But birational maps from smooth (projective) curve to a smooth projective curve is in fact an isomorphism. This says, the cardinality of $X-Y, X'-Y$ are the same. Also note that since $X,X'$ are isomorphic, their genus are same, giving you uniqueness of $(g,r)$.
