Closed sets of a curve I am guessing that the closed sets of a curve (i.e. an algebraic variety of dimension 1) are finite . How would one go about proving this? Is it trivial?!
Thanks for any help.
 A: If $X$ is an (irreducible, which may or may not be part of your definition) variety over a field $k$ of dimension $1$ and $F\subseteq X$ is a proper closed subset, then the dimension of $F$ must be strictly less than that of $X$, i.e., $F$ is zero-dimensional. Therefore the irreducible components of $F$ must be points, and since $F$ has finitely many irreducible components (being a Noetherian topological space), and $F$ is equal to the union of its irreducible components, $F$ must be a finite set.
If $X$ is not irreducible, this is not true. For example, $\mathrm{Spec}(k[x,y]/(xy))$, the union of the coordinate axes in $\mathbf{A}_k^2$, is one-dimensional but has two irreducible components, both of dimension one, which are not finite.  
A: Since any curve is a finite union of affine open subsets, you may assume your curve $C=Spec (A)$ is affine.  
a) Assume  $A$ is a one dimensional noetherian domain and $\mathfrak p\subsetneq  A$ is a prime  ideal.
Then $\mathfrak p$ is maximal (because $0\subsetneq \mathfrak p$ is a  saturated  chain of prime ideals since $dim(A)=1$) and $V(\mathfrak p)$ is a closed point.
Since $A$  is noetherian, a closed subset $V(I)\subset C$ only has  finitely many irreducible component and thus is finite.
b) If $A$  is not reduced, the result is false: a counterexample being the curve $C=Spec(\mathbb C[X,Y] /\langle Y^2\rangle)$ which has $C_\text{red}=V(Y)=Spec(\mathbb C[X,Y] /\langle Y\rangle)\subsetneq C$ as a strict closed subscheme of dimension 1.
c) Even if $A$ is reduced but not a domain the result is not true: consider the one dimensional closed subschemes  $V(X)\subsetneq V(XY)\subsetneq Spec(\mathbb C[X,Y])$ 
