topology question in Hartshorne's Algebraic Geometry On page 15, in Lemma 3.1 it is claimed: "A subset Z of a topological space Y is closed if and only if Y can be covered by open subsets U such that Z $\cap$ U is closed in U for each U."
How do I prove this?
 A: For the $(\Leftarrow)$ direction, let $Z$ be a subset of $Y$ and we will show that $V := Y \setminus Z$ is open.  The given covering of $Y$ has the property that $V \cap U$ is open in $U$ for all $U$.  But $U$ is open in $Y$, so $V \cap U$ is open in $Y$.  And $V = \cup (V \cap U)$ since the $U$'s cover $Y$, so $V$ is open in $Y$, as desired.
A: For the first direction, if $Z$ is closed, then $Y$ is open and $\{Y\}$ is an open cover of $Y$ such that $Z\cap Y$ is closed in $Y$.  
For the converse direction, if $Z$ is not closed, then there is some $z\in \partial Z \cap Z'$.  Let $\{U\}$ be an open cover of $Y$ such that $Z\cap U$ is closed in each $U$.  Consider $U_z \ni z$.  Then in $U_z$, $z\in\partial (Z\cap U)$ but $z\notin (Z\cap U)$, so $Z\cap U$ cannot be closed in $U$ after all.
A: If $Z$ is closed, consider the points of $Y-Z$. For each such point, take an open set containing it disjoint of $Z$. Also, for each point of $Z$, consider any open set containing it. Then you'll have open sets covering $Y$ satisfying the conditions required on the intersections. This proves $(\Rightarrow)$.
