Vakil 5.1 E

Show that in a quasi-compact scheme every point has a closed point in its closure

Solution: Let $X$ be a quasi-compact scheme so that it has a finite cover by open affines $U_i$.

Let $z \in X$, and $\bar z$ its closure. Consider the (finite) sub-collection of $\{U_i\}_{i=1}^N$ that intersect $\bar z$. Because $U_1$ is just the spec of a ring, we can pick a closed point $z_1 \in \bar z \cap U_1$. If $z_1$ is also closed in all other $U_i$ that contain it, we're done, but if it is not closed in some $U_i$ then it is not closed in $X$. However, in that case, we can pick another $z_i \in \bar {z_1} \cap U_i$. Certainly, $z_i$ does not lie in $U_1$, because if it did $z_1$ wouldn't have been a closed point in $U_1$ in the first place.

Now we proceed in the obvious way until we get to a point $z_n$ that is closed in all the open affines that contain it. Notice that this procedure terminates because once we move from $z_i$ to $z_{i+1}$ we can no longer have $z_{i+1}$ contained in any open affine where some $z_{j\le i}$ was closed, because if we did then $z_{i+1} \in \bar{z_j} \cap U_j = \{z_j\}$.