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 $z_{i+1} \in \bar{z_j} \cap U_j = \{z_j\}$.
Best Answer
Maybe an easier answer? Notice it is enough to show that every closed subset $Z$ of $X$ has a closed point. Observe a point $p \in Z$ is closed in $Z$ if and only if it is closed in $X$ so it suffices to show that $Z$ has a closed point. But $Z$ is also a quasicompact scheme so we reduce to the case of showing that a quasicompact sheme $X$ has a closed point. For this, say $X = U_1 \cup \dots \cup U_n$ is an irredundant decomposition of $X$ as a union of open affines. We can then pick a point $p \in U_1$ that is closed in $U_1$ and such that $p \notin U_j$ for $j \neq 1$. Because $p \in (U_2 \cup \dots \cup U_n)^c$ the closure is also in $(U_2 \cup \dots \cup U_n)^c$. It is then easy to check that the closure of $p$ in $X$ is $p$.