Does every nonempty scheme contains a closed point?
Answer: No. There exist nonempty schemes with no closed points.
However, for quasi-compact schemes the statement is true.
Recall the following
Easy Lemma: A scheme $X$ is quasi-compact iff it can be written as a finite union of affine schemes.
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Answer: No. There exist nonempty schemes with no closed points.
However, for quasi-compact schemes the statement is true.
Recall the following
Easy Lemma: A scheme $X$ is quasi-compact iff it can be written as a finite union of affine schemes.
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Quasi-Compact Schemes have Closed Points
Let $X$ be a quasi-compact scheme
Proof 1: (Stack Exchange) As $X$ is quasi-compact, it is a finite union of affine pieces $X = \bigcup_{i=1}^n U_i$. Notice that a subset of $X$ is closed iff it is closed in every $U_i$.
Suppose $X$ does not contain any closed point.
Let $x_1$ be a closed point of $U_1$, then $\overline{\{ x_1\}} \cap U_1 = \{x_1\}$, i.e. $x_1$ is the unique point in $U_1$ of its closure.
As $x$ is not closed, $x$ must not be closed in some $U_i$. WLOG, assume $x_1$ is not closed in $U_2$. Then $\overline{\{ x_1\}} \cap U_2$, being a closed subset of an affine scheme, must contain some closed point $x_2$ of $U_2$ other than $x_1$.
Repeating the argument, we can find $x_3 \in \overline{\{ x_2\}} \subset \overline{\{ x_1\}} $ such that $x_3$ is closed in $U_3$ but $x_3 \not \in U_2 \cup U_1$. Continuing the process we will get $x_{n+1} \not \in U_{n} \cup U_{n-1}\cup \cdots \cup U_1$, a contradiction.
Remark. We constructed a sequence $x_1, \ldots, x_n, \ldots$ satisfying:
Proof 2: (Akhil) Using Quasi-compactness + Zorn's Lemma, we can show that $X$ has a minimal nonempty closed set $Z$. We can put a scheme structure on $Z$ such that the underlying topology is the subspace topology. As $Z$ is a scheme, it contains an affine piece $U$. As $Z - U$ is a closed subset of $Z$, by minimality of $Z$ we must have $Z- U = \emptyset$, i.e. $Z = U$ is affine so it must contains a closed point.
Proof 3: (Mine)
If $X$ has no closed points, then the closure of every point contains a point other than itself.
Thus we have a nested sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is some $x$ such that $\overline{\{x\}}$ does not contain any closed point other than $x$, contradiction.
Suppose $X$ does not contain any closed point.
Let $x_1$ be a closed point of $U_1$, then $\overline{\{ x_1\}} \cap U_1 = \{x_1\}$, i.e. $x_1$ is the unique point in $U_1$ of its closure.
As $x$ is not closed, $x$ must not be closed in some $U_i$. WLOG, assume $x_1$ is not closed in $U_2$. Then $\overline{\{ x_1\}} \cap U_2$, being a closed subset of an affine scheme, must contain some closed point $x_2$ of $U_2$ other than $x_1$.
Repeating the argument, we can find $x_3 \in \overline{\{ x_2\}} \subset \overline{\{ x_1\}} $ such that $x_3$ is closed in $U_3$ but $x_3 \not \in U_2 \cup U_1$. Continuing the process we will get $x_{n+1} \not \in U_{n} \cup U_{n-1}\cup \cdots \cup U_1$, a contradiction.
Remark. We constructed a sequence $x_1, \ldots, x_n, \ldots$ satisfying:
- $x_i$ is a closed point of $U_i$;
- $x_j$ is a specialization of $x_i$ for every $j > i$;
- $x_i$'s are all distinct
- $x_j \not \in U_i$ for all $j < i$.
Proof 2: (Akhil) Using Quasi-compactness + Zorn's Lemma, we can show that $X$ has a minimal nonempty closed set $Z$. We can put a scheme structure on $Z$ such that the underlying topology is the subspace topology. As $Z$ is a scheme, it contains an affine piece $U$. As $Z - U$ is a closed subset of $Z$, by minimality of $Z$ we must have $Z- U = \emptyset$, i.e. $Z = U$ is affine so it must contains a closed point.
Proof 3: (Mine)
If $X$ has no closed points, then the closure of every point contains a point other than itself.
Thus we have a nested sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is some $x$ such that $\overline{\{x\}}$ does not contain any closed point other than $x$, contradiction.
This kind of argument is useful in proving some property (that might not be open) holds for every point of $X$ if it holds for closed point of $X$. For example, we use this argument to check that for a quasi-compact scheme $X$, it suffices to check reducedness at closed points.
Main Usage
We say that property $P$ of points of a scheme is open if whenever a point $x$ satisfies $P$, a neighborhood of it must satisfy $P$ also.
Let $X$ be a quasi-compact scheme. To show that all points of $X$ satisfies $P$. It suffices to show that all \emph{closed} points of $X$ do.
Explanation: As every closed subset of a quasi-compact scheme is quasi-compact, it must contains a closed point. In particular, for every $x \in X$, $\overline{\{x\}}$ contains a closed point $y$. By definition of closure, we have that every neighborhood of $y$ must contain $x$. So any covering of closed points must also be a cover of $X$.
