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 $X$.
I don't think your proof works in full generality, since an affine scheme is Noetherian iff the ring whose spectrum it is isomorphic to is Noetherian. Of course, in most situations the rings we are working over are Noetherian, and my first instinct was to do something like you did in your proof, but I don't think it works, unfortunately.
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