TITLE: Is the set of closed points of a $k$-scheme of finite type dense?
QUESTION [11 upvotes]: Let $k$ be a field.
Let $X$ be a scheme of finite type over $k$.
We denote by $X_0$ the set of closed points of $X$.
Is $X_0$ dense in $X$?
Motivation
See my comment to Martin Brandenburg's answer to this question.

REPLY [7 votes]: Lemma 1
Let $k$ be a field.
Let $A$ be an integral domain which is a $k$-algebra of finite type.
Let $\bar k$ be an algebraic closure of $k$.
Then there exists $k$-algebra homomorphism $\psi\colon A \rightarrow \bar k$.
Proof:
This is a special case of this question(take $b = 1$).
Lemma 2(weak Nullstellensatz)
Let $k$ be a field.
Let $A$ be a $k$-algebra of finite type.
Let $\mathfrak{m}$ be a maximal ideal of $A$.
Then $A/m$ is finite over $k$.
Proof:
Let $\bar k$ be an algebraic closure of $k$.
By Lemma 1, there exists $k$-algebra homomorphism $\psi\colon A/\mathfrak{m} \rightarrow \bar k$.
Since $A/\mathfrak{m}$ is a field, $\psi$ is injective.
Hence $A/\mathfrak{m}$ is algebraic over $k$.
Since it is of finite type over $k$, it is finite over $k$.
QED
Lemma 3
Let $k$ be a field.
Let $X$ be an affine $k$-scheme of finite type.
Let $x \in X$.
Then $k(x)$ is algebraic over $k$ if and only if $x$ is a closed point of $X$.
Proof:
Suppose $X = Spec(A)$, where $A$ is a $k$-algebra of finite type.
Let $P$ be a prime ideal which corresponds to $x$.
We can identify $k(x)$ with the field of fractions of $A/P$.
Suppose $x$ is a closed point of $X$.
Then $P$ is a maximal ideal.
Hence $k(x) = A/P$ is algebraic over $k$ by Lemma 2.
Conversely suppose $k(x)$ is algebraic over $k$.
Then $A/P$ is a finite algebra over $k$.
Since $A/P$ is an integral domain, $A/P$ is a field.
Hence $P$ is a maximal ideal.
QED
Lemma 4
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $U, V$ be affine open subsets of $X$ such that $U\cap V \ne \emptyset$.
Let $x \in U \cap V$.
Suppose $x$ is a closed point of $U$.
Then $x$ is also a closed point of $V$.
Proof:
This follows immediately from Lemma 3.
Lemma 5
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $U$ be a non-empty affine open subset of $X$.
Then any closed point of $U$ is a closed point of $X$.
Proof:
Let $x$ be a closed point of $U$.
Let $Z$ be the closure of $\{x\}$ in $X$.
Let $\{U_i\}$ be an affine open cover of $X$.
Since $Z = Z \cap X = \bigcup_i Z \cap U_i$, it suffices to prove that $Z \cap U_i = \{x\}$ whenever$Z \cap U_i$ is non-empty.
Suppose $Z \cap U_i$ is non-empty.
Then $x \in U_i$.
By Lemma 4, $x$ is a closed point of $U_i$.
Hence $Z \cap U_i = \{x\}$
QED
Proposition
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $X_0$ be the set of closed points of $X$.
Then $X_0$ is dense in $X$.
Proof:
This follows immediately from Lemma 5.