Let $T$ be a topological space. According to the second edition of EGA I, a maximal point of $T$ is a generic point of an irreducible component of $T$. If $T = |X|$ is the topological space associated to an algebraic space $X$, there are at least two notions of maximal points: we can look at maximal points of $T$ viewed as a topological space, or we can look at images of maximal points of $U$ where $U \to X$ is an étale morphism and $U$ is a scheme. The second notion corresponds to the set of points of codimension $0$ (Lemma 66.11.1). The codimension $0$ points are easier to work with for general algebraic spaces; the two notions agree for quasi-separated and more generally decent algebraic spaces (Decent Spaces, Lemma 68.20.1).
Lemma 66.11.1. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. Let $x \in |X|$. Consider étale morphisms $a : U \to X$ where $U$ is a scheme. The following are equivalent
$x$ is a point of codimension $0$ on $X$,
for some $U \to X$ as above and $u \in U$ with $a(u) = x$, the point $u$ is the generic point of an irreducible component of $U$, and
for any $U \to X$ as above and any $u \in U$ mapping to $x$, the point $u$ is the generic point of an irreducible component of $U$.
If $X$ is representable, this is equivalent to $x$ being a generic point of an irreducible component of $|X|$.
Proof. Observe that a point $u$ of a scheme $U$ is a generic point of an irreducible component of $U$ if and only if $\dim (\mathcal{O}_{U, u}) = 0$ (Properties, Lemma 28.10.4). Hence this follows from the definition of the codimension of a point on $X$ (Definition 66.10.2). $\square$
Lemma 66.11.2. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. The set of codimension $0$ points of $X$ is dense in $|X|$.
Proof. If $U$ is a scheme, then the set of generic points of irreducible components is dense in $U$ (holds for any quasi-sober topological space). Thus if $U \to X$ is a surjective étale morphism, then the set of codimension $0$ points of $X$ is the image of a dense subset of $|U|$ (Lemma 66.11.1). Since $|X|$ has the quotient topology for $|U| \to |X|$ we conclude. $\square$
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