In Goertz-Wedhorn, a prevariety is defined to be a connected space with functions that locally is an affine variety (were an affine variety is a space with functions that is isomorphic to the space with functions of an irreducible algebraic set. The functions of an irreducible algebraic set are just the regular functions). As being irreducible is stronger than being connected, one would expect that there are non-irreducible prevarieties. So the question is:
Are there non-irreducible prevarieties?
I think the answer should be no: Every prevariety is homeomorphic to $X(k)$, where $X$ is an integral scheme of finite type over $k$ (see p. $80$ or so in Goertz-Wedhorn, were they prove that there is a category equivalence between prevarieties and integral schemes of finite type.). But then $X$ is irreducible and so also $X(k)$, as $X(k)$ is very dense. The details of the last step are the following: Assume we have open sets $U, V \subseteq X$ such that $U \cap X(k) \neq \emptyset \neq V \cap X(k)$. We have to prove that the intersection is nonempty. We have
$$(U \cap X(k)) \cap (V \cap X(k)) = (U \cap V) \cap X(k).$$
As $U \neq \emptyset \neq V$ and as $X$ is irreducible, we have $U \cap V \neq \emptyset$. But as the map $W \mapsto W \cap X(k)$ is a bijection from the topology of $X$ to that of $X(k)$ (Here we use that $X(k)$ is very dense) it follows that $(U \cap V) \cap X(k) \neq \emptyset$, thus $X(k)$ is irreducible.
So, if my argument works, I ask you: Why are prevarieties not defined to be irreducible? And can one show directly by the definition of a prevariety that it is irreducible?