$\newcommand{\mc}{\mathcal}$ Let $\mc F$ be a presheaf of abelain groups on a topological space $X$. For each open set $U$ of $X$, and for each $s\in \mc F(U)$, write $s_x$ to denote the germ of $s$ at $x$.
We define the plant $P\mc F$ generated by $\mc F$ as the set of all the germs of $\mc F$. We topologize it as follows:
For an open set $U$ in $X$, define $B_U$ as the set $\{s_x: x\in U, s\in \mc F(U)\}$. The first paragraph of Section 2.3 of this article mentions that the collection $\mc B=\{B_U\}_{U\text{ open in } X}$ forms a basis for $P\mc F$.
It is clear to me that $\bigcup_{U\text{ open in } X} B_U = P\mc F$. But I am not able to show the following (which is wehre I am stuck):
For each $U$ and $V$ open in $X$ with the property that $B_U\cap B_V\neq \emptyset$, there is $W$ open in $X$ such that $B_W\subseteq B_U\cap B_V$.
This needs to hold in order for $\mc B$ to actually be a basis (and not just a subbasis).
What I tried was this: Suppose $B_U\cap B_V\neq \emptyset$. Then it follows that $U\cap V$ is non-empty and that there is an open $W\subseteq U\cap V$ with $s_x\in B_U\cap B_V$ for some $s\in \mc F(W)$ and some $x\in W$. So it is natural to expect that $B_W\subseteq B_U\cap B_V$. But I am not able to show this. For if we take a point $p\in W$ and $t\in \mc F(W)$, then if the expectation is well-founded we should have $t_p\in B_U\cap B_W$. So there must exist $t^U\in \mc F(U)$ and $t^V\in \mc F(V)$ such that $t_x=t^U_x=t^V_x$. But I just can't see why this should hold.
Can somebody please help. I have just started to learn about sheaves.
You simply misread how the topology is defined. A basis of the topology is given by $\{B_{s,U}\,|\, U\subseteq X \text{ open}, s\in\mathcal F(U)\}$, where $B_{s,U} = \{s_x\,|\, x\in U\}$, where $s_x$ is the germ of $s$ at $x$.
The property $P\mathcal F = \bigcup_{(s,U)} B_{s,U}$ is trivial from the definitions.
The second property to check is the following:
Okay, so given $g\in B_{s,U}\cap B_{t,V}$, we have $g = s_x = t_x$ for some $x\in U\cap V$. By definition of the stalks, we find $W\subseteq U\cap V$ containing $x$, such that $s\big|_W = t\big|_W$. Therefore, putting $r:= s\big|_W = t\big|_W$, we have $r_y = (s\big|_W)_y = s_y$ and $r_y = (t\big|_W)_y= t_y$ for all $y\in W$. This shows $B_{r,W}\subseteq B_{s,U}\cap B_{t,V}$, and $g = r_x\in B_{r,W}$ is clear.