Example 6.11.5. Let $X$ be a topological space. Let $A_ x$ be a set for each $x \in X$. Consider the sheaf $\mathcal{F} : U \mapsto \prod _{x\in U} A_ x$ of Example 6.7.5. We would just like to point out here that the stalk $\mathcal{F}_ x$ of $\mathcal{F}$ at $x$ is in general *not* equal to the set $A_ x$. Of course there is a map $\mathcal{F}_ x \to A_ x$, but that is in general the best you can say. For example, suppose $x = \mathop{\mathrm{lim}}\nolimits x_ n$ with $x_ n \not= x_ m$ for all $n \not= m$ and suppose that $A_ y = \{ 0, 1\} $ for all $y \in X$. Then $\mathcal{F}_ x$ maps onto the (infinite) set of tails of sequences of $0$s and $1$s. Namely, every open neighbourhood of $x$ contains almost all of the $x_ n$. On the other hand, if every neighbourhood of $x$ contains a point $y$ such that $A_ y = \emptyset $, then $\mathcal{F}_ x = \emptyset $.

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