This is paraphrased from an exercise in Hartshorne.
Let $\mathcal{F}$ be a sheaf on $X$, and for a point $P\in X$ let $\mathcal{F}_P$ denote the stalk of $\mathcal{F}$ at $P$. We define the support of the sheaf $\mathcal{F}$, denoted $\operatorname{Supp}\mathcal{F}$, to be $\{P\in X \mid \mathcal{F}_P \neq 0\}$. Show that $\operatorname{Supp}\mathcal{F}$ is not necessarily closed.
The point of this exercise is to contrast with the fact that the support of a section defined over an open is closed. I imagine that there is a canonical example of such a sheaf $\mathcal{F}$ that I just don't know about.
Best Answer
The most common example is the sheaf $j_!\mathbb{Z}$. Here $j:U\rightarrow X$ is the inclusion of an open set, and $j_!:\operatorname{Sh}(U,\mathbb{Z})\rightarrow \operatorname{Sh}(X,\mathbb{Z})$ is the functor such that $j_!\mathcal{F}$ is the sheaf associated to the presheaf $$V\mapsto\left\{\begin{array}{ll} \mathcal{F}(V) & \text{if $V\subset U$}\\ 0 &\text{ otherwise}\end{array}\right. $$ The sheaf $j_!\mathcal{F}$ has the nice property that $(j_!\mathcal{F})_x=\mathcal{F}_x$ if $x\in U$ and $(j_!\mathcal{F})_x=0$ otherwise. From this it is obvious that the support of $j_!\mathbb{Z}$ is simply $U$ which is open.
This example also shows that there are sheaves with non closed support on any non discrete space !