The support of a sheaf is not necessarily closed 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. 
 A: Let $X$ be a space equipped with a topology of nested open sets $\{U_i\}_{i \in \mathbf{N}}$ telescoping down to a point $Q$. So we have $X = U_0 \supset U_1 \supset U_2 \supset \dotsb\; $ and $\bigcap U_i = \{Q\}$. Define the sheaf $\mathcal{F}$ on $X$ such that $\mathcal{F}(U_i) \cong \bigoplus_{\mathbf{N}}\mathbf{Z}$ for each $U_i$ and such that the restriction maps are given by projections that drop the first graded components of the direct sum. For example the restriction $\mathcal{F}(U_0) \to \mathcal{F}(U_{1})$ looks like 
$$\begin{align}
\mathbf{Z} \oplus \mathbf{Z} \oplus \mathbf{Z} \oplus \dotsb \; &\to \; \mathbf{Z} \oplus \mathbf{Z} \oplus \dotsb
\\
(z_0, z_1, z_2, \dotsc) \; &\mapsto \; (z_1, z_2, \dotsc)\,.
\end{align}$$
In general then, the restriction map $\mathcal{F}(U_i) \to \mathcal{F}(U_j)$ drops the first $j-i$ graded components of $\mathcal{F}(U_i)$.  
So the idea is that since direct sums have only finitely many non-zero components, any section $s \in \mathcal{F}(U_i)$ must eventually map to $0 \in \mathcal{F}(U_j)$ for some $j>i$. So $\mathcal{F}_Q = 0$, but for any other point $P\neq Q$, $\mathcal{F}_P = \bigcap_{U_i \ni P} = U_j$ for some $j$ (it's the $j$ corresponding to the "smallest" $U_j$ that contains $P$), and $U_j \neq 0$. So $\operatorname{Supp}\mathcal{F} = X \setminus \{P\}$, which is not closed.
A: Let $X$ be a topological space where any two non-empty open set has non-empty intersection. Fix a point $p\in X$ and define the following for any open set $U$
$$\mathcal{F}(U)=\left\{\begin{array}{ll} 0 & \text{if $U=\emptyset$ or $p\in U$}\\ A &\text{ otherwise}\end{array}\right. $$
where A is any non-trivial abelian group. Restriction maps are either identity or zero maps. This is kind of opposite of Skyscraper and gives a sheaf (strangely).
The stalks of this sheaf are given by
$$\mathcal{F}_q=\left\{\begin{array}{ll} 0 & \text{if $q\in \overline{\{p\}}$}\\ A &\text{ otherwise}\end{array}\right. $$
Then we have $Supp\mathcal{F}=X\setminus\overline{\{p\}}$ which is not necessarily closed and it is not for nice choice of $X$. For example, if you take $X=\mathbb{A}^1$ equipped with Zariski Topology and $p=0$ then $Supp\mathcal{F}=\mathbb{A}^1\setminus\{0\}$ which is not closed in Zariski topology.
A: 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 !
A: It's worth mentioning that if $\mathcal{F}$ (sheaf of $\mathcal{R}$-$\mathscr{Mod}$) is of finite type (locally generated by finite number of sections) then the support is indeed closed.
Proof. let $x\in X\setminus \text{supp}(\mathcal{F})$ (assume it's nonempty), therefore $\mathcal{F}|_{U} = \sum\mathcal{R}|_U \cdot s_i$, therefore $s_i(x) = 0$ therefore locally exist some $V\subset U$(needs to use finite type condition) such that $s_i(y) = 0$ on $y\in V\subset U$ therefore $\mathcal{F}|_V = 0$.
