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Let $\mathcal{F}$ be a sheaf, and let $s\in\mathcal{F}$ be a section. Show that the support of $s$ is a closed subset of $X$.

Let us consider the sheaf of all measurable functions on $\Bbb{R}$. Then the support of the measurable function:

$f(x)=1$ for $x\in\Bbb{R}\setminus \{0\}$

$f(x)=0$ at $x=0$

is not closed under the usual topology.

Where am I going wrong?

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    What did you find the support of $f$ to be? – 2017-01-02
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    @kobe- Ah I think I confused the whole issue. I get it now. – 2017-01-02

1 Answers 1

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Given $x\in \Bbb R$ and a neighborhood $U$ of $x$, the restriction of $f$ to $U$ is nonzero. Thus, the germ of $f$ at each point of $\Bbb R$ is nonzero. Consequently, the support of $f$ is $\Bbb R$, which is closed in $\Bbb R$.

Here is an argument to show that every section of $\mathcal{F}$ has closed support. Let $V$ be an open subset of $X$, and let $s\in \mathcal{F}(V)$. Given $x\notin \operatorname{Supp}(s)$, $s_x = 0$. Thus, there exists an open neighborhood $U$ of $x$ contained in $V$ such that $s|U = 0$. For all $y\in U$, $s_y = 0$ because $s|U = 0$. Therefore $U \cap \operatorname{Supp}(s) = \emptyset$, and $\operatorname{Supp}(s)$ is closed in $X$.