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Let $f : {\mathbb{R}}^n \to \mathbb{R}$ be the function $f(x) = \exp\left(- \frac{1}{1 - {\|x\|}^2}\right) {\chi}_{B(0 , 1)}(x)$ for all $x \in {\mathbb{R}}^n$, where $B(0 , 1)$ is the open unit ball in ${\mathbb{R}}^n$. I have to show that $f \in {\mathcal{C}}_c^{\infty}({\mathbb{R}}^n)$, but I have no idea because I think that $\mbox{supp} f = B(0 , 1)$ and $B(0 , 1)$ isn't a closed set in ${\mathbb{R}}^n$, so $\mbox{supp} f$ can't be a compact set in ${\mathbb{R}}^n$. Obviously, $f \in {\mathcal{C}}^{\infty}({\mathbb{R}}^n)$, so I only would like to show that $\mbox{supp} f$ is a compact set in ${\mathbb{R}}^n$. Thank you very much.

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    In a topological context, the support of a real-valued function is the *closure* of the set of points in the domain where it is nonzero, – 2017-01-20
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    Typically, we define the support of $f$ to be the closure of the set where $f$ is nonzero: $$\text{supp} f = \overline{\{x \in \mathbb R^n : f(x) \neq 0\}}.$$ – 2017-01-20
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    @murray that's the support of a *continuous* function. Though for the purposes here, this doesn't really matter. – 2017-01-20
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    It's quite odd that you find the fact that $f\in C^{\infty}$ is obvious. It isn't that easy to prove. – 2017-01-20
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    I know but I think that I have already shown. My question now is the next: the definition of support is $$ \mbox{supp} f = \overline{\{x \in {\mathbb{R}}^n : f(x) \neq 0\}} \quad \mbox{ or } \quad \mbox{supp} f = \{x \in {\mathbb{R}}^n : f(x) \neq 0\}\mbox{?} $$ – 2017-01-20
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    @joseabp91 the answer is: it depends. See the wiki page I mentioned in my answer. – 2017-01-20
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    Ok thank you to all the people. – 2017-01-20

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For continuous real valued $f$ the set $f^{-1}(0)$ is always closed (since $\{0\}$ is a closed set) and the set $f^{-1}(\mathbb{R}\backslash \{0\})$ is always open. The support of a continuous function $f$, by definition, is the closure of the complement of the zero set of $f$

If you think the support of a (continuous) function has to be closed than you are most likely working with the definition which says the support is the closure of $f^{-1}(\mathbb{R}\backslash \{0\})$. See this wiki page for more details: https://en.wikipedia.org/wiki/Support_(mathematics)

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    Thank you Thomas. And another question if it is not much to ask, why the two definitions don't match when it comes to a topological space? – 2017-01-20
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    @joseabp91 I don't know for sure, but if you only have a set there is no way to change the definition of support in a natural way. A closure is simply not defined. So you can't do better. As soon as you have a topology you can, and it turns out that a definition in which the support is closed is much better adapted to most questions you encounter. – 2017-01-20