Why do we take the closure of the support? In topology and analysis we define the support of a continuous real function $f:X\rightarrow \mathbb R$ to be $ \left\{ x\in X:f(x)\neq 0\right\}$. This is the complement of the fiber $f^{-1} \left\{0 \right\}$. So it looks like the support is always an open set. Why then do we take its closure?
In algebraic geometry, if we look at elements of a ring as regular functions, then it's tempting to define their support the same way, which yields $\operatorname{supp}f= \left\{\mathfrak p\in \operatorname{Spec}R:f\notin \mathfrak p \right\}$. But these are exactly the basic open sets of the Zariski topology. I'm just trying to understand whether this is not a healthy way to see things because I've been told "supports should be closed".
 A: Given  a scheme $X$ there are two notions of support  $f\in \mathcal O(X)$:
1) The first definition is the set of of points $$\operatorname {supp }(f)= \left\{ x\in X:f_x\neq 0_x\in \mathcal O_{X,x}\right\}$$ where the germ of $f$ at $x$ is not zero.
This  support is automatically closed: no need to take a closure.
2) The second definition is  the good old zero set of $f$ defined by $$V(f)=\{ x\in X:f[x]=\operatorname {class}(f_x)\neq 0\in \kappa (x)=\mathcal O_{X,x}/ \mathfrak m_x\}$$ It is also automatically closed.
3) The relation between these closed subsets is$$ V(f)\subset \operatorname {supp }(f)$$with strict inclusion in general:
For a simple example, take $X=\mathbb A^1_\mathbb C=\operatorname {Spec}\mathbb C[T],\: f=T-17$ .
Then for $a\in \mathbb C$ and $x_a=(T-a)$ we have $f[a]=a-17\in \kappa(x_a)=\mathbb C$ and  for the generic point $\eta=(0)$ we have $f[\eta]=T-a\in \kappa(\eta)=\operatorname {Frac}(\frac  {\mathbb  C[T]}{(0)})=\mathbb C(T)$.
Thus  $f[x_{17}]=0$ and  $f[P]\neq 0$ for all other  $P\in \mathbb A^1_\mathbb C$ , so that  $$V(f)=\{x_{17}\}\subsetneq \operatorname {supp }(f)=\mathbb A^1_\mathbb C$$
