Bump function has a compact support? Sorry for the basic question, but couldn't find the answer. We say that the bump function $\phi(x)=e^{-1/(1-x^2)}$ has a compact support. However, $\phi(x)\neq 0$ only for $x \in [0,1)$, which means the support is $[0,1)$. Is $[0,1)$ compact in $\mathbb{R}$?
 A: Indeed, this function, as other bump functions, has compact support: namely, $[-1,1]$. 
Since the support of a function is closed by definition, and in Euclidean spaces "compact = closed and bounded", one might wonder why we don't say bounded support instead. The main reason is that often the domain under consideration is not all $\mathbb R^n$ but  an open subset $\Omega\subset \mathbb R^n$, which could well be bounded itself. The support of a function $u:\Omega\to\mathbb R$ is a subset of $\Omega$ which is closed in $\Omega$ (that is, with respect to the induced topology on $\Omega$). Now the compactness of support hinges on it being at a distance from the boundary of $\Omega$. And this is essential in analysis, where multiplying something by a bump function is a way to avoid dealing with boundary issues. 
As an aside, bump functions are sometimes considered on infinite-dimensional vector spaces (typically, Banach spaces). In this case they are understood as (smooth or Lipschitz) functions with bounded support. Compactness is rare to find in infinite dimensions.
