Does every function on a bounded subset of $\mathbb{R}^N$ have compact support?

Question

Let $\Omega \subset \mathbb{R}^N$ and let $C^1_c(\Omega)$ be the set of continuous differentiable functions with compact support in $\Omega$, i.e. the set of $C^1(\Omega)$ functions such that $$\mathrm{supp}(f)=\overline{\{x \in \Omega: f(x) \neq 0 \}},$$ is a compact set.
My question is the following: if $\Omega$ is bounded then is superflous to specify "compact support"? Or in other words, in this case $C^1_c(\Omega)=C^1(\Omega)$?
Since $\mathrm{supp}(f) \subset \Omega$ and $\Omega$ is bounded then $\mathrm{supp}(f)$ is bounded and closed by definition then is compact.
Is it correct?

Answer 1

In this context, $$\mathrm{supp}(f)=\overline{\{x \in \Omega: f(x) \neq 0 \}}$$ means you are taking the closure in $\Omega$, not in $\mathbb{R}^N$. (The idea is that you are considering $f$ as just a function on $\Omega$ as a space on its own, so you only care about points of $\Omega$, not the space $\mathbb{R}^N$ in which it is embedded.) So, $\mathrm{supp}(f)$ will be closed in $\Omega$, but may not be closed in $\mathbb{R}^N$, and so may not be compact. For instance, if you take $\Omega$ to be an open ball in $\mathbb{R}^N$ and $f$ to be the constant function with value $1$, $\mathrm{supp}(f)$ will be $\Omega$ (not the closed ball which is the closure of $\Omega$ in $\mathbb{R}^N$) which is not compact.

Answer 2

NO. The function $f(x)=\frac 1 x$ is continuously differentiable on $(0,1)$ but it is not of compact support. There is no compact set $K$ contained in the open interval $(0,1)$ such that $f(x)=0$ for $x \notin K$.