Question: why in below example(see below) $I$ is open interval in $\mathbb{R}$ containing $0$? What happens when we take any other subinterval of $\mathbb{R}$ which does not contains $0$ ?
While solving examples on linear transformations, I came to example : " let $I$ be an open interval of $\mathbb{R}$ containing $0$. Let $D:C^1(I)→C^0(I)$ be differential mapping $D(f)=f'(x)$ then $D$ is linear transformation and $D$ is not injective but $D$ is surjective.
I have solved example completely but I am curious to know, why $I$ is open interval in $\mathbb{R}$ containing $0$? What happens when we take any other subinterval of $\mathbb{R}$ which does not contains $0$ ? I think $D$ still remains linear transformation and with same properties? Please help me