Units of a function ring I am working on the following question :

Let $S$ be the ring of all functions from the closed interval $[0, 1]$ to
  $\mathbb{R}$. I need to find the group of units of $S$, i.e., the set of all
  invertible elements of $S$.

I understand that I need to find all bijections from $[0,1]$ to $\mathbb{R}$. But I am failing to construct even one example. Further, I want to write these bijections as a set. Any help is appreciated. Thanks for you time.
 A: I think the answer is "{f(x)} for all x in [0,1],f(x) is not equal to zero" which means set of all non zero functions with domain in [0,1]. because for units of the ring which you mentioned, "being non zero" is necessary and sufficient condition because you just need to inverse every points in range of function and only 0 in real numbers field does not have inverse.
Also, if you were asked to use a field K for range of functions, All fuctions which does not have 0 of this field K are answers.
You can check that set at the first line with operator "product" is an infinite group.Which means ({f(x)},.) which f(x) is a non zero function with domain in [0,1] is a group.1-It is closed under operator. 2-Operator is associative. 3-The constant function g(x)=1 is unique identity. 4-For all members there exists a unique member which is inverse of the member.
This group is a good opportunity for practicing more because it is infinite abelian group and trying to make a ring with this group and ... is a beneficail practice.
