I am asked to find set of units, $R^{\times}$, of a ring of continuous functions $\mathbb{R} \to \mathbb{R}$ denoted as $C(\mathbb{R})$.
Now, unit is $a \in R$ such that $ab = 1$ for some $b \in R$. Are these just the inverses of all the functions? I am having a hard time trying to grasp the problem. Thanks.
Note that a necessary condition for $f(x) \in C(\Bbb R)$ to have a multiplicative inverse $g(x) \in C(\Bbb R)$ is that
$f(x)g(x) = 1, \; \forall x \in \Bbb R; \tag 1$
this in turn implies
$f(x) \ne 0, \; \forall x \in \Bbb R; \tag 2$
indeed, if (2) failed there would be $y \in \Bbb R$ with
$1 = f(y)g(y) = 0 \cdot g(y) = 0, \; \Longrightarrow \Longleftarrow; \tag 3$
so we see (2) must hold if (1) does; in fact, (2) is sufficient for $f(x) \in R^\times$, since then the function
$g(x) = (f(x))^{-1} = \dfrac{1}{f(x)} \in C(\Bbb R) = R^\times; \tag 4$
in short, the inverses $R^\times$ are the reciprocals of the non-vanishing functions; but if
$\exists y \in \Bbb R, \; f(y) = 0, \tag 5$
then $f(x)$ is not possessed of a multiplicative inverse.