Pointwise Addition on Continuous Real Functions forms Group

Theorem
Let $$C$$ be the set of all continuous real functions on the set of real numbers $$\R$$.

Let $$f, g \in C$$.

Let $$f + g$$ be the pointwise sum of $$f$$ and $$g$$:
 * $$\forall x \in R: \left({f + g}\right) \left({x}\right) = f \left({x}\right) + g \left({x}\right)$$

Then $$\left({C, +}\right)$$, the algebraic structure on $C$ induced by $+$, forms a group.

Proof
Taking the group axioms in turn:

G0: Closure
From the Combination Theorem for Functions, if $$f$$ and $$g$$ are continuous real functions then so is $$f + g$$.

Thus Closure is demonstrated.

G1: Associativity
Let $$f, g, h \in C$$.

From Real Addition is Associative, it follows directly that:
 * $$\forall x \in \R: f \left({x}\right) + \left({g \left({x}\right) + h \left({x}\right)}\right) = \left({f \left({x}\right) + g \left({x}\right)}\right) + h \left({x}\right)$$

so proving Associativity

G2: Identity
The constant function $$f_0$$ defined as:
 * $$\forall x \in \R: f_0 \left({x}\right) = 0$$

fulfils the role of the Identity:


 * $$\forall x \in \R: f_0 \left({x}\right) + f \left({x}\right) = 0 + f \left({x}\right) = f \left({x}\right) = f \left({x}\right) + 0 = f \left({x}\right) = f_0 \left({x}\right)$$

Note that $$f_0 \in C$$ as the Constant Function is Uniformly Continuous, and hence continuous.

G3: Inverses
From the Combination Theorem for Functions, if $$f \left({x}\right)$$ is continuous then so is $$g \left({x}\right)$$ where:
 * $$\forall x \in \R: g \left({x}\right) = - f \left({x}\right)$$.

Then we note that:
 * $$\forall x \in \R: f \left({x}\right) + \left({- f \left({x}\right)}\right) = 0 = \left({- f \left({x}\right)}\right) + f \left({x}\right)$$

So every element has an inverse.

All the group axioms are satisfied, hence the result.