Show C(X) is a vector space over $\mathbb R$ with the following operations? I have a set of continuous functions, $C(X): X \rightarrow R$ on a compact metric space, and definitions of addition & multiplication:
$$(f+g)(x) = f(x)+g(x)$$
$$(\lambda f)(x) = \lambda \centerdot f(x)$$
I'm asked to prove this is a vector space over $\mathbb R$. The problem is I thought proving vector spaces involved showing that the Abelian group axioms and the multiplication operation hold, but this seems clear from the question statement already. What am I missing here?
 A: I'll go through the proof for a couple of the axioms for a vector space.
$C(X)$ has a zero element
Let $0\colon X\to\mathbb{R}$ be given by $0(x)=0$ for all $x\in X$. We see that for any $f\colon X\to\mathbb{R}$ we get $(0+f)(x)=0(x)+f(x)=0+f(x)=f(x)=(f+0)(x)$ and so $f+0=f=0+f$, hence $0$ acts as a zero element of $C(X)$.
The addition in $C(X)$ is associative
Let $f,g,h\colon X\to\mathbb{R}$ be continuous functions and let $x\in X$. We see that $$(f+(g+h))(x) = f(x)+(g+h)(x) = f(x)+g(x)+h(x)\\ = (f+g)(x)+h(x)=((f+g)+h)(x)$$ and so $f+(g+h)=(f+g)+h$ so the addition in $C(X)$ is associative.

You're now required to show that


*

*addition of elements in $C(X)$ is closed (so the addition of two continuous functions is still continuous).

*scalar multiplication of elements in $C(X)$ is closed (so the scalar multiple of a continuous function is still continuous).

*addition is commutative.

*every element in $C(X)$ has an additive inverse (with respect to the $0$ element shown to exist as above).

*scalar multiplication distributes over addition (both scalar addition and vector addition).

*$1f=f$ for all $f\in C(X)$

*$\lambda(\lambda'f)=(\lambda\lambda')f$ for all $\lambda,\lambda'\in \mathbb{R}$ and $f\in C(X)$.

