Is the set of all functions from any set into R closed under addition and multiplication? I am doing a bit of self-study in Linear Algebra and I am having difficulty understanding the following:


*

*If $S$ is a set, then $\textbf{F}^S$ denotes the set of functions from $S$ to $\textbf{F}$.

*For $f, g \in \textbf{F}^S$, the sum $f + g \in \textbf{F}^S$ is the function defined by
$$(f + g)(x) = f(x) + g(x)$$
for all $x \in S$.

*For $\lambda \in \textbf{F}$ and $f \in \textbf{F}^S$, the product $\lambda f \in \textbf{F}^S$ is the function defined by
$$(\lambda f)(x) = \lambda f(x)$$
for all $x \in S$.


Does this mean that, for sums and products defined as above, the set of all functions from any set $S\mapsto R$ is closed under addition and scalar multiplication?
If so, could I use this to quickly prove, for example, that the set of all continuous functions on $[0,1] \subset R$ is closed under addition and scalar multiplication?

Yes, it is true that the set of functions from any set $S$ to any field $\mathbb{F}$ has the structure of an $\mathbb{F}$-vector space. The proof of this fact is contained in your question (you don't check the axioms, but they are extremely easy to check).
It does not follow immediately that the set of continuous functions from $[0,1]$ to $\mathbb{R}$ is a $\mathbb{R}$-vector space, though. If $f$ and $g$ are two functions in this space, you know only that $f+g$ and $\lambda f$ are again functions from $[0,1]$ to $\mathbb{R}$. You need then prove that $f+g$ and $\lambda f$ are continuous. You probably saw this proof in a first calculus class.
