Unpacking definition of $k$-algebra of $k$-valued functions on $X$? Sorry for another question, but here we go.

Given a set $X$, we write $k\{X\}$ for the $k$-algebra of $k$-valued functions on $X$, with the operations of pointwise addition, multiplication, and multiplication by scalars.

I don't understand this statement. Can anybody help me unpack it? What is it actually saying? What should I have in mind/what should be my intuition for working with the $k$-algebra of $k$-valued functions on $X$?
 A: It means really just what it says; $k$ is a field (or maybe just a ring), and $X$ is a set, and we give the name $k\{X\}$ to the set 
$$\{\text{functions }f:X\to k\}$$
and give it operations: for $f,g:X\to k$ and $\lambda\in k$, we define
$$(f+g):X\to k\qquad\text{defined by}(f+g)(x)=f(x)+g(x)\text{ for all }x\in X$$
$$(f\cdot g):X\to k\qquad\text{defined by}(f\cdot g)(x)=f(x)\cdot g(x)\text{ for all }x\in X$$
$$(\lambda \cdot f):X\to k\qquad\text{defined by}(\lambda\cdot  f)(x)=\lambda \cdot f(x)\text{ for all }x\in X$$
A: As a set, $k\{X\}$ is just the set of all functions $f:X\to k$.  Given $f,g:X\to k$, we define $(f+g)(x)=f(x)+g(x)$, $(f\cdot g)(x)=f(x)g(x)$, and $(c\cdot f)(x)=cf(x)$ for $c\in k$.  These operations make $k\{X\}$ a $k$-algebra.
The intuition is very simple: this is just a big product of copies of $k$.  For instance, if $X=\{x,y\}$ has two elements, then $k\{X\}\cong k^2$ by sending $f$ to the pair $(f(x),f(y))$.  All the algebraic operations are defined separately on each coordinate, as usual.  In general, $k\{X\}$ is a product of $|X|$ copies of $k$.
