Question about direct sum of function space I am reading Applied linear algebra: the decoupling principle by Lorenzo Adlai Sadun (btw very recommendable!) 
On page 30 about direct sums on vector spaces it says:

Let $V$ be the space of continuous functions on a domain $U \subset \mathbb R^3$. Then $V \oplus V \oplus V$ is the space of continuous $\mathbb R^3$-valued functions on $U$.

What I don't understand is how this threefold direct sum of $V$ can lead to a statement about the codomain ("-valued" functions). But I just guess that I don't get it altogether - please enlighten me. Thank you!
EDIT
I think what confuses me most is that we start with $\mathbb R^3$ and end up with it. It would come more natural if we started with $\mathbb R$ and after taking the direct sum three times would end in $\mathbb R^3$.
 A: In fact, the domain $U$ does not play an actual role here, specifying that it is a subset of $\mathbb R^3$ is just adding to the confusion, in my opinion.
First let us establish that $V$ is indeed a vector space. $f\in V$ if and only if $f\colon U\to\mathbb R$ and $f$ is continuous.
If $f\in V$ and $\alpha\in\mathbb R$ then $g(x) = \alpha\cdot f(x)$ is indeed a continuous function whose domain is $U$ and codomain $\mathbb R$, and if $f,g\in V$ then the function $h(x) = f(x) + g(x)$ is also a continuous function whose domain is $U$ and codomain is $\mathbb R$.
Now, the definition of $V\oplus V\oplus V$ is simply $\{(f,g,h)\mid f,g,h\in V\}$.
Suppose $(f,g,h)\in V\oplus V\oplus V$, then it defines a function: $h(x) = ( f(x), g(x), h(x))$. The function $h$ is continuous if and only if it is continuous in every coordination, and since $f,g,h$ are all continuous then so is $h$.
Since the domain of $f,g,h$ is $U$, then $h\colon U\to\mathbb R^3$, as it takes a point in $U$ and returns a vector in $\mathbb R^3$.
On the other hand, if $h$ is a continuous function from $U$ to $\mathbb R^3$, then $h(x) = (h_1(x), h_2(x), h_3(x))$, and $h_i$ are continuous functions from $U$ to $\mathbb R$, so $h_1,h_2,h_3\in V$ as needed.
A: The space $V\oplus V \oplus V$ is nothing else than the set of triples
$$ (v_1,v_2,v_3) $$
where
$$v_1,v_2,v_3\in V.$$
So the set $U$ is being mapped to 3-component objects which I will call "vectors". Each element $P$ of $U$ is assigned a 3-component vector $(v_1(P), v_2(P), v_3(P))$. The continuity of a vector-valued function, as Dactyl pointed out, means that each of the components of the vector is a continuous function. Indeed, it means that each $v_i$ among the three components is an element of $V$.
One may either say that it is a definition of the continuity of vector-valued functions.
Equivalently, one may define the continuity in this case by looking at balls in the space ${\mathbb R}^3$ into which we're mapping things. But in the manipulation with $\epsilon$ and $\delta$, balls are equivalent to little cubes - one may squeeze a ball inside a cube or vice versa (at least for finite dimensions such as 3), so we may equivalently replace the balls by the cubes, and then the definition of continuity using balls gets reduced to that of the cubes, i.e. to the component-wise continuity.
At any rate, the continuity of 3-component functions is either defined as - or may be proved to be - the continuity of each component.
A: What is unclear in the exposition is just exactly what the author means by "the space of continuous functions on a domain $U\subseteq \mathbb{R}^3$." 
What he means is that $U$ is the vector space of real-valued continuous functions with domain $U\subseteq\mathbb{R}^3$; that is, the elements of $U$ are function $f\colon U\to\mathbb{R}$ that are continuous, with $U\subseteq \mathbb{R}^3$. (He omits "real valued" when he says "functions"). 
That means that an element of $V\oplus V\oplus V$ consists of a $3$-tuple of real valued functions, $(f_1,f_2,f_3)$, with $f_i\colon U\to\mathbb{R}$. Thus, the tuple is naturally interpreted as a function from $U$ to $\mathbb{R}^3$ in which each component is continuous. 
A: I guess you can identify the space of continuous functions on $U$ that take values in $\mathbb{R}$ with subspaces of the space of continuous functions on $U$ that take values in $\mathbb{R}^3$ by isomorphisms of type $x\mapsto (x,0,0)$, $x\mapsto (0,x,0)$, $x\mapsto (0,0,x)$. In this view the book's assertion may look more natural, because now you're indeed having the direct sum of subspaces of a given space.
A: I know that this question is 9 years old, but the other answers didn't help my tiny brain. If someone comes across like I did, I hope this helps!
Let $C(\mathbb{R}^3)$ denote the space of continuous functions with domain $\mathbb{R}^3$.  Let $V \subset C(\mathbb{R}^3)$ be defined by $V := \{f \in C(\mathbb{R}^3) : f:U\subset \mathbb{R}^3 \rightarrow \mathbb{R}\}$. Then $V$ is simply the subspace of  $C(\mathbb{R}^3)$ where each function is restricted to $U$.
The direct sum $$V\oplus V \oplus V := \{ (f_1,0,0)+ (0,f_2,0) + (0,0,f_3) : f_i \in V\} = \{(f_1,f_2,f_3) : f_i \in V\}$$
is simply the tuple of continuous functions on the domain $V$. In otherwords, you give me an $(x_1,x_2,x_3) \in U \subset \mathbb{R}^3$ and I give you back $(f_1(x_1,x_2,x_3),f_2(x_1,x_2,x_3),f_3(x_1,x_2,x_3))$. Hence, elements of the direct sum are the maps
$$[ (x_1,x_2,x_3) \mapsto(f_1(x_1,x_2,x_3),f_2(x_1,x_2,x_3),f_3(x_1,x_2,x_3))] \in V \oplus V \oplus V$$
This is what is meant by $(f_1,f_2,f_3)$.
