Understanding my mistake when dealing with the direct sum (3 subspaces involved) Given two subspaces $U_1, U_2$, I understand the concept of direct sum
$$ W= U_1 \oplus U_2 \iff W= U_1 + U_2, \quad U_1 \cap U_2 = \{ 0 \}$$
Where $W$ is a subspace of $V$.
I am trying to generalize it for more than $2$ subspaces, say $3$. I thought of the following.
$$ W= U_1 \oplus U_2 \oplus U_3 \iff U_1 \cap U_2 = \{ 0 \}, U_1 \cap U_3 = \{ 0 \}, U_2 \cap U_3 = \{ 0 \}, U_1 + U_2 + U_3 = W $$
It does not seem to have the same structure that for the statement with $k$ subspaces
\begin{align*}
W= U_1 \oplus U_2 \oplus ... \oplus U_k \iff& U_i \cap \left(U_1 + ... + U_{i-1} + U_{i+1} + ... + U_k\right) = \{ 0 \} \\
&U_1 + U_2 + ... + U_k = W
\end{align*}
In particular, the issue lies on the intersection statement. Might you explain why my thought is faulty? I should be able to find a counterexample once I see it :)
Edit 0
Let us use the following definition for the direct sum (Axler, page 21)
$U_1 + U_2 + \ ... \ + U_k$ is a direct sum if $x \in U_1 + U_2 + \ ... \ + U_k$ can be written in a unique way as $x = u_1 + u_2 + \ ... \ + u_k$, where $u_i \in U_i$
I think I found a counterexample:
$$U_1 = \{ (x, 0) | x \in \Bbb R\}, \quad U_2 = \{ (y, y) | y \in \Bbb R\}, \quad U_3 = \{ (0, y) | y \in \Bbb R\}$$
These subspaces satisfy $U_1 \cap U_2 = \{ 0 \}, U_1 \cap U_3 = \{ 0 \}, U_2 \cap U_3 = \{ 0 \}, U_1 + U_2 + U_3 = \Bbb R^2$ but not $\Bbb R^2= U_1 \oplus U_2 \oplus U_3$ because we can write, say, the zero vector in (at least) two ways
$$(0, 0) = (1, 0) + (-1, -1) + (0, 1) \quad \text{and} \quad  (0, 0) = (0, 0) + (0, 0) + (0, 0)$$
Do you agree? :)
 A: I guess the problem is that (for three subspaces) $U_1+U_2$ may intersect $U_3$ in a non-zero vector even though $U_1\cap U_3 = \{0\}$ and $U_2\cap U_3 = \{0\}$ - can you find an example of this? If $U_1+U_2$ intersects $U_3$ in a non-zero vector, then the sum $U_1+U_2+U_3$ cannot be a direct sum. (I guess the way to see this depends on the definition of direct sum that you are using - but basically you can represent a vector in the intersection $\left(U_1+U_2\right)\cap U_3$ as a sum of vectors in $U_1$, $U_2$ and $U_3$ in at least two different ways, so uniqueness of representation fails.)
Hope this helps and let me know if you would like further clarifications!
A: This indicates that $U\cap V=\{0\}, U +V = W$ is not the "right" definition (since it does not generalize nicely). Instead, I would take the following as the definition :

Let $(U_i)_{i\in I}$ a family of subspaces. We say that $\bigoplus_{i\in I} U_i = W$ when, for any $x\in W$, there is a unique family $(x_i)_{i\in I}$ with $x_i \in U_i$, equal to $0$ but for a finite number of them and with  :
$$\sum_{i\in I}x_i = x$$

Then you could show that the properties you mention are equivalent to this, when $I$ is finite or has two elements.
Edit 0
Using this, your counterexample shows that :
$$ W= U_1 \oplus U_2 \oplus U_3 \iff U_1 \cap U_2 = \{ 0 \}, U_1 \cap U_3 = \{ 0 \}, U_2 \cap U_3 = \{ 0 \}, U_1 + U_2 + U_3 = W $$
is not true.
It is easy to show from the definition that direct sum are associative : ie $U_1\oplus U_2\oplus U_3 = W$ iff $U_1$ and $U_2$ are in direct sum and $(U_1\oplus U_2) \oplus U_3 = W$. By induction, you get the right characterization.
A: @SolubleFish's definition is the right way to generalize the definition of direct sum for more than two subspaces, from this definition it is easy to show that
\begin{align*}
W=U_{1}\oplus U_{2}\oplus \dots\oplus U_{k} \iff& U_{1}\cap U_{2}=\{0\}, \ \left( 
U_{1}+ U_{2}\right) \cap U_{3}=\{0\} \\
&\left( U_{1}+ U_{2}+ U_{3}\right)\cap U_{4}=\{0\} \\
&\vdots \\
&\left( U_{1}+ U_{2}+ U_{3}+\cdots+U_{k-1}\right)\cap U_{k}=\{0\} \\
&U_{1}+ U_{2}+ U_{3}+\cdots+U_{k-1}+U_{k}=W
\end{align*}
