When is $V$ a direct sum of a subspace $S$ and the set $(V-S)\cup\{\mathbf{0}\}$? In general, for a vector space V, $\dim(V)< \infty$, with subspace $S \subseteq V$ and the set $W :=(V-S) \cup \{ 0\}$ we find that $V \neq S \oplus W$.
For example, if $V=\mathbb{R}^2$ and $S=\{(x,0) : x \in \mathbb{R} \}$. Then take $(a,b), (c, -b) \in W$, for fixed $a, b, c \in \mathbb{R}$ and $a, b, c$ non-zero, and $a \neq -c$. Then $(a,b)+(c, -b) = (a+c, 0) \in S$, proving W is no subspace.
But, some of the time, $V = S \oplus W$. For example, if $S=V$. (But, that seems trivial.)
For what kind of subspaces, $S \subset V$, is this possible?

UPDATE (to clarify) 
Okay, then if $V = W_1 \oplus W_2 \oplus \cdots \oplus W_n$ and if $W_1, W_2,..., W_n \neq V$ or { $ 0 $ } then, of course, by definition, we have:
$V = W_1 + W_2 + \cdots + W_n$ and 
Wi are pair-wise disjoint (except for the 0 vector)
$W_i \cap (W_1 + W_2 + \cdots + W_{i-1} + W_{i+1} + \cdots + W_n) = \{ 0 \}, \forall i = 1, 2, ..., n$
But you are telling me we also have $V \neq (W_1 \cup W_2 \cup \cdots \cup W_n)$ always.
I could see that $V \neq (W_1 \cup W_2 \cup \cdots \cup W_n)$, some of the time, but it basically never happens unless one of the $W_i$ is $V$!
This really clears things up for me in terms of what a direct sum is really about! It's not like a partition of V it is more like a set of scaffolds for V.
 A: If $S$ is a subspace of $V$, and $W=(V-S)\cup\{0\}$, then $W$ is a subspace of $V$ if and only if $S=V$ or $S=\{0\}$.
Indeed, if $S=V$ then $W=\{0\}$ is a subspace; if $S=\{0\}$, then $W=V$ is a subspace.
Assume then that $S\neq\{0\}$ and $S\neq V$. Then there exists $\mathbf{s}\in S$, $\mathbf{s}\neq\mathbf{0}$, and there exists $\mathbf{v}\in V-S$, $\mathbf{v}\neq\mathbf{0}$. Note that $\mathbf{s}\notin W$.
Then $\mathbf{v}+\mathbf{s}\notin S$, since $\mathbf{v}\notin S$; hence $\mathbf{v}+\mathbf{s}\in W$. Since $\mathbf{v}\in W$, but $\mathbf{s} = (\mathbf{v}+\mathbf{s})-\mathbf{v}$ is not in $W$, it follows that $W$ is not closed under differences, hence is not a subspace.
This works whether $V$ is finite dimensional or not.

Added. While it is possible to express a vector space as a union of proper subspaces (see in particular Pete Clark's recent Monthly article for the precise answer of how many proper subspaces you need; there's a copy on his website), the union never yields a direct sum; and if $V$ is a direct sum of at least two proper nontrivial subspaces, then it is never equal to the union: take $w_1\in W_1$, $w_2\in W_2$, both nonzero; then $w_1+w_2$ cannot lie in any of the direct summands.
A: It can never happen. Note that  $V=S\oplus W$ implies that $W$ is a subspace. Take any $s\in S$, $w\in W$ with $w\ne0$. Then $w\not\in S$, and neither is $s+w$. But if $s+w\not\in S$, this means that $s+w\in W$, which implies that $s\in W$, a contradiction. We conclude that $W=\{0\}$. 
