Let $W_1$ and $W_2$ be subspaces of a vector space $V$
Prove that $V$ is the direct sum of $W_1$ and $W_2$ if and only if each vector in $V$ can be uniquely written as $x_1+x_2$, where $x_1 \in W_1$ and $x_2 \in W_2$
Question 1
"If each vector in $V$ can be uniquely written as $x_1 + x_2$" = $\forall v \in V \exists ! x_1 \in W_1 \exists ! x_2 \in W_2(v=x_1+x_2)$
When $V = W_1 = W_2 = \{0\}$,
$\forall v \in V \exists ! x_1 \in W_1 \exists ! x_2 \in W_2(v=x_1+x_2)$ is true?
$0 \in V = 0 \in W_1 + 0 \in W_2$
$0 \in V = 0 \in W_2 + 0 \in W_1$
So I think it is false but it seems true so very confusing.
Question 2
If $V = W_1 \oplus W_2$ and some vector $y \in V$ can be represented as $y = x_1 + x_2 = x_1' + x_2'$, where $x_1, x_1′ \in W_1$ and $x_2, x_2' \in W_2$, then we have $x_1 − x_1′ \in W_1$ and $x_1 − x_1' = x_2 + x_2′ \in W_2$. But since $W_1 \cap W_2 = \{0\}$, we have $x_1 = x_1'$ and $x_2 = x_2'$
Conversely, if each vector in $V$ can be uniquely written as $x_1 + x_2$, then $V = W_1 + W_2$. Now if $x \in W_1 \cap W_2$ and $x \neq 0$, then we have that $x = x + 0$ with $x \in W_1$ and $0 \in W_2$ or $x = 0+x$ with $0 \in W_1$ and $x \in W_2$, a contradiction.
I think this part
If each vector in $V$ can be uniquely written as $x_1 + x_2$, then $V = W_1 + W_2$.
is wrong because of counterexample: $V=\{0, 1, 2\}, W_1=\{0, 1\}, W_2=\{0, 2\} (3 \notin V)$
I think that part is wrong because
$W_1+W_2=\{x+y:x \in W_1,y \in W_2\}, V=\{v:\forall v \exists ! x_1 \in W_1 \exists ! x_2 \in W_2(v=x_1+x_2)\}$
$W_1+W_2 \neq V$ because $W_1+W_2 \nsubseteq V$
Is this correct?