Show that: $V=U\oplus W \wedge \exists_{u\in U,w\in W}u\neq 0,w\neq 0\Rightarrow \exists \tilde{W}\neq W\subseteq V: \tilde{W}\oplus U=V$

Question

$V$ is a vectorspace over $F$ and $W,\tilde{W},U$ are subspaces of $V$.

And $W\oplus U = V\iff W+U=V \wedge W\cap V=\{0\}$

$W+U=V\iff \forall_{v\in V}\exists_{J\subseteq W}\exists_{I\subseteq U}\text{#}J=n_0\in\mathbb{N}_0 \wedge \text{#}I=n_1\in\mathbb{N_0}\wedge v=\sum_{j\in J}j+\sum_{i\in I}i$

Now the proof says that as a result of a previous theorem $(*)$ we can say that there exists bases $B_U$ of $U$ and $B_W$ of $W$ with $u\in B_U$ and $w\in B_W$. However this implication is not obvious to me and I hope somebody can help me to understand it.

The theorem $(*)$ which I am stating can be seen below. It is the theorem $5.23(d)$

Answer 1

Consider the vector subspace $U$. We know that $U$ is not the trivial subspace since there exists an element $u\in U$ with $u\neq 0$. Using Theorem 5.23 (d) we now have that $B=\{u\}$ (that of course is a set of linearly independent vectors) can be completed to a basis $B_U$ of $U$.