Let $\tau:W\times V \rightarrow K $ be a bilinear map where $V$ and $W$ are vector spaces over a field $K$.
Then, let $U$ be a subspace of $V$ and define $S(U) = \{w \in W\ |\ \tau(w,u)=0\ \forall u \in U\}$
My question is this:
Prove that $U \subset S(S(U))$
Now, I have already managed to prove that $S(U)$ is a subspace of $W$.
It seems to me that $S(S(U)) = \{w \in W\ |\ \tau(w,x)=0\ \forall x \in S(U)\}$ but this seems to get me in trouble because we're dealing with a set containing elements of $W$ not $U$.
If someone could persuade me that the following is true, then I think I can finish the proof from there:
$S(S(U)) = \{u \in U\ |\ \tau(u,x)=0\ \forall x \in S(U)\}$
Thanks for any help in pointing me in the right direction!
It looks like the author skipped the "and analogously for $E \subset W$" part and that led to confusion.
Let's modify the notation a little: for $U \subset V$, define
$$S_W(U) := \left\lbrace w\in W : \bigl(\forall u\in U\bigr)\bigl( \tau(w,u) = 0\bigr) \right\rbrace,$$
and for $Z \subset W$, define analogously
$$S_V(Z) := \left\lbrace v\in V : \bigl(\forall z\in Z\bigr)\bigl( \tau(z,v) = 0\bigr) \right\rbrace.$$
Then your task is to show that for all $U\subset V$ you have
$$U \subset S_V(S_W(U)).$$