0
$\begingroup$

It is pretty obvious why $(U^{0})_0 \supseteq U$.

but how do I show that $(U^{0})_0 \subseteq U$?

For clarification: $U^{0} = \left \{ \phi\in V^{*} \mid \phi(a) = 0, for \; u \in U\right \}$

$(U^{0})_0 = \left \{ u\in V \mid \phi(u) = 0, for \; \phi \in U^{0}\right \}$

$\endgroup$
  • $\begingroup$ @HagenvonEitzen, Sorry, I changed it to $u \in V$ $\endgroup$ – vpam Aug 17 at 15:43
0
$\begingroup$

Hint: Given $v\notin U$, show there exists $\phi\colon V\to k$ with $\phi(v)=1$ and $\phi|_U=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.