# Let $U$ be a subspace of $V$, show that $(U^{0})_0 = U$

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

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

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

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

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

## 1 Answer

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