$U = \{v\in V|\phi(v) = 0\ \text{for every }\phi\in U^0\}$

Question

Is the following Proof Correct?

Theorem. Given that $V$ is finite-dimensional and $U$ is a subspace of $V$ $$U = \{v\in V|\phi(v) = 0\ \text{for every }\phi\in U^0\}$$

Proof. Let $H = \{v\in V|\phi(v) = 0\ \text{for every }\phi\in U^0\}$. The proof of the proposition $U\subseteq H$ is trivial we therefore prove the converse $H\subseteq U$.

Let $u_1,u_2,...,u_n$ be a basis for $U$ which we may extend to yield a basis for $V$ namely $u_1,u_2,...,u_n,u_{n+1},u_{n+2},...,u_m$, furthermore let $\phi_1,\phi_2,...,\phi_n,...,\phi_{n+1},\phi_{n+2},...,\phi_m$ be the corresponding dual basis for $V'$.

Now let $v = c_1u_1+cu_2+\cdot\cdot\cdot+c_nu_n+c_{n+1}u_{n+1}+c_{n+2}+\cdot\cdot\cdot+c_{m}v_m$ be an arbitrary vector in $H$. It is not difficult to see that $U^0 = \operatorname{span}(\phi_{n+1},\phi_{n+1},...,\phi_{m})$.

Since $\phi(v) = 0$ for all $\phi\in H$ then in particular for $\phi_{n+1},\phi_{n+1},...,\phi_{m}$ we have $\phi_{n+1}(v) = c_{n+1} = \phi_{n+2}(v) = c_{n+2} = \cdot\cdot\cdot = \phi_{m}(v) = c_{m} = 0$ consequently $v\in\operatorname{span}(u_1,u_2,...,u_n) = U$.

$\blacksquare$

Answer 1

It looks correct, but:

1. There's a typo. You wrote $\operatorname{span}(\phi_{n+1},\phi_{n+1},\ldots,\phi_{m})$ where you meant $\operatorname{span}(\phi_{n+1},\phi_{n+2},\ldots,\phi_{m})$.
2. You ought to justify the assertion that $U^0=\operatorname{span}(\phi_{n+1},\phi_{n+2},\ldots,\phi_{m})$.