# Dense subset of given space

If $E$ is a Banach space, $A$ is a subset such that $$A^{\perp}:= \{T \in E^{\ast}: T(A)=0\}=0,$$ then $$\overline{A} = E.$$ I don't why this is true. Does $E$ has to be Banach? Thanks

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Did you mean to say that $A$ is a vector subspace, or does $\overline{A}$ mean the closed subspace of $E$ generated by $A$? If $A$ were only assumed to be a subset and $\overline{A}$ means the closure, then this is false. E.g., let $A$ be the unit ball of $E$.
Suppose that the closed subspace of $E$ generated by $A$, $\overline{\mathrm{span}}(A)$, is not $E$. Let $x\in E\setminus \overline{\mathrm{span}}(A)$. Using Hahn-Banach you can show that there is an element $T$ of $E^*$ such that $T(A)=\{0\}$ and $T(x)=1$. (Start by defining $T$ on the subspace $\overline{\mathrm{span}}(A)+\mathbb C x$.) No, this does not depend on $E$ being complete.