I am familiar with the concept of a dual space $V^*$ as the set of all linear functionals $\tilde{\omega}: V \rightarrow \mathbb{R}$. The inner product on $V$ is usually used to define the dual of a vector $\vec{v}\in V$, that is the dual vector $\tilde{v}$ is the unique linear functional which satisfies \begin{equation} \tilde{v}(\vec{w}) := \langle \vec{v}, \vec{w}\rangle, \end{equation} and the dual basis satisfies $\tilde{e}^\alpha(\vec{e}_\beta) = \delta^\alpha_\beta $.
I am used to thinking about the dual space as a distinct object to the original vector space - that vectors in $V$ cannot be expressed as linear combinations of dual vectors. However, in their book General Relativity: An Introduction for Physicists, the authors Hobson, Efstathiou and Lasenby write that any vector may be written as either a sum of basis vectors or as basis covectors:
$$\vec{v} = v^{\alpha}\vec{e}_\alpha = v_\alpha\vec{e}^\alpha $$
Is this reasonable?
If it is not reasonable, then suppose for some basis $\{\vec{e}_\alpha\}$ of a $3D$ vector space, a set of reciprocal vectors $\{\vec{e}^\alpha\}$ are defined as $$ \vec{e}^1= \frac{\vec{e}_2\times \vec{e}_3}{\vec{e}_1\cdot( \vec{e}_2\times\vec{e}_3)} $$ etc. (this can easily be extended to higher dimensions). These vectors satisfy $\langle{\vec{e}^\alpha}, \vec{e}_\beta\rangle = \delta^\alpha_\beta$, so act as dual vectors, but are still members of $V$. Is there any problem with the quoted equation if these reciprocal vectors are used? Are reciprocal vectors covectors?