I'm reading Hoffman and Kunze's Linear Algebra book and on page 284 they prove the following theorem
I didn't understand the proof of the part (ii). Why is the orthogonality condition satisfied by at most one vector in $W$? this means that always there is a best approximation of a vector $\beta$ by vectors in $W$?
If $(\beta-\alpha) \perp W$ and $(\beta-\alpha')\perp W$ for vectors $\alpha,\alpha'\in W$, then $(\alpha-\alpha')\perp W$ because it is the difference of two vectors orthogonal to $W$. However, $\alpha-\alpha'\in W$, which makes $\alpha-\alpha'$ orthogonal to itself and, hence, must be $0$. So $\alpha=\alpha'$.
If $W$ is complete, then there is always a closest point in $W$ to a given $\beta$. This is because if you choose any $\{ \alpha_n \} \subset W$ such that $\|\alpha_n -\beta\| < \inf_{\alpha \in W}\|\alpha-\beta\| +\frac{1}{n}$, then $\{\alpha_n\}$ is a Cauchy sequence. This has to do with the nice properties of norms generated by inner products.
If $W$ is not complete, then there may not be a closest point $\alpha\in W$ to a given $\beta$. For example, suppose $V$ is a Hilbert space with complete orthonormal basis $\{ e_j \}_{j=1}^{\infty}$. If you let $W$ be the linear space of all finite linear combinations of the basis elements, then $\beta = \sum_{j=1}^{\infty}\frac{1}{j}e_j$ is not in $W$, and $$ \inf_{\alpha\in W}\|\alpha-\beta\|=0. $$ But there is no element $\alpha \in W$ for which $\|\alpha-\beta\|=0$.
