I am going over a proof of the classical projection theorem which states the following: Let $H$ be a Hilbert space and $M$ a closed subspace of $H$. Corresponding to any vector $x \in H$, there is a unique vector $m_0 \in M$ such that $||x-m_0|| \le ||x-m||$ for all $m \in M$. I know that the minimizing vector $m_0$ is such that $x-m_0$ is orthogonal to M. The solution boils down to showing that there is an $m_0$ such that $\delta=\text{inf}_{m \in M}||x-m||=||x-m_0||$. So far so good. However, the author goes into a great length showing that a sequence $\{m_i\}$ so that $||x-m_i|| \rightarrow \delta$ is a Cauchy sequence and using the fact that $M$ is a complete space.
What I don't understand is the following: isn't the fact that $M$ is closed implies that the set $M^1=\{x-m|m \in M\}$ is also closed? Therefore, there is an $m_0 \in M$ such that $\delta=||x-m_0||$. Obviously I am missing something in the reasoning here..
Thanks in advance!
The problem with your argument is that in order to conclude from the fact that $M^1$ contains its limit points that there is an $m_0 \in M$ such that $\delta = \|x - m_0\|$ you already need to know that there is a limit point of $M$ such that $\delta = \|x - m_0\|$. This does not come for free from the definition of $\delta$. The approximation property of the $\inf$ tells you that there is a sequence $m_n \in M$ such that $\|x - m_n\| \to \delta$ but this does not tell you that $m_n$ has a convergent subsequence a priori and so you don't get the desired limit point.
The authors argument that $(m_n)_{n \geq 1}$ must be Cauchy is exactly a proof that the desired limit point must exist since Cauchy sequences must converge and the limit of $(m_n)$ is then the desired limit point.