Proof of Closest Point Theorem in Hilbert Space

The theorem:

Let $C$ be a non-empty closed convex subset of a Hilbert space
$X$, and let $x \in X$. Then there exists a unique $y_0 \in C$ such that $||x-y_0|| \le ||x – y||$ all $y \in C$. In other words there is a point in $C$ which is closest to $x$.

I’m reading a proof here (, p.4) and have two questions about the initial steps in the prooof (I can follow the later steps.):

  1. “Let $d = inf\{||x – y||: y \in C\}$” I think that the infimum should exist by the property (axiom) of completeness of the real numbers. $\{||x – y||: y \in C\}$ is a set of real numbers which is bounded below (all norms are $\ge 0$) and therefore has a greatest lower bound – is this correct ?
  2. “Let $(y_n)$ be a sequence of points in C such that $||x – y_n|| \to d$”. I think such a sequence should exist by the following reasoning: With $d = inf\{||x – y||: y \in C\}$ then for any $\epsilon > 0$ there must be a point $y$ in $C$ such that $||x – y|| \lt d + \epsilon$ (otherwise $d$ is not the infimum). If for this point $||x – y|| = d $ then $y_0 = y$ and we are finished: $(y_0)$ is a one element convergent sequence. Otherwise $||x – y|| = d + \epsilon`$ where $\epsilon` \le \epsilon$ and I can repeat the process to find a different point where $||x – y|| \lt d + \epsilon`/2$ and so generate a convergent sequence. I’m not sure this is valid, and it occurs to me that unless $C = \{y_0\}$ there are probably an infinite number of points at each step and the Axiom of choice is then required to extract a convergent sequence. Any help would be appreciated.

Solutions Collecting From Web of "Proof of Closest Point Theorem in Hilbert Space"

1) You are correct. Every set that is bounded below in $\mathbb{R}$ has a least lower bound.

2)The standard why to show that there exists such a sequence $(y_n)$ is as follows. Let $n\in\mathbb{N}$ then as d is the infimium, $d+\frac{1}{n}$ is not a lower bound so there exists $y_n$ such that $d<||x-y_n||<d+\frac{1}{n}$. This gives a sequence of points $(y_n)$. Then it should be obvious that $||x-y_n||\rightarrow d$.