Why does this sequence exist?

Question

We have the following theorem from Stein and Shakarchi's Real Analysis Book 3 that says if $H$ is a Hilbert space, $f \in H$, and $S$ is a closed subspace of $H$, then there is an element $g_0 \in S$ such that $\|f-g_0\| = \inf_{g \in S}\|f-g\|$.

The proof starts out by letting $d = \inf_{g \in S}\|f-g\|$ and considering a sequence $\{g_n\} \subset S$ with the property that

$$\|f-g_n\| \rightarrow d \text{ as }n \to \infty $$

I'm having trouble seeing why such a sequence $\{g_n\}$ should exist. My intuition is telling me that I need to use the epsilon characterization of infimum. That is, if $A$ is a set of real numbers and $d := \inf A$, then

$$\forall \epsilon > 0, \exists x \in A \text{ such that }x < d + \epsilon$$

Answer 1

For every $n\in\mathbb{N}$ there must be an element $g_n\in S$ such that $d\leq ||f-g_n||<d+\frac{1}{n}$, this indeed follows from the epsilon characterization of infimum. By the squeeze theorem $||f-g_n||\to d$.

Answer 2

Given $\epsilon > 0$, we can find $g\in S$ with $||f-g|| < d + \epsilon$. For suppose not; then for all $g \in S$, $||f-g|| \geq d + \epsilon$, which means that $d + \epsilon$ is a lower bound for $\{||f-g|| \big| g \in S\}$. But $d + \epsilon > d$, a contradiction since $d$ is the greatest lower bound.

Now for any $n$, set $\epsilon = 1/n$ and use the above fact to find $g_n$ with $||f-g_n|| < d + 1/n$. Since $d \leq ||f - g_n||$ by definition, we have $||f-g_n||\rightarrow d$.