A Vector Divided by it's Distance to a Subspace Yields a Vector with Distance 1 to Subspace

Question

Let X be a normed vector space, S a subspace of X and x∈X. Distance is defined by: $$|x,S|:=\inf||x-s||, x\in X, s\in S$$ How does one prove that for: $$z:=\frac{x}{|x,S|}\Rightarrow |z,S| = 1$$ This is intuitively obvious in $R^n$ but how does one prove it in any normed vector space? Any pointers would be much appreciated, thanks!

Answer 1

$[z,S]=\inf \{\|\frac x {[x,S]} -y\|:y \in S\}=\inf \{\|\frac x {[x,S]} -\frac y {[x,S]}\|:y \in S\}$ because $S$ is a subpace. Hence $[z,S]=\frac 1 {[x,S]} \inf \{\|x-y\|: y \in S\}=\frac {[x,S]} {[x,S]}=1$.