Kuratowski’s embedding theorem

Let $X$ be a set and $\mathrm{Bou}(X,ℝ)$ be the set of bounded functions $f:X\to ℝ$ with norm  $||f||=\sup \{|f(x)|:x\in X\}$. Kuratowski’s embedding theorem states that every metric space $(X,d)$ can be embedded isometrically into the Banach space  $E=\mathrm{Bou}(X,ℝ)$.

Proof. One can assume that $X\ne \mathrm{\varnothing }$. Fix a point $a_0\in X$ and for every $a\in X$ define a function $f_a:X\to ℝ$ by

$$f_a(x)=d(x,a)-d(x,a_0).$$

Then $|f_a(x)|\le d(a,a_0)$ for every $x\in X$ so $f_a$ is bounded. By setting  $\phi :X\to E$,  $\phi (a)=f_a$, we have the mapping $\phi :X\to E$. It requires to prove that $\phi$ is an isometry.

Let $a,b\in X$. As $x\in X$ we have that

$$|f_a(x)-f_b(x)|=|d(x,a)-d(x,b)|\le d(a,b).$$

Therefore $||f_a-f_b||\le d(a,b)$. On the other hand

$$|f_a(a)-f_b(a)|=|d(a,a)-d(a,a_0)-d(a,b)+d(a,a_0)|=d(a,b).$$

Therefore $||\phi (a)-\phi (b)||=||f_a-f_b||=d(a,b)$.