# Kuratowski’s embedding theorem

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

Proof.  One can assume that $X\neq\emptyset$. Fix a point $a_{0}\in X$ and for every $a\in X$ define a function $f_{a}:X\to\mathbb{R}$ by

 $f_{a}(x)=d(x,a)-d(x,a_{0}).$

Then $|f_{a}(x)|\leq d(a,a_{0})$ for every $x\in X$ so $f_{a}$ is bounded. By setting  $\varphi:X\to E$,  $\varphi(a)=f_{a}$, we have the mapping $\varphi:X\to E$. It requires to prove that $\varphi$ is an isometry.

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

 $|f_{a}(x)-f_{b}(x)|=|d(x,a)-d(x,b)|\leq d(a,b).$

Therefore $||f_{a}-f_{b}||\leq d(a,b)$. 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 $||\varphi(a)-\varphi(b)||=||f_{a}-f_{b}||=d(a,b)$.

## References

• 1 J. VÃÂisÃÂlÃÂ: Topologia II.  2nd corrected issue, Limes ry., Helsinki, Finland (2005), ISBN 951-745-209-8
Title Kuratowski’s embedding theorem KuratowskisEmbeddingTheorem 2013-03-22 18:24:48 2013-03-22 18:24:48 puuhikki (9774) puuhikki (9774) 10 puuhikki (9774) Theorem msc 54-00
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