# totally bounded subset of a metric space is bounded

###### Theorem 1.

Every totally bounded subset of a metric space is bounded.

###### Proof.

Let $K$ be a totally bounded subset of a metric space. Suppose $x,y\in K$. We will show that there exists $M>0$ such that for any x,y we have $d(x,y). From the definition of totally bounded, we can find an $\varepsilon>0$ and a finite subset $\{x_{1},x_{2}...,x_{n}\}$ of $K$ such that $K\subseteq\bigcup_{k=1}^{n}B(x_{k},\varepsilon)$, so $x\in B(x_{i},\varepsilon)$,$y\in B(x_{l},\varepsilon)$, $i,l\in\{1,2,...n\}$. So we have that

 $\displaystyle d(x,y)$ $\displaystyle\leq$ $\displaystyle d(x,x_{i})+d(x_{i},x_{l})+d(x_{l},y)$ $\displaystyle<$ $\displaystyle\varepsilon+\max_{1\leq{s,t}\leq n}d(x_{s},x_{t})+\varepsilon=M$

Title totally bounded subset of a metric space is bounded TotallyBoundedSubsetOfAMetricSpaceIsBounded 2013-03-22 15:25:27 2013-03-22 15:25:27 georgiosl (7242) georgiosl (7242) 12 georgiosl (7242) Theorem msc 54E35