# 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 $$. From the definition of totally bounded, we can find an $\epsilon >0$ and a finite subset $\{{x}_{1},{x}_{2}\mathrm{\dots},{x}_{n}\}$ of $K$ such that $K\subseteq {\bigcup}_{k=1}^{n}B({x}_{k},\epsilon )$, so $x\in B({x}_{i},\epsilon )$,$y\in B({x}_{l},\epsilon )$, $i,l\in \{1,2,\mathrm{\dots}n\}$. So we have that

$d(x,y)$ | $\le $ | $d(x,{x}_{i})+d({x}_{i},{x}_{l})+d({x}_{l},y)$ | ||

$$ | $\epsilon +\underset{1\le s,t\le n}{\mathrm{max}}d({x}_{s},{x}_{t})+\epsilon =M$ |

∎

Title | totally bounded subset of a metric space is bounded |
---|---|

Canonical name | TotallyBoundedSubsetOfAMetricSpaceIsBounded |

Date of creation | 2013-03-22 15:25:27 |

Last modified on | 2013-03-22 15:25:27 |

Owner | georgiosl (7242) |

Last modified by | georgiosl (7242) |

Numerical id | 12 |

Author | georgiosl (7242) |

Entry type | Theorem |

Classification | msc 54E35 |