(adsbygoogle = window.adsbygoogle || []).push({}); Compact --> bounded

In lecture 8 of Francis Su'sReal Analysisonline lecture series, he has a proof that a compact subset of a metric space is bounded: Given a metric space (X,d), if A is a compact subset of X, then every open cover of A has a finite subcover. Let B be a set of open balls of radius r, one centered on each point of A. This is an open cover of A, so it has a finite subcover {B_{i}}. Let C be the set of distances between pairs of center points of the elements of the subcover {B_{i}}. Let s = max C. Then B(x,s+2r) is an open ball that includes A.

My question: why the 2? Isn't B(x,s+r) also an open ball that includes A? If u is a point in A, then

d(x,u) <= d(x,w) + d(w,u) <= s + r

where w is the center of an open ball in {B_{i}} that contains u.

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Compact -> bounded

Loading...

Similar Threads for Compact bounded | Date |
---|---|

I Compact subspace in metric space | Yesterday at 10:15 AM |

A Questions about Covering maps, manifolds, compactness | Oct 26, 2017 |

I Compact Disk | Oct 3, 2017 |

Closed and bounded in relation to compact | Nov 13, 2012 |

Spivak's proof of A closed bounded subset of R^n is compact | Sep 20, 2012 |

**Physics Forums - The Fusion of Science and Community**