proof of Heine-Cantor theorem
We prove this theorem in the case when and are metric spaces.
Suppose is not uniformly continuous. Then
In particular by letting we can construct two sequences and such that
Since is compact the two sequence have convergent subsequences i.e.
Since we have . Being continuous we hence conclude which is a contradiction being .
Title | proof of Heine-Cantor theorem |
---|---|
Canonical name | ProofOfHeineCantorTheorem |
Date of creation | 2013-03-22 13:31:26 |
Last modified on | 2013-03-22 13:31:26 |
Owner | paolini (1187) |
Last modified by | paolini (1187) |
Numerical id | 5 |
Author | paolini (1187) |
Entry type | Proof |
Classification | msc 46A99 |