PDA

View Full Version : product of compact sets compact in box topology?


spicychicken
Jul20-11, 11:33 AM
So Tychonoff theorem states products of compact sets are compact in the product topology.

is this true for the box topology? counterexample?

micromass
Jul21-11, 08:08 AM
A counterexample is \prod_{n\in \mathbb{N}}{[0,1]}. Can you show why?

spicychicken
Jul21-11, 01:17 PM
if S_n is the set with empty sets in each index except n where for index n you have [0,1], then {S_n} is an open cover with no finite subcover...i think

micromass
Jul21-11, 02:24 PM
if S_n is the set with empty sets in each index except n where for index n you have [0,1], then {S_n} is an open cover with no finite subcover...i think

Such a sets will always be empty. Try to consider a cover by all sets of the form

\prod_{n\in \mathbb{N}}{A_i}

Where Ai=[0,0.6[ or Ai=]0.5,1]