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Product of compact sets compact in box topology?

by spicychicken
Tags: compact, product, sets, topology
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spicychicken
#1
Jul20-11, 11:33 AM
P: 3
So Tychonoff theorem states products of compact sets are compact in the product topology.

is this true for the box topology? counterexample?
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micromass
#2
Jul21-11, 08:08 AM
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A counterexample is [itex]\prod_{n\in \mathbb{N}}{[0,1]}[/itex]. Can you show why?
spicychicken
#3
Jul21-11, 01:17 PM
P: 3
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
#4
Jul21-11, 02:24 PM
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Product of compact sets compact in box topology?

Quote Quote by spicychicken View Post
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

[tex]\prod_{n\in \mathbb{N}}{A_i}[/tex]

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


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