Why is the closed unit interval compact?

Asked Oct 25 '17 at 12:17

Active Oct 25 '17 at 12:23

Viewed 243 times

Looking at the definition for compactness, i.e. "every open cover has a finite subcover", it seems like $[0,1]$ wouldn't be compact, since you can't construct a cover with a finite subcover that fills the bounded space.

This is obviously wrong because of Heine-Borel. Any counter-examples?

edited Oct 25 '17 at 12:23

asked Oct 25 '17 at 12:17

Researcher_Witty

Can you give an example of an open cover you think has no finite subcover? – Joppy Oct 25 '17 at 12:20
4

I can construct a cover of $[0,1]$ with a finite sub cover: ${[0,3/4),(1/4,1]}$. Not only does it have a finite subcover, it's actually finite itself. – Lee Mosher Oct 25 '17 at 12:20
Also, you copied the definition incorrectly, "every open cover has a finite subcover". – Lee Mosher Oct 25 '17 at 12:21
Your cover has no subcovers. Since neither of the two intervals covers the whole of $[0,1]$. – Researcher_Witty Oct 25 '17 at 12:24
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@user167289 These is not required for a cover, only that $[0, 1]$ is contained in the union of the covering sets. – lisyarus Oct 25 '17 at 12:26
You can do something similar for R however it is not compact – Researcher_Witty Oct 25 '17 at 17:19

Why is the closed unit interval compact?

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