# Why is the closed unit interval compact? [duplicate]

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?

• Can you give an example of an open cover you think has no finite subcover? – Joppy Oct 25 '17 at 12:20
• 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 $[0,1]$. – user167289 Oct 25 '17 at 12:24
• @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