# Prove that interval $(0,1]$ is not compact

I'm trying to prove that interval $(0,1]$ is not compact by showing it doesn't have Heine-borel property.

I know a set is compact if a set is closed and bounded or has BW property or has Heine-borel property. But I'm trying to use heine-borel property to prove that it is not compact. I know I have to use the definition of open cover to prove this, but I don't know how to begin.

my guess: in order to prove $(0,1]$ is not compact by showing it doesn't have heine-borel property, is to show that there exists open cover $(0,1]$ that cannot be reduced to a finite subcover. but then what would be $\mathscr{U}$?

• take the obvious open cover: the collection $(1/n,2)$, say, for positive integers $n$. Oct 24, 2016 at 3:18
• sorry i don't understand open cover too well. how is $(1/n,2)$ an obvious open cover? @symplectomorphic Oct 24, 2016 at 3:19
• sorry, what is the Heine Borel property? Oct 24, 2016 at 3:20
• state the definition of an open cover and think about my example. Oct 24, 2016 at 3:21
• @symplectomorphic ah i was dumb. i can even make $(1/2^n,2)$ Oct 24, 2016 at 3:32

You want to construct an open cover for which no finite subcover can still cover all of the interval $(0,1]$. One way you might do this is to take a collection of covers $U_n = (a_n, 2)$, where $a_n \to 0$.
If you create $a_n$ such that all $a_n > 0$, then clearly no finite subcover will still cover the interval.
Define $$I_{n} = (\frac{1}{n}, 2)$$
Then $$\bigcup_{n \in \mathbb{N}} I_{n}$$ covers $$(0,1]$$ but it's easy to see that finitely many $$I_{n}$$ can't cover the interval. This is because if a finite cover $$\{I_{n}\}_{n=1}^{k}$$ existed, then this would not cover $$(0, \frac{1}{k})$$ and thus wouldn't be a cover at all.