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 $\mathscr{U}$?

  • $\begingroup$ take the obvious open cover: the collection $(1/n,2)$, say, for positive integers $n$. $\endgroup$ – symplectomorphic Oct 24 '16 at 3:18
  • $\begingroup$ sorry i don't understand open cover too well. how is $(1/n,2)$ an obvious open cover? @symplectomorphic $\endgroup$ – Allie Oct 24 '16 at 3:19
  • $\begingroup$ sorry, what is the Heine Borel property? $\endgroup$ – Jorge Fernández Hidalgo Oct 24 '16 at 3:20
  • $\begingroup$ state the definition of an open cover and think about my example. $\endgroup$ – symplectomorphic Oct 24 '16 at 3:21
  • $\begingroup$ @symplectomorphic ah i was dumb. i can even make $(1/2^n,2)$ $\endgroup$ – Allie Oct 24 '16 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.

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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.

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  • $\begingroup$ Can the downvoter(s) explain? $\endgroup$ – gtoques Mar 19 at 5:48

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