The problem is: "Prove that $[0,1]$ is compact using the definition of compactness"

So we cannot use the Heine Borel Theorem which states that any closed bounded set of $\mathbb{R}^n$ is compact. We have to use the definition of compactness which is that for any open cover of the set, there exists a finite subcover.

Consider an open cover $G$ of $[0,1]$. Then $0$ and $1$ are interior points of an open set. Hence there exist neighborhoods $N_0$ and $N_1$ around $0$ and $1$ respectively (with radius $\varepsilon$) such that $N_0 \subset G$ and $N_1 \subset G$. Then consider the set $ E = (\varepsilon/2, 1 - \varepsilon/2)$. The union $N_0 \cup E \cup N_1$ is then a open cover of $[0,1]$ which is also a subset of $G$, hence it is a finite sub cover of $G$. Therefore $[0,1]$ is compact.

But I have looked online of proofs and I get completely different arguments. Am I wrong here? I feel like I am. Can anyone point to the incorrectness?

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    $\begingroup$ It looks as though you do not understand the definition of an open cover. If $G$ is an open cover of $[0,1]$, then $G$ is a collection of open sets whose union contains $[0,1]$. $G$ is not itself an open set. $\endgroup$ – Ben Grossmann Feb 7 '20 at 21:33
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    $\begingroup$ You say "$N_0\subset G$". $G$ is a collection of sets, so actually you are choosing $N_0, N_1\in G$. Now note that there is no reason to believe that $E$ is a set that is an element of $G$. I think you are getting mixed up between the points of $[0,1]$, which are covered by the elements of $G$, and the subsets of $[0,1]$ that are members of $G$. $\endgroup$ – rogerl Feb 7 '20 at 21:33
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    $\begingroup$ For the set to be compact you need to prove that for any cover of $[0,1]$ you can extract a finite cover. Not just from one cover in particular. Extracting means $E$ needs to belong to the original cover. $\endgroup$ – zwim Feb 7 '20 at 21:34
  • $\begingroup$ +1 what zwim said. You don't get to pick what open sets to use, e.g. you can't say $E$ is one of them. $\endgroup$ – rschwieb Feb 7 '20 at 21:35
  • $\begingroup$ @zwim I am pretty confused on what a sub cover actually is... I have read that it is just another open cover of the set in question but also a subset of the original open cover. Is that incorrect? Also wouldn't any open cover of $[0,1]$ also have to include $0$ and $1$, hence my argument would pertain to any open cover? $\endgroup$ – benwalker Feb 7 '20 at 21:42

Hint: if $G$ is an open cover for $[0,1]$, and let $A=\{a \in [0,1]: [0,a] \text{ has a finite subcover from } G\}$. Trivially $0 \in A$, as $0$ is covered by some element of $G$. So $a_0 = \sup A$ exists. (lub property of $\Bbb R$). Try to reason why $a_0 < 1$ cannot happen.

  • $\begingroup$ Okay, so if $a_0 < 1$, and $a_0$ is in the set $[0,a_0]$, since there is an open finite subcover $U$ from $G$ that covers $[0,a_0]$, we know that $a_0$ has some neighborhood $N_\delta (a_0) \subset U$ (with radius $\delta$). Then there is another point $b \in (a_0,a_0 + \delta) \subset U$ greater than $a_0$, within this open cover, and hence there exists an interval $[0,b]$ that is also covered from a finite sub cover extracted from $G$. This contradicts that $a_0$ is the supremum of $A$, and hence $a_0 = 1$. How does this sound? $\endgroup$ – benwalker Feb 7 '20 at 22:05
  • $\begingroup$ @benwalker not quite. Be more precise. You know $a_0$ is covered but not that it is in $A$. But $a_1:= a-\frac{\delta}{2}$ is by the properties of $\sup$. So $U$ plus the finite subcover for $[0,a_1]$ that must exist show $a_0 \in A$ and even $a_0 + \frac{\delta}{2} \in A$ now and we have a contradiction with being the sup. $\endgroup$ – Henno Brandsma Feb 7 '20 at 22:11
  • $\begingroup$ Ahh okay that makes sense. I see where I was being imprecise. Thank you. $\endgroup$ – benwalker Feb 7 '20 at 22:23

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