# Need help understanding why my proof that [0,1] is compact is wrong?

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?

• 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. – Ben Grossmann Feb 7 '20 at 21:33
• 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 $G$. – rogerl Feb 7 '20 at 21:33
• 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. – zwim Feb 7 '20 at 21:34
• +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. – rschwieb Feb 7 '20 at 21:35
• @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? – 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.
• 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? – benwalker Feb 7 '20 at 22:05
• @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. – Henno Brandsma Feb 7 '20 at 22:11