# Compactness of the open and closed unit intervals

In the article by Tao it's explained that the compactness can be formulated in the most general way as:

(All open covers have finite subcovers) If $V_\alpha:\alpha\in\mathcal{a}$ is any collection of open sets which covers $X$, then there must exist a finite sub-collection $V_{\alpha_1},V_{\alpha_2}...V_{\alpha_k}$ of these sets which still cover $X$.

Question: How is it possible to show (both intuitively and rigorously) that such coverage by a finite collection of open sets is possible for $X=[0,1]$ and not possible for $X=(0,1)$?

Edit: a related question and another one on open covers.

• For the open interval, consider $\bigcup_{n=2}^\infty(1/n,1)$. – Akiva Weinberger Jan 8 '17 at 23:43

Take the closed interval $[0, 1]$, and let $\mathcal{V} = \{ V_1 , V_2 , \ldots \}$ be an open cover. We want to say that $\mathcal{V}$ has a finite subcover, so suppose to the contrary it doesn't. Then we can express $[0, 1]$ as $[0, 1/2] \cup [1/2, 1]$. If both $[0, 1/2]$ and $[1/2, 1]$ admit finite subcovers from $\mathcal{V}$, then so does $[0, 1]$, so we know (at least) one of these intervals doesn't. We'll then perform the same process on that interval (if it was $[0, 1/2]$, then we'll look at $[0, 1/4]$ and $[1/4, 1/2]$). This yields a sequence $(I_N)$ of nested closed intervals whose intersection is a singleton $\{ x \}$. We know there exists $n$ such that $x \in V_n$, and since $V_n$ is open, we know there's some $\epsilon > 0$ such that the $\epsilon$-ball about $x$ is contained in $V_n$. But if we choose $N$ large enough that $2^{- N} < \epsilon$, we know that the $N$th closed interval $I_N$ is contained in $V_n$. So in fact $I_N$ not only admitted a finite subcover from $\mathcal{V}$, but a singleton cover, namely $\{ V_n \} \subset \mathcal {V}$, a contradiction.
EDIT: A similar method can be used to show the cube $[0, 1]^n \subset \mathbb{R}^n$. Another modification yet will yield the most general form of the Heine-Borel theorem: That a subset of a complete metric space is compact if and only if it's closed and totally bounded.
showing it is not possible for $(0,1)$ is easy, just take the open cover $(0,1-\frac{1}{n})$.
Showing it is possible for $[0,1]$ is not so easy, you may want to look here for proofs that do not go through the whole Heine-Borel theorem.
• Thank you for the reply! So for $(0,1)$ we select a "special" cover by an infinite number of sets without a finite sub-collection. Is there any simple intuition behind $[0,1]$ case or there is just a technical proof? – Konstantin Jan 8 '17 at 19:33
I think the most rigorous way of showing this more generally is using the Heine-Borel theorem, which says that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. Maybe this is less intuitive but if you understand the proof of the theorem it gives you a large insight.