Is set $[0,1)$ compact in $\mathbb{R}$?

Question

I know that $[0,1]$ is compact in $\mathbb{R}$ using nested interval theorem. But I am stuck for the case of $[0,1)$. How to do this?

Answer 1

No, Heine-Borel Theorem characterizes compact sets in $\mathbb{R}$: a set $E \subset \mathbb{R}$ is compact if and only if it is closed and bounded. But this is not the case.

Note: more generally, this is true for $\mathbb{R}^n$.

Answer 2

You can check that

$$[0,1)\subseteq \bigcup_{n=1}^{\infty}A_n$$ where $$A_1 = (-\frac{1}{2},+\frac{1}{2})$$ $$A_n = (0,1-\frac{1}{n})\text{ for } n\geq 2$$

is an open cover for $[0,1)$. But you can't cover $[0,1)$ with only finitely many of these covering sets.

To see why you can't cover $[0,1)$ with only finitely many of them, suppose that we have chosen the covering sets indexed by $n_1,\cdots,n_k$ to cover $[0,1)$. Set $m=\max(n_1,\cdots,n_k)$. Then any number $\frac{m-1}{m} \leq x < 1$ will not be covered by $A_{n_1},\cdots,A_{n_k}$ which shows that a finite subcover does not exist and $[0,1)$ is not compact.