How to show that $[-2,2)$ is not compact?

Question

How to show that $[-2,2)$ is not compact?

I can show that $(-2,2)$ is not compact since $K=\bigcup_{n\in\mathbb{N}}(-2,2-\frac{1}{n})$ has no finite subcover.

However I'm not sure how I can write a union of open sets which will include $-2$?

Answer 1

A subset of $\mathbb{R}$ is compact iff it is closed and bounded. Your set is bounded, but not closed as $2 \in \overline{[-2,2[}$ (because for example $(2-1/n)_{n=1}^\infty$ is a sequence in $[-2,2[$ converging to $2$).

Alternatively,

$$\{[-2,2-1/n[\}_{n=1}^\infty$$

is an open cover of $[-2,2[$ (with the relative topology!) without finite subcover.

Answer 2

Compact implies sequentially compact. Consider $x_n=2-1/n$. The limit, $2$, is not in the set.

Answer 3

Using the relative topology of the subset [$-2,2$) where the open sets are [$-2$,b) so:

$\cup^{\infty}_{i=1}A_i$ where $A_i$ are the open sets $A_i$ = [$-2$,$2-\frac{1}{i}$) and with this covering it is impossible to obtain a finite sub cover.