ok, suppose we have the set $U_1=[a,\frac{a+b}{2}) \cup (\frac{a+2}{2},b]$ where $a,b$ are rational. It is easy to see that there exists a countable cover which consists of intervals that converges towards, a,b and $\frac{a+b}{2}$. Therefore $U_1$ is not compact. Now we can construct $U_2$ by taking the midpoint of each half open interval of $U_1$ and we can similarly construct a countable cover that has no finite subcover.

By induction on the naturals, we eventually end up with the set $\Bbb{I} \cap [a,b]$. Thus this set is not compact

"The following statements are equivalent for $\alpha,\beta,\gamma\in[0,\pi]$:

1) The matrix \begin{pmatrix} 1 & \cos \alpha & \cos \beta \\ \cos\alpha & 1 & \cos \gamma \\ \cos \beta& \cos \gamma & 1\end{pmatrix} is positive semidefinite.

2) There exist unit vectors $v_1,v_2,v_3\in\mathbb{R}^3$ such that $\alpha=\arccos(v_1^T v_2), \beta=\arccos(v_2^T v_3),$ and $\gamma=\arccos(v_3^R v_1)$.

3) $\alpha\leq \beta+\gamma$, $\beta\leq \alpha+\gamma$, $\gamma\leq \alpha+\beta$ and $\alpha+\beta+\gamma\leq 2\pi$."