A continuous-time sinusoid $x(t) = \sin(2 \pi f_0 t) = \sin(\omega_0 t) $ is periodic with period $T_0 = \frac{1}{f_0} = \frac{2\pi}{\omega_0}$ where $f_0$ is frequency in Hz, and $\omega_0$ is frequency in radians per second. Since $f_0$, $\omega_0$, and $T_0$ are all continuous variables, there is no difficulty on their interpretation: larger the $f_0$ shorter will $T_0$ be.
For the discrete-time case, however, a fundamental difference exists: the period $N$ is an integer. And the smallest integer period is $N_{min} = 1$ which corresponds to a DC signal, and the minimim period for a sinusoid is effectively $N_{min} = 2$. Furthermore, discrete-time frequency $\omega$ is effectively periodic with $2\pi$, which limits the analysis to the range $\omega \in [0,2\pi)$.
The integer period $N$ (if it exists) of the sinuodid $x[n] = \sin(\omega_0 n)$ is found from the relation $$x[n] = x[n+N]$$ which yields:
$$\sin(\omega_0 n) = \sin(\omega_0 (n+N)) = \sin(\omega_0 n + \omega_0 N) = \sin(\omega_0 n + 2\pi m) $$ resulting in:
$$ \omega_0 N = 2\pi m \longrightarrow \omega_0 = \frac{ 2 \pi m}{ N} $$
Putting $m=1$ for $N=2$ gives the highest frequency as $$\omega_{max} = \frac{ 2 \pi m}{N_{min}} = \frac{2 \pi}{2} = \pi $$
On the other hand, the minimum frequency goes to zero as the period goes to infinity $N_{max} \rightarrow \infty$ hence
$$ \omega_{min} = \frac{ 2\pi m}{N_{max}} \rightarrow 0 $$ for any finite $m$.
This way, it can bee seen that discrete-time frequencies begin from $\omega=0$ at the minimum, and reach up to a maximum of $\omega = \pi$, and then decrease again to a minimum into $\omega = 2\pi$, which must be a minimum by the observation that $\omega$ is periodic with $2\pi$. Hence if $\omega = 0$ is a minimum, then $\omega = 0 + 2\pi = 2\pi$ is also an effective minimum, and so on.