So, we had a very brief introduction to the Nyquist-Shannon sampling theorem (the discrete time version).
While discussing this, we have seen that multiplying a discrete time signal $x[n]$ by an impulse train $p[n]$, we get a sampled signal $x_p[n]$, and if certain conditions are fulfilled, then it is possible to recover the original signal.
In the frequency domain, what it means (informally) is that we are taking $X(e^{j\omega})$ and shifting it by integer multiples of $2\pi/N$ (i.e the sampling frequency) and "adding" all of this to the plot of $X(e^{j\omega})$ giving us the plot of $X_p(e^{j\omega})$ (in other words we have replicas of $X(e^{j\omega})$)
After looking on the internet and doing a couple of drawings, I've seen that one important condition to recover the initial signal $x[n]$ is that $\omega_s > 2\cdot \omega_m$, where $\omega_s$ is the sampling frequency and $\omega_m$ is the frequency of the original signal (i.e the frequency of $x[n]$).
However, the teacher, in his lecture only mentioned the following condition:
"If the original signal satisfies the following condition: $$X(e^{j\omega}) = 0, \text{ for } \frac{\pi}{N} < |\omega| \leq \pi$$ then the sampled signal $x_p[n]$ contains all the information about the original signal and we can easily recover $x[n]$ from $x_p[n]$".
This makes a little sense to me, as I understand that if $X(e^{j\omega}) \neq 0 $ for all $\omega$, then we do, indeed, have overlapping (i.e aliasing) and it is not possible to recover the signal. But I do not understand why $X(e^{j\omega})$ has to be $0$ specifically for $\frac{\pi}{N} < |\omega| \leq \pi$.
Also, how come the teacher did not speak of the other condition (namely that $\omega_s > 2\cdot \omega_m$) ? Has he forgotten it, or is there something I am missing ?