How the increasing of sampling frequency in OFDM didn't cause increase on the required channel bandwidth
You can say the null subcarriers have their own bandwidth if you define an alphabet including "zeros" and use null carriers to transmit these "zeros". As the comment of MBaz, 


  If we start with the discrete-time OFDM symbol s[n], and increase the
  sampling rate (defined by the time interval between successive samples
  of s[n]), then the OFDM symbol becomes "shorter" and therefore its
  bandwidth will increase.


But it is not the case. The OFDM symbol $T_u = N_{dft} T_s = 1/\Delta f$.

In this bandwidth interpretation, I think the term "virtual subcarrier" is more appropriate.



What you send over by electromagetic wave is data in $N$ active subcarriers $&lt; N_{dft}$, or the frequency positions of null/virtual subcarriers can be used for other systems; because when you revert back the time domain signal to the frequency domain, what you need is the frequency positions of these active subcarriers.

It is easier to think OFDM as FDM. In this case you have $N$ narrowbands of bandwidth $B_{nb}$ and a sample rate $F_s &gt; N B_{nb}$ is required for practical implementations, such as easily-realizable anti-aliasing filters. OFDM is simply one implementation of this system by using $\Delta f = B_{nb}$ and $F_s = N_{dft} \Delta f = N_{dft} B_{nb} &gt; N B_{nb}$.