ISI Nyquist criterion and spectral efficiency While reading Bernard Sklar's Digital Communications book I've got confused with ISI Nyquist criterion (not Nyquist rate criterion). There is stated (Sklar, 3.3 section) for symbol rate $R_s$ minimum bandwidth $W = R_s/2$ and for practical cases we have equation:

$W = 1/2 \cdot (R_s + r \cdot R_s)$, 

where $r = 0...1$. So theoretically we have 2 symbols per Hz. For BPSK it is 2 bits/s per Hz. For QAM-64 it is 12 bits/s per Hz. So this statement I can't understand, because through my practice I used to think if we use BPSK and we want to achive e.g. 6 Mbps we need to allocate about 6 MHz of bandwidth (actually more because of rounding and synchronization/code redundancy). Where is my mistake in understanding of Nyquist theory?

Upd. Here is the picture from the book, where Nyquist channel is pictured


It seems the bandwidth of such a channel is calculated from $0$ to $1/2T$ and negative frequencies are neglected. Why in the baseband case we need to determine bandwidth in such a way? This can be an answer for the question above.
I don't have Sklar's text but I guess he is referring to the baseband case. For bandpass signals the required bandwidth is doubled. So in the bandpass case you get a maximum spectral efficiency of 1 bit/s/Hz for BPSK, and $\log M$ bits/s/Hz for $M$-ary QAM.

Just as a side note, spectral efficiency is measured in bits/s/Hz (and not in bits/Hz as you seem to suggest).