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.