To get an npn (say) BJT in the cut-off region, you set the base current to 0 and the voltage less than 0.7V. This means that \$V\_c > V\_b\$ and \$V\_e > V\_b\$. However, the emitter voltage is 0 (ground) normally, and the base voltage would be less than 0.7V, which is not less than 0, so \$V\_e < V\_b\$.

So when people say BJT's in cut-off have both junctions in reverse bias, what do they mean?