Electronic – “pull the collector below ground” and saturation

I can't come up with any suitable title. I am reading about transistor from a book
Here's the 4 properties of an NPN transistor (for PNP, it is reversed)

1. The collector must be more positive
   than the emitter.
2. The base-emitter and base-collector
   circuits behave like diodes (Fig. 2.2).
   Normally the base-emitter diode is con-
   ducting and the base-collector diode is re-
   verse-biased, i.e., the applied voltage is
   in the opposite direction to easy current
   flow.
3. Any given transistor has maximum
   values of Ic, IB, and VCE that cannot
   be exceeded without costing the exceeder
   the price of a new transistor (for typical
   values, see Table 2.1). There are also other
   limits, such as power dissipation (revCE),
   temperature, VBE, etc., that you must keep
   in mind.
4. When rules 1-3 are obeyed, Ic is rough-
   ly proportional to IB and can be written as
   \$I_C = h_FEI_B = {\beta}I_B\$.
   Where \$h_{FE}\$, the current gain (also called
   beta), is typically about 100. Both \$I_C\$
   and \$I_E\$ flow to the emitter. Note: The
   collector current is not due to forward
   conduction of the base-collector diode; that diode is reverse-biased. Just think of
   it as "transistor action."

and then here is the picture

it say when the switch is close then

When the switch is closed, the base rises to 0.6 volt (base-emitter diode is in forward conduction). The drop across the base resistor is 9.4 volts, so the base current is 9.4mA. Blind application of rule 4 gives Ic = 940mA (for a typical beta
of 100). That is wrong. Why? Because rule 4 holds only if rule 1 is obeyed; at a collector current of lOOmA the lamp has 10 volts across it. To get a higher current you would have to pull the collector below ground. A transistor can't do this, and the result is what's called saturation – the collector goes as close to ground as it can (typical saturation voltages are about 0.05- 0.2V, see Appendix G) and stays there. In this case, the lamp goes on, with its rated 10 volts across it.

What I have understood from this explanation is that here the lamp act as resistor that make the voltage at collector terminal 0 while the voltage at base is 0.6V (because \$V_B=V_E+0.6\$ and \$V_E\$ here equal 0) so it doesn't obey rule 1 and then rule 4 doesn't hold. But the after explanation is unclear. Why the collector must have higher current and what does "pull the collector below ground" mean and then how can collector can have 0.5 or 0.2 voltage while the lamp used up all voltage. So what is the use of the transistor in this case?