Consider the circuit below. Will the LED light up if the switch S is closed or open?
When the switch S is closed, a potential of 2 V is applied to the resistor R3 (which is there to protect the transistor from overload). A current of 2 mA will flow from the base to the emitter, enabling a flow of current from collector to emitter. The current will bypass the LED, and so it will not light up.
On the other hand, when the switch S is open, no current will flow through the base to the emitter, and the transistor is effectively an insulator. No current will be able to flow from the collector to the emitter, and thus it must pass through the LED circuit, lighting it up.