Wiring on-off switch with resistor to SoC
Figure 1. Incorrect pull-up.

Your pull-up resistor is connected to the wrong rail. It is pulling down. Your button just pulls it down better.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 2. (a) What you intended. (b) The way you wired it. (c) With internal pull-up enabled.

The wire link on the right end of your resistor is wired to GND instead of V+.

If you can program the GPIO with internal pull-up then omit the resistor. It is not required. In the present configuration it is competing with the pull-up and the input voltage can never rise above that set by the potential divider created by the pull-up and pull-down.



Extreme clarification:



Figure 2. The pull-up resistor jumper wire needs to be moved to terminate in position 2 - not position 1.



From the comments:


  OK, so I will initialize my pin at the software layer via gpioController.provisionDigitalOutputPin(RaspiPin.GPIO_23, 'RunningLed', PinState.HIGH), which will cause the GPIO pin #23's internal resistor to "pull up" (PinState.HIGH), so that when someone pushes the switch (which in turn closes the circuit) the internal resistor will kick in, yes? 


That will work correctly but your description of the action is incorrect.







simulate this circuit

Figure 3. Internal pull-up with external pull-down switch.


Once enabled the pull-up resistor will be in-circuit as though it is wired in place. It doesn't "kick-in" at any point but it will be effective under certain conditions.
Fig. 3a: With SW1 open the pull-up, R1, connects the internal buffer to V+. The input is now "pulled-up" to Vcc. It will be read as "high" by your program.
Fig. 3b: With SW2 closed the pull-up is still trying to pull but is overcome by the much, much lower resistance of the switch. The input voltage will drop to 0 V. The input will be read as "low" by your program.


Note that earlier in the discussion you were going to have an external pull-down and an internal pull-up. This would not work as the voltage (with the switch open) would be neither fully high nor fully low and would probably be in an undefined state alternating between the two when read by your program.