# Mapping the pinout of an illuminated switch to a circuit

I'm trying to figure out how to make the following momentary switch behave as follows:

• The switch is normally open
• When pushed down, it closes; when you release it, it goes back to being open (I assume there's a spring inside of it?)
• When closed, it sends an output signal out of 1+ of its pins, and it also illuminates the LED inside of it

According to the wiring:

It looks like I want to connect this switch to a circuit using Pins 3 and 4 (for being open at rest) as well as hooking pins X1 and X2 up to a circuit to turn the LED on when closed.

I'm wondering what this wiring would actually look like. Say we start with the switch at rest (so, open). Now I push the switch down, closing it:

What is happening to all 4 pins (3, 4, X1 and X2) when I close the switch and what do I wire these pins to? For instance:

1. When closed, does either X1 or X2 need to be connected to ground (GND)? Let's say I wire X2 to GND, what does the other one wire to, 6VDC power (VCC)?
2. When closed, what specifically is happening to both Pins 3 and 4? Does Pin 4 need to be connected to the 6VDC power, and Pin 3 will emit a signal when the circuit is closed? Or is it vice versa, or something else entirely?