How does a pull-down prevent a "false 1" in a pin?
If you don't have a pull-down, and you push the button to give the MCU 4V which it understands as logic one, everything is fine.
Now, as the MCU input is a CMOS input, it has extremely high impedance, assume it's infinite, and assume there is some parasitic capacitance that is now charged to 4V as well.
When you stop pushing the button, the MCU input is disconnected from 4V supply, but since there is nothing to discharge that 4V from the parasitic capacitance, it will stay at 4V forever, and will never read as logic 0, no matter how long you wait.
When you have the pull-down resistor there, as soon as you stop pushing the button, it will be a DC path to discharge the 4V, so the voltage will end up to 0V and the MCU is able to read that as logic 0.
The pull-down resistor has to be low enough in value, that any interference that might be there, will not be able to make the voltage rise above logic 1 threshold.
So if you pick a too high value as the pull-down resistor, like 100 Megaohms, it too high to prevent voltage fluctuations if you touch the pin with a finger. If you pick too low value as the pull-down resistor, like 1 ohm, you are just wasting power with too high current, and the pushbutton may not be rated for it.
It is considered a weak pull-down because it is not strong. It's weak in the sense that it allows the pushbutton to control the voltage without too much current, and it's strong enough to keep the voltage at 0V when pushbutton is not pushed. For argument's sake assume it is 10 kilo-ohms.