Single LED connected in parallel to conducting wire does not light up
Ok, let's take this step-by-step up Wittgenstein's ladder.

Step 1:


Current is lazy and always takes the path of least resistance.


Given the circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

For the current to get from point A to point B it's going to go straight down the simple wire rather than take the more difficult route through the LED. So no current flows through the LED, it just goes straight past.

Step 2:


Kirchhoff's Current Law:



  At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node 


Ok, but on the circuit above point A and point B are connected directly together, so they are in effect the same point.  The circuit is basically the same as:







simulate this circuit

(imagine the little link bit isn't there - the editor won't let you do diagonal lines).

The current I that flows in must equal the current I that flows out of point A. So if all that's going in is going straight out, there's none left to go up to the LED.

Step 3:


Wires aren't perfect.


No wire has absolute zero resistance.  The same with breadboards.  So the actual circuit is more like this:







simulate this circuit

Ok, so we have an fixed voltage \$V_{CC}\$. Let's say this is, for simplicity. 5V.  The LED has a fixed forward voltage drop. Let's assume for the sake of argument that it's 2V.

Ok. Let's take the LED out of the circuit initially and just work out the voltages dropped across the resistors R1, R4 and R5.

The total resistance for that section will be 100.002Ω (simply add them together).  So the current through them would be \$\frac{5}{100.002} = 49.999mA\$.

Therefore the voltage dropped across R4 would be \$0.049999 \times 0.001 = 49.999{\mu}V\$.

Now if you attach the LED across that resistor it's only going to be getting 49.999µV, which is considerably less than the required forward voltage needed to turn it on.  So it won't be conducting as it's not on, so the current through the resistors R2 and R3 will be zero.

Now there are more potential steps in Wittgenstein's ladder, but from here we're getting into the realms of subatomic physics, and even quantum theory, so we'll leave it there for now.